Issuu

CHAPTER

Contents

Preface

Problems Solved in Student Solutions Manual

vii

Matrices, Vectors, and Vector Calculus

Newtonian Mechanics—Single Particle

Oscillations

Nonlinear Oscillations and Chaos

Gravitation

Some Methods in The Calculus of Variations

Hamilton’s Principle—Lagrangian and Hamiltonian Dynamics

Central-Force Motion

Dynamics of a System of Particles

Motion in a Noninertial Reference Frame

Dynamics of Rigid Bodies

Coupled Oscillations

Continuous Systems; Waves

435

Special Theory of Relativity

461

79 127

149 165 181

233 277 333

353

397

iii

CONTENTS

CHAPTER

Preface

This Instructor’s Manual contains the solutions to all the end-of-chapter problems (but not the appendices) from Classical Dynamics of Particles and Systems, Fifth Edition, by Stephen T. Thornton and Jerry B. Marion. It is intended for use only by instructors using Classical Dynamics as a textbook, and it is not available to students in any form. A Student Solutions Manual containing solutions to about 25% of the end-of-chapter problems is available for sale to students. The problem numbers of those solutions in the Student Solutions Manual are listed on the next page. As a result of surveys received from users, I continue to add more worked out examples in the text and add additional problems. There are now 509 problems, a significant number over the 4th edition. The instructor will find a large array of problems ranging in difficulty from the simple “plug and chug” to the type worthy of the Ph.D. qualifying examinations in classical mechanics. A few of the problems are quite challenging. Many of them require numerical methods. Having this solutions manual should provide a greater appreciation of what the authors intended to accomplish by the statement of the problem in those cases where the problem statement is not completely clear. Please inform me when either the problem statement or solutions can be improved. Specific help is encouraged. The instructor will also be able to pick and choose different levels of difficulty when assigning homework problems. And since students may occasionally need hints to work some problems, this manual will allow the instructor to take a quick peek to see how the students can be helped. It is absolutely forbidden for the students to have access to this manual. Please do not give students solutions from this manual. Posting these solutions on the Internet will result in widespread distribution of the solutions and will ultimately result in the decrease of the usefulness of the text. The author would like to acknowledge the assistance of Tran ngoc Khanh (5th edition), Warren Griffith (4th edition), and Brian Giambattista (3rd edition), who checked the solutions of previous versions, went over user comments, and worked out solutions for new problems. Without their help, this manual would not be possible. The author would appreciate receiving reports of suggested improvements and suspected errors. Comments can be sent by email to stt@virginia.edu, the more detailed the better. Stephen T. Thornton Charlottesville, Virginia

PREFACE

CHAPTER

Matrices, Vectors, and Vector Calculus

1-1. x2 = x2′ x1′ 45˚ x1 45˚

x3′

Axes x′1 and x′3 lie in the x1 x3 plane. The transformation equations are:

x1′ = x1 cos 45° − x3 cos 45° x2′ = x2 x3′ = x3 cos 45° + x1 cos 45° x1′ =

1 1 x1 − x3 2 2

x2′ = x2 x3′ =

1 1 x1 − x3 2 2

So the transformation matrix is:

 1  2   0  1   2

0 − 1 0

1  2  0  1   2 

CHAPTER 1

1-2. a) x3

D E

γ β

α A

From this diagram, we have

OE cos α = OA OE cos β = OB

(1)

OE cos γ = OD Taking the square of each equation in (1) and adding, we find 2

OE cos 2 α + cos 2 β + cos 2 γ  = OA + OB + OD

(2)

But 2

OA + OB = OC

(3)

and OC + OD = OE

(4)

Therefore, 2

OA + OB + OD = OE

(5)

Thus, cos 2 α + cos 2 β + cos 2 γ = 1

(6)

b) x3 D E D′

O A A′ x1

E′

B′

B C

C′

First, we have the following trigonometric relation: 2 2 2 OE + OE′ − 2OE OE′ cos θ = EE′

(7)

MATRICES, VECTORS, AND VECTOR CALCULUS

But, 2

2       EE′ = OB′ − OB  + OA′ − OA  + OD′ − OD       

    = OE′ cos β ′ − OE cos β  + OE′ cos α ′ − OE cos α        + OE′ cos γ ′ − OE cos γ   

(8)

or, 2 2 2 EE′ = OE′ cos 2 α ′ + cos 2 β ′ + cos 2 γ ′  + OE  cos 2 α + cos 2 β + cos 2 γ 

− 2OE′ OE cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′  = OE′ 2 + OE2 − 2OE OE′ cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′ 

(9)

Comparing (9) with (7), we find cos θ = cos α cos α ′ + cos β cos β ′ + cos γ cos γ ′

(10)

1-3. x3 e3 O e1

e2′

e1′

e3′

Denote the original axes by x1 , x2 , x3 , and the corresponding unit vectors by e1 , e2 , e3 . Denote the new axes by x′1 , x′2 , x′3 and the corresponding unit vectors by e1′ , e2′ , e3′ . The effect of the rotation is e1 → e3′ , e2 → e1′ , e3 → e′2 . Therefore, the transformation matrix is written as:

 cos ( e1′ , e1 ) cos ( e1′ , e2 ) cos ( e1′ , e3 )  0 1 0    λ = cos ( e2′ , e1 ) cos ( e′2 , e2 ) cos ( e′2 , e3 )  = 0 0 1    1 0 0    cos ( e3′ , e1 ) cos ( e′3 , e2 ) cos ( e′3 , e3 )  

1-4. a) Let C = AB where A, B, and C are matrices. Then,

Cij = ∑ Aik Bkj

(1)

( C ) = C = ∑ A B =∑ B A t

CHAPTER 1

( )

Identifying Bki = Bt ik and Ajk = At kj ,

(C ) = ∑ (B ) ( A ) t

(2)

or, C t = ( AB) = Bt At

(3)

( AB) B−1 A−1 = I = ( B−1 A−1 ) AB

(4)

( AB) B−1 A−1 = AIA−1 = AA−1 = I

(5)

( B A ) ( AB) = B IB = B B = I

(6)

b) To show that ( AB) = B−1 A−1 , −1

That is,

−1

1-5. Take λ to be a two-dimensional matrix:

λ =

λ11 λ12 = λ11λ 22 − λ12 λ 21 λ 21 λ 22

(1)

Then, 2 2 2 2 2 2 λ = λ112 λ 22 − 2λ11λ 22 λ12 λ 21 + λ12 λ 21 + ( λ11 λ 21 + λ122 λ 22 ) − ( λ112 λ 212 + λ122 λ 222 ) 2

(

)

(

) (

2 2 2 2 2 = λ 22 + 2λ11λ 22λ12 λ 21 + λ12 λ112 + λ122 + λ 21 λ112 + λ122 − λ112 λ 21 λ 22

(

)(

)

2 2 2 2 = λ11 + λ12 + λ 21 − ( λ11λ 21 + λ12 λ 22 ) λ 22

) (2)

But since λ is an orthogonal transformation matrix, ∑ λ ij λ kj = δ ik . j

Thus, 2 2 λ112 + λ122 = λ 21 + λ 22 =1

(3)

λ11λ 21 + λ12 λ 22 = 0 Therefore, (2) becomes

λ =1 2

(4)

1-6. The lengths of line segments in the x j and x ′j systems are

∑ x ; L′ = ∑ x ′ 2 j

(1)

MATRICES, VECTORS, AND VECTOR CALCULUS

If L = L ′ , then

∑ x = ∑ x′ 2 j

(2)

xi′ = ∑ λ ij x j

(3)

The transformation is j

Then, 





∑ x = ∑  ∑ λ x   ∑ λ x  2 j

(4)

  = ∑ xk x  ∑ λik λi   i  k,

But this can be true only if

∑λ λ = δ ik

(5)

which is the desired result. 1-7. x3 (0,0,1)

(1,0,1)

(0,0,0)

(0,1,1)

(1,1,1)

(1,0,0)

(0,1,0)

(1,1,0)

There are 4 diagonals: D1 , from (0,0,0) to (1,1,1), so (1,1,1) – (0,0,0) = (1,1,1) = D1 ; D 2 , from (1,0,0) to (0,1,1), so (0,1,1) – (1,0,0) = (–1,1,1) = D 2 ; D 3 , from (0,0,1) to (1,1,0), so (1,1,0) – (0,0,1) = (1,1,–1) = D 3 ; and D 4 , from (0,1,0) to (1,0,1), so (1,0,1) – (0,1,0) = (1,–1,1) = D 4 . The magnitudes of the diagonal vectors are

D1 = D 2 = D 3 = D 4 = 3 The angle between any two of these diagonal vectors is, for example,

D1 ⋅ D 2 (1,1,1) ⋅ ( −1,1,1) 1 = cos θ = = 3 3 D1 D 2

CHAPTER 1

so that 1 θ = cos−1   = 70.5°  3

Similarly, D1 ⋅ D 3 D ⋅D D ⋅D 1 D ⋅D D ⋅D = 1 4 = 2 3 = 2 4 = 3 4 =± 3 D1 D 3 D1 D 4 D2 D3 D2 D 4 D3 D 4 1-8. Let θ be the angle between A and r. Then, A ⋅ r = A2 can be written as

Ar cos θ = A2

or, r cos θ = A

(1)

This implies QPO =

π 2

Therefore, the end point of r must be on a plane perpendicular to A and passing through P. 1-9.

A = i + 2j − k

B = −2i + 3j + k

A − B = 3i − j − 2k A − B = ( 3) + ( −1) + (−2)2  2

A − B = 14 b) B θ A component of B along A

The length of the component of B along A is B cos θ.

A ⋅ B = AB cos θ B cos θ =

A ⋅ B −2 + 6 − 1 3 6 = = or A 2 6 6

The direction is, of course, along A. A unit vector in the A direction is 1 ( i + 2j − k ) 6

(2)

MATRICES, VECTORS, AND VECTOR CALCULUS

So the component of B along A is 1 ( i + 2j − k ) 2 cos θ =

A⋅B 3 3 3 = = ; θ = cos −1 AB 6 14 2 7 2 7

θ 71° i

2 −1 1 −1 1 2 −j +k A × B = 1 2 −1 = i 3 1 −2 1 −2 3 −2 3 1

A × B = 5i + j + 7 k A + B = −i + 5 j

A − B = 3i − j − 2k

( A − B) × ( A + B) = 3 −1 −2 −1

( A − B) × ( A + B) = 10i + 2j + 14k 1-10.

r = 2b sin ω t i + b cos ω t j

v = r = 2bω cos ω t i − bω sin ω t j a)

a = v = −2bω 2 sin ω t i − bω 2 cos ω t j = −ω 2 r speed = v =  4b 2ω 2 cos 2 ω t + b 2ω 2 sin 2 ω t  = bω  4 cos 2 ω t + sin 2 ω t 

speed = bω  3 cos 2 ω t + 1 b)

At t = π 2ω , sin ω t = 1 , cos ω t = 0

So, at this time, v = −bω j , a = −2bω 2 i So, θ

90°

CHAPTER 1

1-11. a) Since ( A × B) i = ∑ ε ijk Aj Bk , we have jk

(A × B) ⋅ C = ∑∑ ε ijk Aj Bk Ci i

j,k

= C1 ( A2 B3 − A3 B2 ) − C2 ( A1B3 − A3 B1 ) + C3 ( A1B2 − A2 B1 ) C1 = A1 B1

C2 A2 B2

C3 A1 A3 = − C1 B3 B1

A2 C2 B2

A3 A1 C3 = B1 B3 C1

(A × B) ⋅ C = − B1 A1

B2 A2

B3 = C1 A3 A1

C2 A2

A2 B2 C2

(1)

A3 B3 = A ⋅ ( B × C ) C3

We can also write C3 = B ⋅ ( C × A ) A3

(2)

We notice from this result that an even number of permutations leaves the determinant unchanged. b) Consider vectors A and B in the plane defined by e1 , e2 . Since the figure defined by A, B, C is a parallelepiped, A × B = e3 × area of the base, but e3 ⋅ C = altitude of the parallelepiped. Then,

C ⋅ ( A × B) = ( C ⋅ e3 ) × area of the base = altitude × area of the base = volume of the parallelepiped

1-12. O

c C

a–c b

b–a

c–b

The distance h from the origin O to the plane defined by A, B, C is

MATRICES, VECTORS, AND VECTOR CALCULUS

= =

a ⋅ ( b − a ) × ( c − b)

( b − a ) × ( c − b)

a ⋅ ( b × c − a × c + a × b) b×c−a×c+a×b a⋅b× c a×b+b×c+c×a

(1)

The area of the triangle ABC is: A=

1-13.

1 1 1 ( b − a ) × ( c − b) = ( a − c ) × ( b − a ) = ( c − b ) × ( a − c ) 2 2 2

Using the Eq. (1.82) in the text, we have A × B = A × ( A × X ) = ( X ⋅ A ) A − ( A ⋅ A ) X = φ A − A2 X

from which X=

( B × A ) + φA A2

1-14. a)

 1 2 −1  2 1 0   1 −2 1       AB =  0 3 1  0 −1 2  =  1 −2 9   2 0 1   1 1 3   5 3 3 

Expand by the first row. AB = 1

−2 9 1 9 1 −2 +2 +1 3 3 5 3 5 3 AB = −104

 1 2 −1  2 1   9 7       AC = 0 3 1   4 3  = 13 9   2 0 1   1 0   5 2   9 7   AC = 13 9   5 2 

(2)

CHAPTER 1

5  1 2 −1  8    ABC = A ( BC ) = 0 3 1   −2 −3   2 0 1   9 4 

 −5 −5    ABC =  3 −5   25 14 

AB − Bt At = ?

 1 −2 1    AB =  1 −2 9   5 3 3 

(from part a )

 2 0 1  1 0 2  1 1 5      B A =  1 −1 1   2 3 0  =  −2 −2 3   0 2 3   −1 1 1   1 9 3  t

 0 −3 −4    AB − B A =  3 0 6  4 −6 0  t

1-15.

If A is an orthogonal matrix, then At A = 1 1 0 0 1 0 0  1 0 0      0 a a  0 a − a  = 0 1 0   0 − a a  0 a a  0 0 1  1 0  2 0 2a 0 0 a=

1 2

0  1 0 0    0  = 0 1 0  2a 2  0 0 1 

MATRICES, VECTORS, AND VECTOR CALCULUS

1-16. x3

P r

θ a

x2 x1

r cos θ

r ⋅ a = constant ra cos θ = constant It is given that a is constant, so we know that r cos θ = constant But r cos θ is the magnitude of the component of r along a. The set of vectors that satisfy r ⋅ a = constant all have the same component along a; however, the component perpendicular to a is arbitrary. Thus the surface represented by r ⋅ a = constant is a plane perpendicular to a.

1-17. a B

Consider the triangle a, b, c which is formed by the vectors A, B, C. Since

C= A−B C = ( A − B) ⋅ ( A − B) 2

(1)

= A 2 − 2A ⋅ B + B 2 or, C = A2 + B2 − 2 AB cos θ 2

which is the cosine law of plane trigonometry. 1-18.

Consider the triangle a, b, c which is formed by the vectors A, B, C. a α

γ c

β A

(2)

CHAPTER 1

C=A−B

(1)

C × B = ( A − B) × B

(2)

so that

but the left-hand side and the right-hand side of (2) are written as:

C × B = BC sin α e3

(3)

( A − B) × B = A × B − B × B = A × B = AB sin γ e3

(4)

and

where e3 is the unit vector perpendicular to the triangle abc. Therefore, BC sin α = AB sin γ

(5)

or, C A = sin γ sin α Similarly, C A B = = sin γ sin α sin β

(6)

which is the sine law of plane trigonometry. 1-19. x2 a

b2 α

β a1

a) We begin by noting that

a − b = a 2 + b 2 − 2ab cos (α − β ) 2

(1)

We can also write that

a − b = ( a1 − b1 ) + ( a2 − b2 ) 2

= ( a cos α − b cos β ) + ( a sin α − b sin β ) 2

(

)

(

)

= a 2 sin 2 α + cos 2 α + b 2 sin 2 β + cos 2 β − 2 ab ( cos α cos β + sin α sin β ) = a 2 + b 2 − 2ab ( cos α cos β + sin α sin β )

(2)

MATRICES, VECTORS, AND VECTOR CALCULUS

Thus, comparing (1) and (2), we conclude that cos (α − β ) = cos α cos β + sin α sin β b)

(3)

Using (3), we can find sin (α − β ) :

sin (α − β ) = 1 − cos 2 (α − β ) = 1 − cos 2 α cos 2 β − sin 2 α sin 2 β − 2cos α sin α cos β sin β

(

)

(

)

= 1 − cos 2 α 1 − sin 2 β − sin 2 α 1 − cos 2 β − 2cos α sin α cos β sin β = sin 2 α cos 2 β − 2sin α sin β cos α cos β + cos 2 α sin 2 β =

( sin α cos β − cos α sin β )2

(4)

so that sin (α − β ) = sin α cos β − cos α sin β

(5)

1-20. a) Consider the following two cases:

When i ≠ j

δ ij = 0

but ε ijk ≠ 0 .

When i = j

δ ij ≠ 0

but ε ijk = 0 .

Therefore,

∑ε δ = 0 ijk

b) We proceed in the following way:

When j = k, ε ijk = ε ijj = 0 . Terms such as ε j11 ε 11 = 0 . Then,

∑ε ε ijk

= ε i12 ε 12 + ε i13 ε 13 + ε i 21 ε 21 + ε i 31 ε 31 + ε i 32 ε 32 + ε i 23 ε 23

Now, suppose i = = 1 , then,

∑= ε jk

123

ε123 + ε 132 ε 132 = 1 + 1 = 2

(1)

CHAPTER 1

for i = = 2 , ∑ = ε 213 ε 213 + ε 231 ε 231 = 1 + 1 = 2 . For i = = 3 , ∑ = ε 312 ε 312 + ε 321 ε 321 = 2 . But i = 1, jk

= 2 gives ∑ = 0 . Likewise for i = 2,

= 1 ; i = 1,

= 3 ; i = 3,

= 1 ; i = 2,

= 3 ; i = 3,

= 2.

Therefore,

∑ε ε ijk

= 2δ i

(2)

j,k

∑ε ε = ε ijk

ijk

123

ε 123 + ε 312 ε 312 + ε 321 ε 321 + ε 132 ε 132 + ε 213 ε 213 + ε 231 ε 231

ijk

= 1 ⋅ 1 + 1 ⋅ 1 + ( −1) ⋅ ( −1) + ( −1) ⋅ ( −1) + ( −1) ⋅ ( −1) + (1) ⋅ (1) or,

∑ε ε = 6 ijk

ijk

(3)

ijk

1-21.

( A × B)i = ∑ ε ijk Aj Bk

(1)

( A × B) ⋅ C = ∑ ∑ ε ijk Aj Bk Ci

(2)

By an even permutation, we find

ABC = ∑ ε ijk Ai Bj Ck ijk

1-22.

To evaluate ∑ ε ijk ε mk we consider the following cases: k

i = j : ∑ ε ijk ε mk = ∑ ε iik ε mk = 0 for all i , , m k

i = : ∑ ε ijk ε mk = ∑ ε ijk ε imk = 1 for j = m and k ≠ i , j k

= 0 for j ≠ m c)

i = m : ∑ ε ijk ε mk = ∑ ε ijk ε ik = 0 for j ≠ k

= −1 for j = and k ≠ i , j d)

j = : ∑ ε ijk ε mk = ∑ ε ijk ε jmk = 0 for m ≠ i k

= −1 for m = i and k ≠ i , j

(3)

MATRICES, VECTORS, AND VECTOR CALCULUS

j = m : ∑ ε ijk ε mk = ∑ ε ijk ε jk = 0 for i ≠ k

= 1 for i = and k ≠ i , j f)

= m : ∑ ε ijk ε mk = ∑ ε ijk ε k = 0 for all i , j , k k

i ≠ or m : This implies that i = k or i = j or m = k.

Then, ∑ ε ijk ε mk = 0 for all i , j , , m k

j ≠ or m : ∑ ε ijk ε mk = 0 for all i , j , , m k

Now, consider δ i δ jm − δ im δ j and examine it under the same conditions. If this quantity behaves in the same way as the sum above, we have verified the equation

∑ε ε ijk

= δ i δ jm − δ im δ j

i = j : δ i δ im − δ im δ i = 0 for all i , , m

i = : δ ii δ jm − δ im δ ji = 1 if j = m, i ≠ j , m = 0 if j ≠ m

i = m : δ i δ ji − δ ii δ j = −1 if j = , i ≠ j , = 0 if j ≠

j = : δ i δ m − δ im δ = −1 if i = m, i ≠ = 0 if i ≠ m

j = m : δ i δ mm − δ im δ m = 1 if i = , m ≠ = 0 if i ≠

= m : δ i δ j − δ il δ j = 0 for all i , j ,

i ≠ , m : δ i δ jm − δ im δ j = 0 for all i , j , , m

j ≠ , m : δ i δ jm − δ im δ i = 0 for all i , j , , m

Therefore,

∑ε ε ijk

= δ i δ jm − δ im δ j

Using this result we can prove that

A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B) C

(1)

CHAPTER 1

First ( B × C ) i = ∑ ε ijk Bj Ck . Then, jk

[ A × (B × C) ] = ∑ ε mn Am ( B × C ) n = ∑ ε mn Am ∑ ε njk BjCk mn

= ∑ ε mn ε njk Am Bj Ck = ∑ ε mn ε jkn Am Bj Ck jkmn

jkmn

  = ∑  ∑ ε lmn ε jkn  Am Bj Ck  jkm  n

(

)

= ∑ δ jlδ km − δ k δ jm Am Bj Ck jkm

    = ∑ Am B Cm − ∑ Am BmC = B  ∑ AmCm  − C  ∑ Am Bm   m   m  m m = ( A ⋅ C ) B − ( A ⋅ B) C Therefore,

A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B) C

1-23.

Write

( A × B) j = ∑ ε j m A Bm m

( C × D) k = ∑ ε krs Cr Ds rs

Then,

(2)

MATRICES, VECTORS, AND VECTOR CALCULUS







[ ( A × B) × (C × D)]i = ∑ ε ijk  ∑ ε j m A Bm   ∑ ε krs Cr Ds  jk

= ∑ ε ijk ε j m ε krs A Bm Cr Ds jk mrs

  = ∑ ε j m  ∑ ε ijk ε rsk  A BmCr Ds  k  j mrs

(

)

= ∑ ε j m δ ir δ js − δ is δ jr A Bm Cr Ds j mrs

(

= ∑ ε j m A BmCi Dj − A Bm Di C j j m

)

    =  ∑ ε j m Dj A Bm  Ci −  ∑ ε j m C j A Bm  Di jm  jm  = (ABD)Ci − (ABC)Di Therefore,

[( A × B) × (C × D) ] = (ABD)C − (ABC)D 1-24.

Expanding the triple vector product, we have e × ( A × e) = A ( e ⋅ e) − e ( A ⋅ e)

(1)

A ( e ⋅ e) = A

(2)

A = e ( A ⋅ e) + e × ( A × e)

(3)

But,

Thus,

e(A · e) is the component of A in the e direction, while e × (A × e) is the component of A perpendicular to e.

CHAPTER 1

1-25. er eφ θ φ eθ

The unit vectors in spherical coordinates are expressed in terms of rectangular coordinates by eθ = ( cos θ cos φ , cos θ sin φ , − sin θ )    eφ = ( − sin φ , cos φ , 0 )   er = ( sin θ cos φ , sin θ sin φ , cos θ ) 

(1)

Thus,

(

eθ = −φ cos θ sin φ − θ sin θ cos φ , φ cos θ cos φ − θ sin θ sin φ , − θ cos θ

)

= −θ er + φ cos θ eφ

(2)

Similarly,

(

eφ = −φ cos φ , − φ sin φ , 0

)

= −φ cos θ eθ − φ sin θ er

(3)

er = φ sin θ eφ + θ eθ

(4)

Now, let any position vector be x. Then, x = rer

(

(5)

)

x = rer + rer = r φ sin θ eφ + θ eθ + rer = rφ sin θ eφ + rθ eθ + rer

(

(6)

(

)

x = rφ sin θ + rθφ cos θ + rφ sin θ eφ + rφ sin θ eφ + rθ + rθ eθ + rθ eθ + rer + rer

(

)

= 2rφ sin θ + 2rθφ cos θ + rφ sin θ eφ + r − rφ 2 sin 2 θ − rθ 2 er

(

)

+ 2rθ + rθ − rφ 2 sin θ cos θ eθ or,

(7)

MATRICES, VECTORS, AND VECTOR CALCULUS

( )

1 d 2  r θ − rφ 2 sin θ cos θ  eθ x = a =  r − rθ 2 − rφ 2 sin 2 θ  er +   r dt   1  d 2 + r φ sin 2 θ  eφ  r sin θ dt 

(

1-26.

)

(8)

When a particle moves along the curve r = k (1 + cos θ )

(1)

we have    2 r = − k θ cos θ + θ sin θ   Now, the velocity vector in polar coordinates is [see Eq. (1.97)] r = − kθ sin θ

(2)

v = rer + rθ eθ

(3)

so that v 2 = v = r 2 + r 2θ 2 2

(

)

= k 2θ 2 sin 2 θ + k 2 1 + 2 cos θ + cos 2 θ θ 2 = k 2θ 2 2 + 2 cos θ 

(4)

and v 2 is, by hypothesis, constant. Therefore,

θ=

v2 2k (1 + cos θ )

(5)

v 2kr

(6)

Using (1), we find

θ=

Differentiating (5) and using the expression for r , we obtain

θ=

v 2 sin θ v 2 sin θ = 2 4r 2 4 k 2 (1 + cos θ )

(7)

The acceleration vector is [see Eq. (1.98)]

(

)

(

)

a = r − rθ 2 er + rθ + 2rθ eθ

so that

(8)

CHAPTER 1

a ⋅ er = r − rθ 2

(

)

= − k θ 2 cos θ + θ sin θ − k (1 + cos θ ) θ 2   θ 2 sin 2 θ = − k θ 2 cos θ + + (1 + cos θ ) θ 2  2 (1 + cos θ )     1 − cos 2 θ = − kθ 2  2 cos θ + + 1 2 (1 + cos θ )   3 = − kθ 2 (1 + cos θ ) 2

(9)

or, 3 v2 4 k

(10)

3 v 2 sin θ 4 k 1 + cos θ

(11)

a =

( a ⋅ er ) 2 + ( a ⋅ eθ ) 2

(12)

a =

3 v2 4 k

(13)

a ⋅ er = −

In a similar way, we find a ⋅ eθ = −

From (10) and (11), we have

or,

1-27.

2 1 + cos θ

Since r × (v × r) = (r ⋅ r) v − (r ⋅ v) r

we have d d r × ( v × r )] = [(r ⋅ r ) v − ( r ⋅ v) r ] [ dt dt = (r ⋅ r) a + 2 (r ⋅ v) v − (r ⋅ v) v − ( v ⋅ v) r − (r ⋅ a) r

(

= r 2a + ( r ⋅ v ) v − r v2 + r ⋅ a

)

(1)

Thus,

(

d [ r × ( v × r )] = r 2a + ( r ⋅ v ) v − r r ⋅ a + v 2 dt

)

(2)

MATRICES, VECTORS, AND VECTOR CALCULUS

1-28.

∂ ( ln r ) ei i ∂x i

(1)

∑x

(2)

grad ( ln r ) = ∑

where

r =

2 i

Therefore, ∂ 1 ln r ) = ( ∂x i r

∑x

2 i

(3)

so that

grad ( ln r ) =

 1  xe 2 ∑ i i  r  i

(4)

r r2

(5)

or,

grad ( ln r ) =

1-29. Let r 2 = 9 describe the surface S1 and x + y + z 2 = 1 describe the surface S2 . The angle θ between S1 and S2 at the point (2,–2,1) is the angle between the normals to these surfaces at the point. The normal to S1 is

(

)

(

grad ( S1 ) = grad r 2 − 9 = grad x 2 + y 2 + z 2 − 9 = ( 2xe1 + 2 ye2 + 2ze3 ) x = 2, y = 2, z =1

) (1)

= 4e1 − 4e2 + 2e3 In S2 , the normal is:

(

)

grad ( S2 ) = grad x + y + z 2 − 1

= ( e1 + e2 + 2ze3 ) x = 2, y =−2, z =1 = e1 + e2 + 2e3 Therefore,

(2)

CHAPTER 1

cos θ =

grad ( S1 ) ⋅ grad ( S2 ) grad ( S1 ) grad ( S2 )

( 4e1 − 4e2 + 2e3 ) ⋅ (e1 + e2 + 2e3 )

(3)

6 6

or, cos θ =

(4)

6 6

from which

θ = cos −1

1-30.

grad ( φψ ) = ∑ ei i =1

(5)

 ∂ψ ∂ φ  ∂ ( φψ ) ψ = ∑ ei φ + ∂x i i  ∂x i ∂x i 

= ∑ ei φ i

6 = 74.2° 9

∂ψ ∂φ ψ + ∑ ei ∂x i ∂x i i

Thus, grad ( φψ ) = φ grad ψ + ψ grad φ

1-31. a) 12   ∂rn ∂  n 2  ∑ xj   = ∑ ei grad r = ∑ ei ∂x i ∂xi  j   i =1 

2 n = ∑ ei 2xi  ∑ x 2j  2 j  i n

 2 = ∑ ei xi n  ∑ x 2j   j  i

−1

= ∑ ei x i n r ( n − 2)

(1)

Therefore, grad r n = nr ( n − 2) r

(2)

MATRICES, VECTORS, AND VECTOR CALCULUS

b) 3

grad f ( r ) = ∑ ei i =1

∂f ( r ) 3 ∂ f ( r ) ∂ r = ∑ ei ∂x i ∂ r ∂x i i =1 12

∂f ( r ) ∂r

−1 2

∂f ( r ) ∂r

 ∂  = ∑ ei x 2j  ∑  ∂x i  j  i   = ∑ ei xi  ∑ x 2j   j  i = ∑ ei i

x i ∂f r dr

(3)

Therefore, grad f ( r ) =

r ∂f ( r ) r ∂r

(4)

 ∂ 2 ln r ∂2    ln  ∑ x 2j  = ∇ ( ln r ) = ∑ ∑ 2 2 ∂x i ∂x i   j  i  2

   

−1 2 1     ⋅ 2xi  ∑ x 2j   2     ∂ j =∑   12   i ∂x i   2    ∑ xj   j   

 ∂    xi  ∑ x 2j  =∑  i ∂x i    j

−1

   

  = ∑ ( − xi )( 2xi )  ∑ x 2j   j  i

(

=−

 ∂x  + ∑ i  ∑ x 2j   i ∂x i  j

−1

) ( ) + 3  r1 

= ∑ −2x 2j r 2 i

−2

2r 2 3 1 + 2 = 2 4 r r r

(5)

or, ∇ 2 ( ln r ) =

1 r2

(6)

24 1-32.

CHAPTER 1

Note that the integrand is a perfect differential:

d d (r ⋅ r) + b (r ⋅ r) dt dt

(1)

∫ ( 2ar ⋅ r + 2br ⋅ r ) dt = ar + br + const.

(2)

d  r  rr − rr r rr = = − 2 dt  r  r2 r r

(1)

2ar ⋅ r + 2br ⋅ r = a Clearly,

1-33.

Since

we have r

rr 

d r 

∫  r − r  dt = ∫ dt  r  dt 2

(2)

from which r

rr 

∫  r − r  dt = r + C 2

(3)

where C is the integration constant (a vector). 1-34.

First, we note that

(

)

d A×A =A×A+A×A dt

(1)

But the first term on the right-hand side vanishes. Thus,

∫ ( A × A) dt = ∫ dt ( A × A) dt

(2)

∫ ( A × A) dt = A × A + C

(3)

so that

where C is a constant vector.

MATRICES, VECTORS, AND VECTOR CALCULUS

1-35. y

We compute the volume of the intersection of the two cylinders by dividing the intersection volume into two parts. Part of the common volume is that of one of the cylinders, for example, the one along the y axis, between y = –a and y = a:

( )

V1 = 2 π a 2 a = 2π a 3

(1)

The rest of the common volume is formed by 8 equal parts from the other cylinder (the one along the x-axis). One of these parts extends from x = 0 to x = a, y = 0 to y = a 2 − x 2 , z = a to

z = a 2 − x 2 . The complementary volume is then a2 − x 2

V2 = 8 ∫ dx ∫ 0

dy ∫

a2 − x 2 a

= 8 ∫ dx a 2 − x 2  a 2 − x 2 − a  0   a

 x 3 a3 x = 8  a2 x − − sin −1  3 2 a 0  =

16 3 a − 2π a 3 3

(2)

Then, from (1) and (2): V = V1 + V2 =

16 a 3 3

(3)

CHAPTER 1

1-36. z

y c2 = x2 + y2

The form of the integral suggests the use of the divergence theorem.

∫ A ⋅ da = ∫ ∇ ⋅ A dv S

(1)

Since ∇ ⋅ A = 1 , we only need to evaluate the total volume. Our cylinder has radius c and height d, and so the answer is

∫ dv = π c d 2

(2)

1-37. z

R y

To do the integral directly, note that A = R3er , on the surface, and that da = daer .

∫ A ⋅ da = R ∫ da = R × 4π R = 4π R 3

(1)

To use the divergence theorem, we need to calculate ∇ ⋅ A . This is best done in spherical coordinates, where A = r 3er . Using Appendix F, we see that ∇⋅A =

1 ∂ 2 r A r = 5r 2 r 2 ∂r

(

)

(2)

Therefore, π

2π

∫ ∇ ⋅ A dv = ∫ sin θ dθ ∫ V

( )

dφ ∫ r 2 5r 2 dr = 4π R5 0

Alternatively, one may simply set dv = 4π r 2 dr in this case.

(3)

MATRICES, VECTORS, AND VECTOR CALCULUS

1-38. z

z = 1 – x2 – y2

y C

x2 + y2 = 1

By Stoke’s theorem, we have

∫ (∇ × A) ⋅ da = ∫ A ⋅ ds S

(1)

The curve C that encloses our surface S is the unit circle that lies in the xy plane. Since the element of area on the surface da is chosen to be outward from the origin, the curve is directed counterclockwise, as required by the right-hand rule. Now change to polar coordinates, so that we have ds = dθ eθ and A = sin θ i + cos θ k on the curve. Since eθ ⋅ i = − sin θ and eθ ⋅ k = 0 , we have

∫ A ⋅ ds = ∫ ( − sin θ ) dθ = −π 2π

1-39. a) Let’s denote A = (1,0,0); B = (0,2,0); C = (0,0,3). Then AB = (−1, 2, 0) ; AC = (−1, 0, 3) ; and

AB × AC = (6, 3, 2) . Any vector perpendicular to plane (ABC) must be parallel to AB × AC , so the unit vector perpendicular to plane (ABC) is n = (6 7 , 3 7 , 2 7 ) b) Let’s denote D = (1,1,1) and H = (x,y,z) be the point on plane (ABC) closest to H. Then DH = ( x − 1, y − 1, z − 1) is parallel to n given in a); this means

x −1 6 = =2 y−1 3

and

x −1 6 = =3 z−1 2

Further, AH = ( x − 1, y , z) is perpendicular to n so one has 6( x − 1) + 3 y + 2 z = 0 . Solving these 3 equations one finds H = ( x , y , z) = (19 49 , 34 49 , 39 49) and DH =

5 7

1-40. a)

At the top of the hill, z is maximum; 0=

∂z = 2 y − 6 x − 18 ∂x

and

∂z = 2 x − 8 y + 28 ∂y

(2)

CHAPTER 1

so x = –2 ; y = 3, and the hill’s height is max[z]= 72 m. Actually, this is the max value of z, because the given equation of z implies that, for each given value of x (or y), z describes an upside down parabola in term of y ( or x) variable. b) At point A: x = y = 1, z = 13. At this point, two of the tangent vectors to the surface of the hill are

t1 = (1, 0,

∂z ) = (1, 0, −8) ∂x (1,1)

and

t2 = (0,1,

∂z ) = (0,1, 22) ∂y (1,1)

Evidently t1 × t2 = (8, −22,1) is perpendicular to the hill surface, and the angle θ between this and Oz axis is cos θ =

(0, 0,1) ⋅ (8, −22,1) 8 + 22 + 1 2

1 23.43

so θ = 87.55 degrees.

c) Suppose that in the α direction ( with respect to W-E axis), at point A = (1,1,13) the hill is steepest. Evidently, dy = (tan α )dx and

dz = 2xdy + 2 ydx − 6 xdx − 8 ydy − 18 dx + 28 dy = 22(tan α − 1)dx

then tan β =

dx 2 + dy 2 dz

dx cos α −1 = 22(tan α − 1)dx 22 2 cos (α + 45)

The hill is steepest when tan β is minimum, and this happens when α = –45 degrees with respect to W-E axis. (note that α = 135 does not give a physical answer). 1-41.

A ⋅ B = 2a( a − 1) then A ⋅ B = 0 if only a = 1 or a = 0.

CHAPTER

Newtonian Mechanics— Single Particle

2-1. The basic equation is

F = mi xi a)

F ( xi , t ) = f ( xi ) g ( t ) = mi xi : Not integrable

F ( xi , t ) = f ( xi ) g ( t ) = mi xi mi

(2)

dxi = f ( xi ) g ( t ) dt

g (t) dxi = dt : Integrable f ( xi ) mi c)

2-2.

(1)

F ( xi , xi ) = f ( xi ) g ( xi ) = mi xi : Not integrable

(3) (4)

Using spherical coordinates, we can write the force applied to the particle as F = Fr er + Fθ eθ + Fφ eφ

(1)

But since the particle is constrained to move on the surface of a sphere, there must exist a reaction force − Fr er that acts on the particle. Therefore, the total force acting on the particle is Ftotal = Fθ eθ + Fφ eφ = mr

(2)

r = Re r

(3)

The position vector of the particle is

where R is the radius of the sphere and is constant. The acceleration of the particle is a = r = Re r

(4)

CHAPTER 2

We must now express er in terms of er , eθ , and eφ . Because the unit vectors in rectangular coordinates, e1 , e2 , e3 , do not change with time, it is convenient to make the calculation in terms of these quantities. Using Fig. F-3, Appendix F, we see that er = e1 sin θ cos φ + e2 sin θ sin φ + e3 cos θ   eθ = e1 cos θ cos φ + e2 cos θ sin φ − e3 sin θ    eφ = −e1 sin φ + e2 cos φ 

(5)

Then

(

)

er = e1 −φ sin θ sin φ + θ cos θ cos φ + e2 θ cos θ sin φ + φ sin θ cos φ − e3 θ sin θ

(6)

= eφ φ sin θ + eθ θ

Similarly, eθ = −er θ + eφ φ cos θ

(7)

eφ = −er φ sin θ − eθ φ cos θ

(8)

And, further,

(

)

(

)

er = −er φ 2 sin 2 θ + θ 2 + eθ θ − φ 2 sin θ cos θ + eφ 2θφ cos θ + φ sin θ

)

(9)

which is the only second time derivative needed. The total force acting on the particle is Ftotal = mr = mRer

(10)

and the components are

(

Fθ = mR θ − φ 2 sin θ cos θ

(

)

Fφ = mR 2θφ cos θ + φ sin θ

)

(11)

NEWTONIAN MECHANICS—SINGLE PARTICLE

2-3. y

v0 α

ℓ

The equation of motion is

F=ma

(1)

The gravitational force is the only applied force; therefore, Fx = mx = 0

  Fy = my = − mg 

(2)

Integrating these equations and using the initial conditions, x ( t = 0 ) = v0 cos α   y ( t = 0 ) = v0 sin α 

(3)

We find x ( t ) = v0 cos α

  y ( t ) = v0 sin α − gt 

(4)

So the equations for x and y are

x ( t ) = v0 t cos α

   1 y ( t ) = v0 t sin α − gt 2  2 

(5)

Suppose it takes a time t0 to reach the point P. Then, cos β = v0 t0 cos α

  1 2  sin β = v0 t0 sin α − gt0  2 

(6)

Eliminating between these equations,   2v sin α 2v0 1 + cos α tan β  = 0 gt0  t0 − 0 2 g g   from which

(7)

CHAPTER 2

t0 =

2v0 ( sin α − cos α tan β ) g

(8)

2-4. One of the balls’ height can be described by y = y0 + v0 t − gt 2 2 . The amount of time it takes to rise and fall to its initial height is therefore given by 2v0 g . If the time it takes to cycle the ball through the juggler’s hands is τ = 0.9 s , then there must be 3 balls in the air during that

time τ. A single ball must stay in the air for at least 3τ, so the condition is 2v0 g ≥ 3τ , or v0 ≥ 13.2 m ⋅ s −1 . 2-5. flightpath N er mg

plane point of maximum acceleration

(

)

a) From the force diagram we have N − mg = mv 2 R er . The acceleration that the pilot feels is

(

)

N m = g + mv R er , which has a maximum magnitude at the bottom of the maneuver. 2

b) If the acceleration felt by the pilot must be less than 9g, then we have

(

−1

3 ⋅ 330 m ⋅ s v2 R≥ = 8g 8 ⋅ 9.8 m ⋅ s −2

)

12.5 km

(1)

A circle smaller than this will result in pilot blackout. 2-6.

Let the origin of our coordinate system be at the tail end of the cattle (or the closest cow/bull). a) The bales are moving initially at the speed of the plane when dropped. Describe one of these bales by the parametric equations

x = x 0 + v0 t

(1)

NEWTONIAN MECHANICS—SINGLE PARTICLE

y = y0 −

1 2 gt 2

(2)

where y0 = 80 m , and we need to solve for x0 . From (2), the time the bale hits the ground is

τ = 2 y0 g . If we want the bale to land at x (τ ) = −30 m , then x0 = x (τ ) − v0τ . Substituting v0 = 44.4 m ⋅ s -1 and the other values, this gives x0 210 m behind the cattle.

−210 m . The rancher should drop the bales

b) She could drop the bale earlier by any amount of time and not hit the cattle. If she were late by the amount of time it takes the bale (or the plane) to travel by 30 m in the x-direction, then she will strike cattle. This time is given by ( 30 m ) v0 0.68 s .

2-7.

Air resistance is always anti-parallel to the velocity. The vector expression is:

1 1  v cw ρ Av 2  −  = − cw ρ Avv 2 2  v

(1)

Including gravity and setting Fnet = ma , we obtain the parametric equations x = −bx x 2 + y 2

(2)

y = −by x 2 + y 2 − g

(3)

where b = cw ρ A 2m . Solving with a computer using the given values and ρ = 1.3 kg ⋅ m -3 , we find that if the rancher drops the bale 210 m behind the cattle (the answer from the previous problem), then it takes 4.44 s to land 62.5 m behind the cattle. This means that the bale should be dropped at 178 m behind the cattle to land 30 m behind. This solution is what is plotted in the figure. The time error she is allowed to make is the same as in the previous problem since it only depends on how fast the plane is moving. 80

y (m)

60 40 20 0

–200

–180

–160

–140

–120 x (m)

With air resistance No air resistance

–100

–80

–60

–40

CHAPTER 2

2-8. y v0 P

From problem 2-3 the equations for the coordinates are x = v0 t cos α y = v0 t sin α −

1 2 gt 2

(1) (2)

In order to calculate the time when a projective reaches the ground, we let y = 0 in (2): v0 t sin α − t=

1 2 gt = 0 2

2v0 sin α g

(3) (4)

Substituting (4) into (1) we find the relation between the range and the angle as x= The range is maximum when 2α =

π 2

v02 sin 2α g

, i.e., α =

π 4

. For this value of α the coordinates become

    v 1 x = 0 t − gt 2   2 2 x=

(5)

v0 t 2

(6)

Eliminating t between these equations yields x2 −

v02 v2 x+ 0 y=0 g g

(7)

We can find the x-coordinate of the projectile when it is at the height h by putting y = h in (7): x2 −

v02 v2 h x+ 0 =0 g g

(8)

This equation has two solutions:  v02 v02 − v02 − 4 gh  2g 2g   v02 v02  2 x2 = v0 − 4 gh  + 2g 2g 

x1 =

(9)

NEWTONIAN MECHANICS—SINGLE PARTICLE

where x1 corresponds to the point P and x2 to Q in the diagram. Therefore, d = x2 − x1 =

v0 g

v02 − 4 gh

(10)

2-9. a) Zero resisting force ( Fr = 0 ):

The equation of motion for the vertical motion is: F = ma = m

dv = − mg dt

(1)

Integration of (1) yields v = − gt + v0

(2)

where v0 is the initial velocity of the projectile and t = 0 is the initial time. The time tm required for the projectile to reach its maximum height is obtained from (2). Since tm corresponds to the point of zero velocity, v ( tm ) = 0 = v0 − gtm ,

(3)

we obtain tm =

v0 g

(4)

b) Resisting force proportional to the velocity ( Fr = − kmv ) :

The equation of motion for this case is: F=m

dv = − mg − kmv dt

(5)

where –kmv is a downward force for t < tm′ and is an upward force for t > tm′ . Integrating, we obtain v (t) = −

g kv0 + g − kt e + k k

(6)

For t = tm , v(t) = 0, then from (6),

(

g ktm e −1 k

)

(7)

 kv  ktm = ln 1 + 0  g  

(8)

v0 = which can be rewritten as

Since, for small z (z

1) the expansion

CHAPTER 2

1 1 ln (1 + z ) = z − z 2 + z 3 2 3

(9)

is valid, (8) can be expressed approximately as 2  v0  kv0 1  kv0   tm = 1− +   − … g  2g 3  2g   

(10)

which gives the correct result, as in (4) for the limit k → 0. 2-10. The differential equation we are asked to solve is Equation (2.22), which is x = − kx . Using the given values, the plots are shown in the figure. Of course, the reader will not be able to distinguish between the results shown here and the analytical results. The reader will have to take the word of the author that the graphs were obtained using numerical methods on a computer. The results obtained were at most within 10 −8 of the analytical solution. v vs t

v (m/s)

t (s) x vs t

x (m)

100

15 t (s)

v vs x

v (m/s)

100

x (m)

2-11.

The equation of motion is m

d2 x = − kmv 2 + mg dt 2

This equation can be solved exactly in the same way as in problem 2-12 and we find

(1)

NEWTONIAN MECHANICS—SINGLE PARTICLE

 g − kv02  1 log  2 2k  g − kv 

(2)

where the origin is taken to be the point at which v = v0 so that the initial condition is x ( v = v0 ) = 0 . Thus, the distance from the point v = v0 to the point v = v1 is

 g − kv02  1 s ( v0 → v1 ) = log  2  2k  g − kv1 

2-12.

(3)

The equation of motion for the upward motion is d2 x = − mkv 2 − mg 2 dt

(1)

d 2 x dv dv dx dv = = =v 2 dt dt dx dt dx

(2)

v dv = − dx kv 2 + g

(3)

(

(4)

m Using the relation

we can rewrite (1) as

Integrating (3), we find

)

1 log kv 2 + g = − x + C 2k where the constant C can be computed by using the initial condition that v = v0 when x = 0:

(

)

1 log kv02 + g 2k

kv 2 + g 1 log 02 kv + g 2k

(6)

d2 x = − mkv 2 + mg dt 2

(7)

v dv = dx − kv 2 + g

(8)

(5)

Therefore,

Now, the equation of downward motion is m

This can be rewritten as

Integrating (8) and using the initial condition that x = 0 at v = 0 (w take the highest point as the origin for the downward motion), we find

CHAPTER 2

g 1 log g − kv 2 2k

(9)

At the highest point the velocity of the particle must be zero. So we find the highest point by substituting v = 0 in (6): xh =

kv 2 + g 1 log 0 g 2k

(10)

Then, substituting (10) into (9), kv 2 + g 1 g 1 = log 0 log 2k 2k g g − kv 2

(11)

g 2 v0 k v= g v02 + k

(12)

Solving for v,

We can find the terminal velocity by putting x → ∞ in(9). This gives g k

vt =

(13)

Therefore, v=

2-13.

v0 vt

(14)

v02 + vt2

The equation of motion of the particle is

(

)

(1)

∫ v ( v + a ) = − k ∫ dt

(2)

dv = − mk v 3 + a 2 v dt

Integrating, dv

and using Eq. (E.3), Appendix E, we find  v2  1 ln  2 = − kt + C 2 2 2a a + v 

(3)

v2 = C ′ e − At a2 + v 2

(4)

Therefore, we have

NEWTONIAN MECHANICS—SINGLE PARTICLE

where A ≡ 2a 2 k and where C′ is a new constant. We can evaluate C′ by using the initial condition, v = v0 at t = 0: C′ =

v02 a 2 + v02

(5)

Substituting (5) into (4) and rearranging, we have  a 2C ′e − At  v= − At   1 − C ′e 

dx dt

(6)

Now, in order to integrate (6), we introduce u ≡ e − At so that du = –Au dt. Then, x=

∫

=−

 a 2C ′e − At   1 − C e − At  ′  

a C′ A

∫

a dt = A

∫

 C ′u   1 − C ′u 

du u

(7)

−C ′u2 + u

Using Eq. (E.8c), Appendix E, we find x=

a sin −1 (1 − 2C ′u) + C ′′ A

(8)

Again, the constant C″ can be evaluated by setting x = 0 at t = 0; i.e., x = 0 at u = 1: C ′′ = −

a sin −1 (1 − 2C ′ ) A

(

)

(9)

Therefore, we have a sin −1 −2C ′ e − At + 1 − sin −1 ( −2C ′ + 1)   A

Using (4) and (5), we can write x=

2 2  1  −1  −v 2 + a 2  −1  − v0 + a  sin sin −   2  2 2  2  2ak   v +a   v0 + a  

(10)

From (6) we see that v → 0 as t → ∞. Therefore,  −v 2 + a2  π lim sin −1  2 = sin −1 (1) = 2  t →∞ 2  v +a 

(11)

Also, for very large initial velocities,  −v 2 + a2  π lim sin −1  20 2  = sin −1 ( −1) = − v0 →∞ 2  v0 + a  Therefore, using (11) and (12) in (10), we have

(12)

CHAPTER 2

x (t → ∞) =

(13)

2ka

and the particle can never move a distance greater than π 2ka for any initial velocity.

2-14.

d y x

a) The equations for the projectile are

x = v0 cos α t y = v0 sin α t −

1 2 gt 2

Solving the first for t and substituting into the second gives y = x tan α −

gx 2 1 2 v02 cos 2 α

Using x = d cos β and y = d sin β gives d sin β = d cos β tan α −

gd 2 cos 2 β 2v02 cos 2 α

 gd cos 2 β  − cos β tan α + sin β  0 = d 2 2  2v0 cos α  Since the root d = 0 is not of interest, we have d=

2 ( cos β tan α − sin β ) v02 cos 2 α g cos 2 β 2v02 cos α ( sin α cos β − cos α sin β ) g cos 2 β d=

2v02 cos α sin (α − β ) g cos 2 β

(1)

NEWTONIAN MECHANICS—SINGLE PARTICLE

b) Maximize d with respect to α

2v02 d  − sin α sin (α − β ) + cos α cos (α − β )  cos ( 2α − β ) ( d) = 0 = dα g cos 2 β  cos ( 2α − β ) = 0 2α − β =

α=

π 4

π 2

β 2

c) Substitute (2) into (1)

dmax =

2v02  π β π β cos  +  sin  −   2     4 2 g cos β  4 2

Using the identity sin A − sin B = 2 cos

1 1 ( A + B) sin ( A − B) 2 2

we have

sin − sin β v 2  1 − sin β  2v02 2 dmax = ⋅ = 0  2 g cos β 2 g  1 − sin 2 β  dmax =

v02 g (1 + sin β )

2-15. mg sin θ

The equation of motion along the plane is m

dv = mg sin θ − kmv 2 dt

(1)

Rewriting this equation in the form 1 dv = dt k g sin θ − v 2 k

(2)

CHAPTER 2

We know that the velocity of the particle continues to increase with time (i.e., dv dt > 0 ), so that

( g k ) sin θ > v2 . Therefore, we must use Eq. (E.5a), Appendix E, to perform the integration. We

find

v   tanh −1  =t+C g g  sin θ sin θ  k  k 

1 k

(3)

The initial condition v(t = 0) = 0 implies C = 0. Therefore, v=

g sin θ tanh k

( gk sin θ t) = dx dt

(4)

We can integrate this equation to obtain the displacement x as a function of time: x=

g sin θ ∫ tanh k

( gk sin θ t) dt

Using Eq. (E.17a), Appendix E, we obtain x=

(

)

ln cosh gk sin θ t g sin θ + C′ k gk sin θ

(5)

The initial condition x(t = 0) = 0 implies C′ = 0. Therefore, the relation between d and t is d=

1 ln cosh k

( gk sin θ t)

(6)

From this equation, we can easily find t=

( )

cosh −1 e dk

gk sin θ

(7)

2-16. The only force which is applied to the article is the component of the gravitational force along the slope: mg sin α. So the acceleration is g sin α. Therefore the velocity and displacement along the slope for upward motion are described by:

v = v0 − ( g sin α ) t x = v0 t −

1 ( g sin α ) t 2 2

(1) (2)

where the initial conditions v ( t = 0 ) = v0 and x ( t = 0 ) = 0 have been used. At the highest position the velocity becomes zero, so the time required to reach the highest position is, from (1), t0 = At that time, the displacement is

v0 g sin α

(3)

NEWTONIAN MECHANICS—SINGLE PARTICLE

x0 =

1 v02 2 g sin α

(4)

For downward motion, the velocity and the displacement are described by v = ( g sin α ) t

(5)

1 ( g sin α ) t 2 2

(6)

where we take a new origin for x and t at the highest position so that the initial conditions are v(t = 0) = 0 and x(t = 0) = 0. We find the time required to move from the highest position to the starting position by substituting (4) into (6): t′ =

v0 g sin α

(7)

2v0 g sin α

(8)

Adding (3) and (7), we find

for the total time required to return to the initial position. 2-17. v0 35˚

60 m 0.7 m

Fence

The setup for this problem is as follows: x = v0 t cos θ y = y0 + v0 t sin θ −

(1) 1 2 gt 2

(2)

where θ = 35 and y0 = 0.7 m . The ball crosses the fence at a time τ = R ( v0 cos θ ) , where R = 60 m. It must be at least h = 2 m high, so we also need h − y0 = v0τ sin θ − gτ 2 2 . Solving for v0 , we obtain v02 = which gives v0

25.4 m ⋅ s −1 .

gR 2

2 cos θ  R sin θ − ( h − y0 ) cos θ 

(3)

CHAPTER 2

2-18. a) The differential equation here is the same as that used in Problem 2-7. It must be solved for many different values of v0 in order to find the minimum required to have the ball go over the fence. This can be a computer-intensive and time-consuming task, although if done correctly is easily tractable by a personal computer. This minimum v0 is 35.2 m ⋅ s −1 , and the trajectory is

shown in Figure (a). (We take the density of air as ρ = 1.3 kg ⋅ m −3 .) 15

y (m)

x (m)

With air resistance No air resistance fence height fence range

b) The process here is the same as for part (a), but now we have v0 fixed at the result just obtained, and the elevation angle θ must be varied to give the ball a maximum height at the fence. The angle that does this is 0.71 rad = 40.7°, and the ball now clears the fence by 1.1 m. This trajectory is shown in Figure (b). 20

y (m)

30 x (m)

Flight Path fence height fence range

NEWTONIAN MECHANICS—SINGLE PARTICLE

2-19.

The projectile’s motion is described by x = ( v0 cos α ) t

   1 y = ( v0 sin α ) t − gt 2   2

(1)

where v0 is the initial velocity. The distance from the point of projection is r = x2 + y2

(2)

Since r must always increase with time, we must have r > 0 :

xx + yy >0 r

(3)

Using (1), we have xx + yy =

1 2 3 3 g t − g ( v0 sin α ) t 2 + v02t 2 2

(4)

Let us now find the value of t which yields xx + yy = 0 (i.e., r = 0 ): t=

3 v0 sin α v0 ± 9 sin 2 α − 8 g 2 2g

(5)

For small values of α, the second term in (5) is imaginary. That is, r = 0 is never attained and the value of t resulting from the condition r = 0 is unphysical. Only for values of α greater than the value for which the radicand is zero does t become a physical time at which r does in fact vanish. Therefore, the maximum value of α that insures r > 0 for all values of t is obtained from 9 sin 2 α max − 8 = 0

(6)

or, 2 2 3

(7)

α max ≅ 70.5°

(8)

sin α max = so that

2-20.

If there were no retardation, the range of the projectile would be given by Eq. (2.54):

v02 sin 2θ 0 g The angle of elevation is therefore obtained from R=

(1)

CHAPTER 2

sin 2θ 0 =

Rg v02

(1000 m ) × ( 9.8 m/sec 2 ) = (140 m/sec ) 2 = 0.50

(2)

so that

θ 0 = 15°

(3)

Now, the real range R′, in the linear approximation, is given by Eq. (2.55):  4 kV  R′ = R 1 −  3g   =

v02 sin 2θ  4 kv0 sin θ  1 −  3g g  

(4)

Since we expect the real angle θ to be not too different from the angle θ 0 calculated above, we can solve (4) for θ by substituting θ 0 for θ in the correction term in the parentheses. Thus,

sin 2θ =

g R′  4kv0 sin θ 0  v 1 −  3g  

(5)

2 0

Next, we need the value of k. From Fig. 2-3(c) we find the value of km by measuring the slope of the curve in the vicinity of v = 140 m/sec. We find km ≅ (110 N ) ( 500 m/s ) ≅ 0.22 kg/s . The curve is that appropriate for a projectile of mass 1 kg, so the value of k is k

0.022 sec −1

(6)

Substituting the values of the various quantities into (5) we find θ = 17.1° . Since this angle is somewhat greater than θ 0 , we should iterate our solution by using this new value for θ 0 in (5). We then find θ = 17.4° . Further iteration does not substantially change the value, and so we conclude that

θ = 17.4° If there were no retardation, a projectile fired at an angle of 17.4° with an initial velocity of 140 m/sec would have a range of R=

(140 m/sec ) 2 sin 34.8° 9.8 m/sec 2 1140 m

NEWTONIAN MECHANICS—SINGLE PARTICLE

2-21. x3 v0 α x2

Assume a coordinate system in which the projectile moves in the x2 − x3 plane. Then, x2 = v0 t cos α

  1 2  x3 = v0 t sin α − gt  2 

(1)

or, r = x 2 e2 + x3 e3

1   = ( v0 t cos α ) e2 +  v0 t sin α − gt 2  e3   2

(2)

The linear momentum of the projectile is p = mr = m ( v0 cos α ) e2 + ( v0 sin α − gt ) e3 

(3)

and the angular momentum is

(

)

L = r × p = ( v0 t cos α ) e2 + v0 t sin α − gt 2 e3  × m ( v0 cos α ) e2 + ( v0 sin α − gt ) e3 

(4)

Using the property of the unit vectors that ei × e j = e3 ε ijk , we find

(

)

1 mg v0 t 2 cos α e1 2

(5)

L = − ( mg v0 t cos α ) e1

(6)

This gives

Now, the force acting on the projectile is F = − mg e3

so that the torque is 1     N = r × F = ( v0 t cos α ) e2 +  v0 t sin α − gt 2  e3  ( − mg ) e3    2  = − ( mg v0 t cos α ) e1 which is the same result as in (6).

(7)

CHAPTER 2

2-22. z B E

y e x

Our force equation is F = q ( E + v × B)

(1)

a) Note that when E = 0, the force is always perpendicular to the velocity. This is a centripetal acceleration and may be analyzed by elementary means. In this case we have also v ⊥ B so that v × B = vB .

mv 2 = qvB r

macentripetal =

(2)

Solving this for r r=

mv v = qB ω c

(3)

with ω c ≡ qB/m . b) Here we don’t make any assumptions about the relative orientations of v and B, i.e. the velocity may have a component in the z direction upon entering the field region. Let r = xi + yj + zk , with v = r and a = r . Let us calculate first the v × B term.

v × B = x y z = B ( yi − xj) 0 0 B

(4)

(

(5)

The Lorentz equation (1) becomes

)

F = mr = qByi + q Ey − Bx j + qEz k

Rewriting this as component equations: x=

qB y = ωc y m

(6)

qEy Ey   qB = −ω c  x −  x+ m m B 

(7)

y=− z=

qEz m

(8)

NEWTONIAN MECHANICS—SINGLE PARTICLE

The z-component equation of motion (8) is easily integrable, with the constants of integration given by the initial conditions in the problem statement. z ( t ) = z0 + z0 t +

qEz 2 t 2m

(9)

c) We are asked to find expressions for x and y , which we will call vx and vy , respectively.

Differentiate (6) once with respect to time, and substitute (7) for v y Ey   vx = ω c vy = −ω c2  vx −  B 

(10)

or vx + ω c2 vx = ω c2

(11)

This is an inhomogeneous differential equation that has both a homogeneous solution (the solution for the above equation with the right side set to zero) and a particular solution. The most general solution is the sum of both, which in this case is vx = C1 cos (ω c t ) + C2 sin (ω c t ) +

Ey B

(12)

where C1 and C2 are constants of integration. This result may be substituted into (7) to get v y v y = −C1ω c cos (ω c t ) − C2ω c sin (ω c t )

(13)

vy = −C1 sin (ω c t ) + C2 cos (ω c t ) + K

(14)

where K is yet another constant of integration. It is found upon substitution into (6), however, that we must have K = 0. To compute the time averages, note that both sine and cosine have an average of zero over one of their periods T ≡ 2π ω c . x =

Ey B

y =0

(15)

d) We get the parametric equations by simply integrating the velocity equations.

x= y=

ωc

sin (ω c t ) − cos (ω c t ) +

ωc

cos (ω c t ) +

Ey B

t + Dx

sin (ω c t ) + Dy

(16) (17)

where, indeed, Dx and Dy are constants of integration. We may now evaluate all the C’s and D’s using our initial conditions x ( 0 ) = − A ω c , x ( 0 ) = Ey B , y ( 0 ) = 0 , y ( 0 ) = A . This gives us C1 = Dx = Dy = 0 , C2 = A and gives the correct answer x (t) =

−A

ωc

cos (ω c t ) +

Ey B

(18)

CHAPTER 2

y(t) =

ωc

sin (ω c t )

(19)

These cases are shown in the figure as (i) A > Ey B , (ii) A < Ey B , and (iii) A = Ey B . (i)

(ii)

(iii)

F ( t ) = ma ( t ) = kte − at

2-23.

(1)

with the initial conditions x(t) = v(t) = 0. We integrate to get the velocity. Showing this explicitly, k

∫ v(0)a (t ) dt = m ∫ 0te v( t )

−α t

(2)

Integrating this by parts and using our initial conditions, we obtain v(t) =

k  1 1  1  −α t  − + t    e  m  α 2 α  α  

(3)

By similarly integrating v(t), and using the integral (2) we can obtain x(t). x (t) =

k  2 1 1  2  − 3 + 2 + 2  t +  e −α t     α α α m α 

(4)

To make our graphs, substitute the given values of m = 1 kg, k = 1 N ⋅ s −1 , and α = 0.5 s −1 . x ( t ) = te − t 2

(5)

v ( t ) = 4 − 2 ( t + 2) e − t 2

(6)

α ( t ) = −16 + 4t + 4 ( t + 4 ) e − t 2

(7)

NEWTONIAN MECHANICS—SINGLE PARTICLE

x(t)

100

10 t

v(t)

a(t)

0.5

2-24. Ff

sθ

N′

sin

d = length of incline s = distance skier travels along level ground

While on the plane:

∑ Fy = N − mg cosθ = my = 0 ∑ Fx = mg sin θ − Ff ;

so N = mg cos θ

Ff = µ N = µ mg cos θ

mg sin θ − µ mg cos θ = mx So the acceleration down the plane is: a1 = g ( sin θ − µ cos θ ) = constant

CHAPTER 2

While on level ground: N ′ = mg ; Ff = − µ mg So ∑ Fx = mx becomes − µ mg = mx The acceleration while on level ground is a2 = − µ g = constant For motion with constant acceleration, we can get the velocity and position by simple integration: x=a v = x = at + v0 x − x 0 = v0 t +

(1)

1 2 at 2

(2)

Solving (1) for t and substituting into (2) gives: v − v0 =t a x − x0 =

v0 ( v − v0 ) a

1 ( v − v0 ) + ⋅ 2 a

or 2a ( x − x0 ) = v 2 − v02 Using this equation with the initial and final points being the top and bottom of the incline respectively, we get: 2a1d = VB2

VB = speed at bottom of incline

Using the same equation for motion along the ground: 2a2 s = −VB2

(3)

Thus a1d = − a2 s

a1 = g ( sin θ − µ cos θ )

So gd ( sin θ − µ cos θ ) = µ gs Solving for µ gives

µ=

d sin θ d cos θ + s

Substituting θ = 17°, d = 100 m, s = 70 m gives

µ = 0.18 Substituting this value into (3):

a2 = − µ g

NEWTONIAN MECHANICS—SINGLE PARTICLE

−2µ gs = −VB2 VB = 2µ gs VB = 15.6 m/sec 2-25. a)

At A, the forces on the ball are: N

The track counters the gravitational force and provides centripetal acceleration N − mg = mv 2 R Get v by conservation of energy: Etop = Ttop + Utop = 0 + mgh EA = TA + U A =

1 mv 2 + 0 2

Etop = EA → v = 2 gh So N = mg + m2 gh R 2h   N = mg  1 +   R b)

At B the forces are: N

45˚ mg

N = mv 2 R + mg cos 45° = mv 2 R + mg Get v by conservation of energy. From a), Etotal = mgh . At B, E =

1 mv 2 + mgh ′ 2

(1)

CHAPTER 2

R cos 45˚ = R

R 45˚

h′

R + h′ 2

1   h′ = R  1 −   2

So Etotal = TB + UB becomes: 1  1  2 mgh = mgR  1 −  + mv  2 2 Solving for v 2  1   2 2  gh − gR  1 −   = v  2   Substituting into (1):  2h  3  − 2  N = mg  +   R  2 c) From b) vB2 = 2 g  h − R + R

2 

(

v =  2 g h − R + R

)

2 

d) This is a projectile motion problem 45˚ 45˚ B A

Put the origin at A. The equations: x = x0 + vx 0 t y = y0 + v y 0 t −

1 2 gt 2

become x=

R v + B t 2 2

y = h′ + Solve (3) for t when y = 0 (ball lands).

vB 1 t − gt 2 2 2

(2) (3)

NEWTONIAN MECHANICS—SINGLE PARTICLE

gt 2 − 2 vBt − 2h ′ = 0 2 vB ± 2vB2 + 8 gh′

We discard the negative root since it gives a negative time. Substituting into (2): x=

2 R v  2 vB ± 2vB + 8 gh ′   + B  2g 2 2  

Using the previous expressions for vB and h′ yields x= e)

(

)

3   2 − 1 R + h +  h 2 − R2 + 2 R2  2  

U ( x) = mgy( x ) , with y(0) = h , so U ( x) has the shape of the track.

2-26. All of the kinetic energy of the block goes into compressing the spring, so that mv 2 2 = kx 2 2 , or x = v m k 2.3 m , where x is the maximum compression and the given values have been substituted. When there is a rough floor, it exerts a force µk mg in a direction

that opposes the block’s velocity. It therefore does an amount of work µk mgd in slowing the block down after traveling across the floor a distance d. After 2 m of floor, the block has energy mv 2 2 − µk mgd , which now goes into compressing the spring and still overcoming the friction on the floor, which is kx 2 2 + µk mgx . Use of the quadratic formula gives x=−

µ mg k

2µ mgd mv  µ mg  +  + −   k  k k 2

Upon substitution of the given values, the result is

(1)

1.12 m.

2-27.

0.6 m

To lift a small mass dm of rope onto the table, an amount of work dW = ( dm) g ( z0 − z ) must be done on it, where z0 = 0.6 m is the height of the table. The total amount of work that needs to be done is the integration over all the small segments of rope, giving z0

µ gz02

W = ∫ (µ dz) g( z0 − z) =

When we substitute µ = m L = ( 0.4 kg ) ( 4 m ) , we obtain W

0.18 J .

(1)

CHAPTER 2

2-28. v4

m v

v M

before collision

after collision

The problem, as stated, is completely one-dimensional. We may therefore use the elementary result obtained from the use of our conservation theorems: energy (since the collision is elastic) and momentum. We can factor the momentum conservation equation m1v1 + m2 v2 = m1v3 + m2 v4

(1)

out of the energy conservation equation 1 1 1 1 m1v12 + m2 v22 = m1v32 + m2 v42 2 2 2 2

(2)

v1 + v3 = v2 + v4

(3)

and get

This is the “conservation” of relative velocities that motivates the definition of the coefficient of restitution. In this problem, we initially have the superball of mass M coming up from the ground with velocity v = 2gh , while the marble of mass m is falling at the same velocity. Conservation of momentum gives Mv + m ( −v ) = Mv3 + mv4

(4)

and our result for elastic collisions in one dimension gives v + v3 = ( − v ) + v 4 solving for v3 and v4 and setting them equal to

(5)

2 ghitem , we obtain

3 −α  hmarble =  h  1 + α 

(6)

 1 − 3α  hsuperball =  h  1 + α 

(7)

where α ≡ m M . Note that if α < 1 3 , the superball will bounce on the floor a second time after the collision.

NEWTONIAN MECHANICS—SINGLE PARTICLE

2-29. N

Ff mg cos θ

y mg

mg sin θ

θ = tan −1 0.08 = 4.6°

∑ F = N − mg cos θ y

= my = 0 N = mg cos θ

∑ F = mg sin θ − F x

= mx Ff = µ N = µ mg cos θ so mx = mg sin θ − µ mg cos θ x = g ( sin θ − µ cos θ ) Integrate with respect to time x = gt ( sin θ − µ cos θ ) + x0

(1)

Integrate again: x = x0 + x0 t +

1 2 gt ( sin θ − µ cos θ ) 2

Now we calculate the time required for the driver to stop for a given x0 (initial speed) by solving Eq. (1) for t with x = 0 . t′ = −

x0 −1 sin θ − µ cos θ ) ( g

Substituting this time into Eq. (2) gives us the distance traveled before coming to a stop.

( x ′ − x0 ) = x0t ′ + 2 gt ′ 2 ( sin θ − µ cos θ ) 1

∆x = − ∆x =

x02 1 x2 −1 −1 sin θ − µ cos θ ) + g 02 ( sin θ − µ cos θ ) ( 2 g g

x02 −1 µ cos θ − sin θ ) ( 2g

(2)

CHAPTER 2

We have θ = 4.6° , µ = 0.45 , g = 9.8 m/sec 2 . For x0 = 25 mph = 11.2 m/sec , ∆x = 17.4 meters . If the driver had been going at 25 mph, he could only have skidded 17.4 meters. Therefore, he was speeding How fast was he going? ∆x ≥ 30 meters gives x0 ≥ 32.9 mph . T = t1 + t2

2-30.

where

T = total time = 4.021 sec. t1 = the time required for the balloon to reach the ground. t2 = the additional time required for the sound of the splash to reach the first student.

We can get t1 from the equation y = y0 + y 0 t −

1 2 gt ; 2

y0 = y 0 = 0

When t = t1 , y = –h; so (h = height of building) −h = − t=

1 2 gt1 2

t1 =

2h g

distance sound travels h = speed of sound v

Substituting into (1): T=

2h h + g v

h 2h + −T =0 v g

This is a quadratic equation in the variable − h=

Substituting

2 4T + g v

2 ± g 2 v

h . Using the quadratic formula, we get: 2 gT  v   −1 ± 1 +  V  2g 

V = 331 m/sec g = 9.8 m/sec 2 T = 4.021 sec

and taking the positive root because it is the physically acceptable one, we get:

(1)

NEWTONIAN MECHANICS—SINGLE PARTICLE

h = 8.426 m1 2 h = 71 meters 2-31. For x0 ≠ 0 , example 2.10 proceeds as is until the equations following Eq. (2.78). Proceeding from there we have

α B = x0 ≠ 0 α A = z0 so

( x − x0 ) = α0 cos α t + α0 sin α t

( y − y0 ) = y 0 t x

( z − z0 ) = − α0 cos α t + α0 sin α t Note that

( x − x 0 ) + ( z − z0 ) = 2

z02

+ 2

x02

α2

Thus the projection of the motion onto the x–z plane is a circle of radius

(x + z ) . α 1

2 0

2 12 0

So the motion is unchanged except for a change in the 12 m 2 radius of the helix. The new radius is x0 + z02 . qB0

(

)

2-32.

The forces on the hanging mass are T

The equation of motion is (calling downward positive) mg − T = ma The forces on the other mass are

T = m ( g − a)

(1)

CHAPTER 2

N y

2mg cos θ

Ff 2mg

2mg sin θ

The y equation of motion gives N − 2mg cos θ = my = 0 or N = 2mg cos θ

(

The x equation of motion gives Ff = µ k N = 2µk mg cos θ

)

T − 2mg sin θ − 2µ k mg cos θ = ma

(2)

Substituting from (1) into (2) mg − 2mg sin θ − 2µ k mg cos θ = 2ma When θ = θ 0 , a = 0. So g − 2 g sin θ 0 − 2µk g cos θ 0 = 0

1 = sin θ 0 + µ k cos θ 0 2

(

= sin θ 0 + µ k 1 − sin 2 θ 0

)

Isolating the square root, squaring both sides and rearranging gives

(1 + µ ) sin θ − sin θ  14 − µ  = 0 2 k

2 k

Using the quadratic formula gives sin θ 0 =

2-33.

1 ± µk 3 + 4 µ k2

(

2 1 + µ k2

)

The differential equation to solve is  v  2  cW ρ Av 2 y= − g = g    − 1 2m   vt  

where vt = 2mg cw ρ A is the terminal velocity. The initial conditions are y0 = 100 m , and v0 = 0 . The computer integrations for parts (a), (b), and (c) are shown in the figure.

(1)

y (m)

NEWTONIAN MECHANICS—SINGLE PARTICLE

100

v (m/s)

t (s)

5 t (s)

t (s)

–20

–5

0 –5 –10

a (m/s2)

t (s)

–5

4 t (s)

t (s)

d) Taking ρ = 1.3 kg ⋅ m −3 as the density of air, the terminal velocities are 32.2, 8.0, and 11.0 (all

m ⋅ s -1 ) for the baseball, ping-pong ball, and raindrop, respectively. Both the ping-pong ball and the raindrop essentially reach their terminal velocities by the time they hit the ground. If we rewrite the mass as average density times volume, then we find that vt ∝ ρmaterial R . The differences in terminal velocities of the three objects can be explained in terms of their densities and sizes. e) Our differential equation shows that the effect of air resistance is an acceleration that is inversely proportional to the square of the terminal velocity. Since the baseball has a higher terminal velocity than the ping-pong ball, the magnitude of its deceleration is smaller for a given speed. If a person throws the two objects with the same initial velocity, the baseball goes farther because it has less drag. f) We have shown in part (d) that the terminal velocity of a raindrop of radius 0.004 m will be larger than for one with radius 0.002 m ( 9.0 m ⋅ s -1 ) by a factor of 2 . 2-34. FR

y mg

Take the y-axis to be positive downwards. The initial conditions are y = y = 0 at t = 0.

CHAPTER 2

FR = α v

The equation of motion is my = m

dv = mg − α v dt

m dv = dt mg − α v

Integrating gives: −

ln ( mg − α v ) = t + C

Evaluate C using the condition v = 0 at t = 0: − −

So or −

ln ( mg − α v ) +

ln ( mg ) = C

ln ( mg ) = t

 mg − α v   αv  = ln    = ln 1 − m  mg   mg 

αt

Take the exponential of both sides and solve for v: e −α r m = 1 −

αv mg

= 1 − e −α t m

dy =

(1 − e

(1 − e α

−α t m

)

(1)

) dt

Integrate again: y+C =

mg  m −α t m  t + e  α  α

y = 0 at t = 0, so: C= y=

mg  m  = m2 g α 2 α  α 

mg  m m  − + t + e −α t m   α  α α 

(2)

Solve (3) for t and substitute into (4): 1−

αv mg

= e −α t m

(3)

NEWTONIAN MECHANICS—SINGLE PARTICLE

t=−

 αv  ln 1 −  α  mg  m

mg  m m  α v  m  α v   mg  v m  α v    − − ln 1 −  − − ln 1 −   + 1 −  = α  α α  mg  α  mg   α  g α  mg   y=−

(4)

mg  α v   m ln 1 − v +  α  α  mg  

FR = β v 2

The equation of motion becomes: m

dv = mg − β v 2 dt

m dv = dt mg − β v 2

Integrate and apply the initial condition v = 0 at t = 0: dv

∫ mg β From integral tables ∫

−v

= 2

m∫

dx 1 x = tanh −1 ; so 2 a −x a a 2

mg 1 v β tanh −1 = t + C where a ≡ β a a m

tanh −1 0 = 0 = 0 + C

so: 1 v β tanh −1 = t a a m Solving for v: v = a tanh

aβ t m

dy aβt = a tanh dt m

From integral tables ∫ tanh u du = ln cosh u So y + C =

ln cosh

aβ t m

(5)

CHAPTER 2

Apply the conditions at y = 0 and t = 0 C=

ln ( cosh 0 ) =

ln 1 = 0

So y=

ln cosh

aβ t m

(6)

Solving (5) for t:

αβ

tanh −1

v a

Substituting into (6): y=

Use the identity: tanh −1 u = cosh −1

v  ln cosh  tanh −1  β a 

1 1 − u2

(In our case u < 1 as it should be because

, where u < 1 . v βv2 = ; and the condition that u < 1 just says that a mg

gravity is stronger than the retarding force, which it must be.) So y=

 ln cosh  cosh −1  β

y=−

 m −1 2 = ln 1 − β v 2 mg  1 − β v 2 mg  β

(

m ln 1 − β v 2 mg 2β

)

NEWTONIAN MECHANICS—SINGLE PARTICLE

2-35. 14 12

y (km)

10 8 6 4 2 0

x (km)

Range (km)

30 20 10 0

0.02

0.04

0.06

0.08

k (1/s)

We are asked to solve Equations (2.41) and (2.42), for the values k = 0, 0.005, 0.01, 0.02, 0.04, and 0.08 (all in s−1 ), with initial speed v0 = 600 m ⋅ s −1 and angle of elevation θ = 60° . The first figure is produced by numerical solution of the differential equations, and agrees closely with Figure 2-8. Figure 2-9 can be most closely reproduced by finding the range for our values of k, and plotting them vs. k. A smooth curve could be drawn, or more ranges could be calculated with more values of k to fill in the plot, but we chose here to just connect the points with straight lines.

CHAPTER 2

2-36. y θ h x R

Put the origin at the initial point. The equations for the x and y motion are then x = v0 ( cos θ ) t y = v0 ( sin θ ) t −

1 2 gt 2

Call τ the time when the projectile lands on the valley floor. The y equation then gives − h = v0 ( sin θ ) τ −

1 2 gτ 2

Using the quadratic formula, we may find τ v02 sin 2 θ + 2 gh v0 sin θ + τ= g g

(We take the positive since τ > 0 .) Substituting τ into the x equation gives the range R as a function of θ. R=

v02   cos θ  sin θ + sin 2 θ + x 2  g

(1)

where we have defined x 2 ≡ 2 gh v02 . To maximize R for a given h and v0 , we set dR dθ = 0 . The equation we obtain is cos 2 θ − sin 2 θ − sin θ sin 2 θ + x 2 +

sin θ cos 2 θ sin 2 θ + x 2

(2)

Although it can give x = x (θ ) , the above equation cannot be solved to give θ = θ ( x ) in terms of the elementary functions. The optimum θ for a given x is plotted in the figure, along with its respective range in units of v02 g . Note that x = 0, which among other things corresponds to h = 0, gives the familiar result θ = 45° and R = v02 g .

NEWTONIAN MECHANICS—SINGLE PARTICLE

50 45˚

30 20 10 0

R/(v02/g)

1 0

2-37.

v=α x

dv dx = −α x 2

Since

dv dv dx dv = = v then dt dx dt dx F=m

dv dv α   α  = mv = m   − 2  dt dx  x  x  F ( x ) = − mα 2 x 3

2-38.

v ( x ) = ax − n

F=m

(

)(

dv dv dx dv =m = mv = m ax − n − nax − n −1 dt dx dt dx

)

F ( x ) = − mna 2 x − ( 2 n + 1) b)

v ( x) =

dx = ax − n dt

x n dx = adt

Integrate: x n+1 = at + C n+1

C = 0 using given initial conditions

CHAPTER 2

x n + 1 = ( n + 1) at

x = [ ( n + 1) at ]

1 ( n + 1)

c) Substitute x(t) into F(x):

F ( t ) = − mna 2 {( n + 1) at} 

1 ( n + 1)

F ( t ) = − mna 2 [ ( n + 1) at ]

 

− ( 2 n + 1)

− ( 2 n + 1) ( n + 1)

2-39. a)

F = −α e β v dv α = − e βv dt m

∫e −

− βv

dv = −

e − βv = −

m∫

α m

t+C

v = v0 at t = 0, so − −

(e β 1

e − β v0 = C

− βv

)

− e − β v0 = −

α m

Solving for v gives v (t) = −

 αβ t − β v0  ln  +e  β  m 1

b) Solve for t when v = 0

αβ t m t=

+ e − β v0 = 1

1 − e − β v0 

αβ 

c) From a) we have

dx = −

 αβ t − β v0  +e ln   dt β  m 1

NEWTONIAN MECHANICS—SINGLE PARTICLE

Using ∫ ln ( ax + b ) dx =

ax + b ln ( ax + b ) − x we obtain a   αβ t   αβ t − β v0   + e − β v0  ln  +e     1  m m   − t x+C =−  β αβ m   

Evaluating C using x = 0 at t = 0 gives C=

v0 m

αβ

e − β v0

So x=−

mv0

αβ

e − β v0 +

−

m  αβ t − β v0   αβ t − β v0  +e  ln  m + e  αβ 2  m

Substituting the time required to stop from b) gives the distance required to stop x=

m 1 1  − β v0  v0 +    −e  αβ  β β  

2-40. y (x(t),y(t)) an

at x

Write the velocity as v(t) = v(t)T(t). It follows that

a (t) =

dv dv dT T+v = = at T + an N dt dt dt

(1)

where N is the unit vector in the direction of dT dt . That N is normal to T follows from 0 = d dt ( T ⋅ T ) . Note also an is positive definite. a) We have v = x 2 + y = Aα 5 − 4 cos α t . Computing from the above equation, 2

at =

dv 2 Aα 2 sin α t = dt 5 − 4 cos α t

(2)

We can get an from knowing a in addition to at . Using a = x 2 + y = Aα 2 , we get 2

an = a 2 − at2 = Aα 2

2 cos α t − 1 5 − 4 cos α t

b) Graphing an versus t shows that it has maxima at α t = nπ , where an = Aα 2 .

(3)

CHAPTER 2

2-41. a)

As measured on the train: Ti = 0 ; Tf =

1 mv 2 2

∆T = b)

1 mv 2 2

As measured on the ground: Ti =

1 1 2 mu 2 ; T f = m ( v + u ) 2 2 ∆T =

1 mv 2 + mvu 2

c) The woman does an amount of work equal to the kinetic energy gain of the ball as measured in her frame.

1 mv 2 2

d) The train does work in order to keep moving at a constant speed u. (If the train did no work, its speed after the woman threw the ball would be slightly less than u, and the speed of the ball relative to the ground would not be u + v.) The term mvu is the work that must be supplied by the train.

W = mvu

2-42.

Rθ b

R θ

From the figure, we have h(θ ) = (R + b 2) cos θ + Rθ sin θ , and the potential is U (θ ) = mgh(θ ) . Now compute: dU  b  = mg  − sin θ + Rθ cos θ  dθ  2 

(1)

d 2U b   = mg  R −  cos θ − Rθ sin θ  2 2 dθ  

(2)

NEWTONIAN MECHANICS—SINGLE PARTICLE

The equilibrium point (where dU dθ = 0 ) that we wish to look at is clearly θ = 0. At that point, we have d 2U dθ 2 = mg ( R − b 2) , which is stable for R > b 2 and unstable for R < b/ 2 . We can

use the results of Problem 2-46 to obtain stability for the case R = b 2 , where we will find that the first non-trivial result is in fourth order and is negative. We therefore have an equilibrium at θ = 0 which is stable for R > b 2 and unstable for R ≤ b 2 .

2-43.

F = − kx + kx 3 α 2 U ( x ) = − ∫ F dx =

1 2 1 x4 kx − k 2 2 4 α

To sketch U(x), we note that for small x, U(x) behaves like the parabola behavior is determined by −

1 2 kx . For large x, the 2

1 x4 k 4 α2 U(x) E0 E1 E2 x4 x5 x1 x2 x3

x E3 = 0

1 mv 2 + U ( x ) 2

For E = E0 , the motion is unbounded; the particle may be anywhere. For E = E1 (at the maxima in U(x)) the particle is at a point of unstable equilibrium. It may remain at rest where it is, but if perturbed slightly, it will move away from the equilibrium. What is the value of E1 ? We find the x values by setting

dU =0. dx

0 = kx − kx 3 α 2 x = 0, ± α are the equilibrium points U ( ±α ) = E1 =

1 2 1 2 1 2 kα − kα = kα 2 4 4

For E = E2 , the particle is either bounded and oscillates between − x2 and x2 ; or the particle comes in from ±∞ to ± x3 and returns to ±∞.

CHAPTER 2

For E3 = 0 , the particle is either at the stable equilibrium point x = 0, or beyond x = ± x4 . For E4 , the particle comes in from ±∞ to ± x5 and returns. 2-44. θ

T m1

m1 g

m2 g

From the figure, the forces acting on the masses give the equations of motion m1 x 1 = m1 g − T

(1)

m2 x 2 = m2 g − 2T cos θ

(2)

where x2 is related to x1 by the relation 2 b − x1 ) ( − d2 x = 2

(3)

and cos θ = d ( b − x1 ) 2  . At equilibrium, x 1 = x 2 = 0 and T = m1 g . This gives as the equilibrium values for the coordinates x10 = b −

4 m1 d

(4)

4 m12 − m22 m2 d

x20 =

(5)

4 m12 − m22

We recognize that our expression x10 is identical to Equation (2.105), and has the same requirement that m2 m1 < 2 for the equilibrium to exist. When the system is in motion, the descriptive equations are obtained from the force laws: m1 ( x 1 − g ) =

m2 ( b − x1 ) 4 x2

(x 2 − g)

(6)

To examine stability, let us expand the coordinates about their equilibrium values and look at their behavior for small displacements. Let ξ1 ≡ x1 − x10 and ξ2 ≡ x2 − x20 . In the calculations, take terms in ξ1 and ξ2 , and their time derivatives, only up to first order. Equation (3) then becomes ξ2 −(m1 m2 )ξ1 . When written in terms of these new coordinates, the equation of motion becomes

ξ1 = −

(

g 4 m12 − m22

)

4 m1m2 ( m1 + m2 ) d

ξ1

(7)

NEWTONIAN MECHANICS—SINGLE PARTICLE

which is the equation for simple harmonic motion. The equilibrium is therefore stable, when it exists. 2-45. and 2-46.

Expand the potential about the equilibrium point 1  di u  i  i x i = n + 1 i !  dx  0

(1)

dU 1  d( n + 1)U  = −  ( n + 1)  x n dx n!  dx 0

(2)

∞

U ( x) = ∑ The leading term in the force is then F( x ) = −

The force is restoring for a stable point, so we need F ( x > 0 ) < 0 and F ( x < 0 ) > 0 . This is never true when n is even (e.g., U = kx 3 ), and is only true for n odd when  d( n + 1)U dx( n + 1)  0 < 0 .

2-47.

We are given U ( x) = U 0 ( a x + x a ) for x > 0 . Equilibrium points are defined by

dU dx = 0 , with stability determined by d 2U dx 2 at those points. Here we have dU  a 1 = U0  − 2 +  dx a  x

(1)

 d 2U   2U0   dx 2  =  a 3  > 0  a

(2)

which vanishes at x = a. Now evaluate

indicating that the equilibrium point is stable. 25

U(x)/U0

20 15 10 5 0

0.5

1 x/a

1.5

CHAPTER 2

2-48. In the equilibrium, the gravitational force and the eccentric force acting on each star must be equal

mG πd ⇒τ = = 2d v

Gm2 mv 2 = ⇒v= d2 d /2 2-49.

2π d 3 2 mG

The distances from stars to the center of mass of the system are respectively r1 =

dm2 m1 + m2

and

r2 =

dm1 m1 + m2

At equilibrium, like in previous problem, we have Gm1m2 m1v12 = ⇒ v1 = d2 r1

2π r1 2π d 3 2 Gm2 2 ⇒τ = = d(m1 + m2 ) v1 G(m1 + m2 )

The result will be the same if we consider the equilibrium of forces acting on 2nd star. 2-50. t

m0 v d  m0 v   m0 v  = F ⇒∫d = = Ft ⇒ v(t) =  dt  v2  v2  v2 0  1− 2   1− 2  1− 2   c  c  c t

⇒ x(t) = ∫ v(t)dt = 0

Ft m02 +

 c2  F 2t 2 2 + − m0  m  0 2  F c

b) v

c) From a) we find

vm0 F 1−

Now if

F = 10 , then m0

F 2t 2 c2

v2 c2

NEWTONIAN MECHANICS—SINGLE PARTICLE

when v = c 2 , we have t =

c = 0.55 year 10 3

when v = 99% c, we have t =

99c = 6.67 years 10 199

2-51. a)

mv0 dv dv b = − bv 2 ⇒ ∫ 2 = − ∫ dt ⇒ v(t) = dt v m btv0 + m

Now let v(t) = v0/1000 , one finds t =

999m = 138.7 hours . v0 b

t t

x(t ) = ∫ vdt = 0

m  btv0 + m  ln   b  m 

We use the value of t found in question a) to find the corresponding distance x(t ) =

m ln(1000) = 6.9 km b

2-52. a)

F( x ) = −

4U x  dU x2  = − 20  1 − 2  dx a  a 

b) U

When F = 0, there is equilibrium; further when U has a local minimum (i.e. dF dx < 0 ) it is stable, and when U has a local maximum (i.e. dF dx > 0 ) it is unstable.

CHAPTER 2

So one can see that in this problem x = a and x = –a are unstable equilibrium positions, and x = 0 is a stable equilibrium position. c)

Around the origin, F ≈ −

4U 0 x ≡ − kx ⇒ ω = a2

k = m

4U 0 ma 2

d) To escape to infinity from x = 0, the particle needs to get at least to the peak of the potential, 2 2U 0 mvmin = U max = U 0 ⇒ vmin = m 2

e) From energy conservation, we have 2 mv 2 U 0 x 2 mvmin dx + 2 = ⇒ =v= a dt 2 2

2U 0  x2   1 − 2  a m

We note that, in the ideal case, because the initial velocity is the escape velocity found in d), ideally x is always smaller or equal to a, then from the above expression,

  8U 0   −1 a  exp  t  ma 2    m dx ma a+x ln = ⇒ x(t) = 2U 0 ∫0  8U 0 a − x  x2   8U 0    1 − 2  exp  t +1  a  ma 2    x

2-53.

F is a conservative force when there exists a non-singular potential function U(x) satisfying F(x) = –grad(U(x)). So if F is conservative, its components satisfy the following relations ∂Fx ∂Fy = ∂y ∂x and so on. a) In this case all relations above are satisfied, so F is indeed a conservative force.

Fx = −

bx 2 ∂U = ayz + bx + c ⇒ U = − ayzx − − cx + f1 ( y , z) 2 ∂x

(1)

where f1 ( y , z) is a function of only y and z Fy = −

∂U = axz + bz ⇒ U = − ayzx − byz + f 2 ( x , z) ∂y

(2)

NEWTONIAN MECHANICS—SINGLE PARTICLE

where f 2 ( x , z) is a function of only x and z Fz = −

∂U = axy + by + c ⇒ U = − ayzx − byz + f 3 ( y , z) ∂z

(3)

then from (1), (2), (3) we find that U = − axyz − byz − cx 2 −

bx 2 +C 2

where C is a arbitrary constant. b) is

Using the same method we find that F in this case is a conservative force, and its potential U = − z exp( − x) − y ln z + C

c) Using the same method we find that F in this case is a conservative force, and its potential is ( using the result of problem 1-31b):

U = − a ln r 2-54. a) Terminal velocity means final steady velocity (here we assume that the potato reaches this velocity before the impact with the Earth) when the total force acting on the potato is zero.

mg = kmv

and consequently

v = g k = 1000 m/s .

b) x

dx dv vdv dv F = dt = − ⇒ ∫ dx = − ∫ ⇒ = = −( g + kv) ⇒ dt m v g + kv g + kv 0 v0 xmax =

g v0 g + 2 ln = 679.7 m k k g + kv0

2-55.

Let’s denote vx 0 and vy 0 the initial horizontal and vertical velocity of the pumpkin.

where v0 is the initial velocity of the potato.

Evidently, vx 0 = vy 0 in this problem. m

vx 0 − vxf dvx dv dx = Fx = − mkvx ⇒ − = − dt = x ⇒ x f = dt vx kvx k

(1)

where the suffix f always denote the final value. From the second equality of (1), we have − dt =

dvx − kt ⇒ vxf = vx 0 e f kvx

(2)

vx 0 ( − kt 1− e f ) k

(3)

Combining (1) and (2) we have xf =

CHAPTER 2

Do the same thing with the y-component, and we have m

dvy dt

= Fy = − mg − mkvy ⇒ −

dvy g + kvyf vy 0 − vyf dy g = − dt = ⇒ 0 = y f = 2 ln + vy g + kvy k g + kvy 0 k

and

− dt =

dvy g + kvy

⇒ g + kvyf = ( g + kv0 f ) e

− kt f

(4)

(5)

From (4) and (5) with a little manipulation, we obtain 1− e

− kt f

gkt f g + kvy 0

(3) and (6) are 2 equations with 2 unknowns, t f and k. We can eliminate t f , and obtain an equation of single variable k. xf = Putting x f = 142 m and vx 0 = vy 0 =

vx 0 ( − kt g + kv ( gv ) 1 − e f ( y0 ) x0 ) k

v0 = 38.2 m/s we can numerically solve for k and obtain 2 K= 0.00246 s −1 .

(6)

CHAPTER

Oscillations

3-1.

1 ν0 = 2π

k 1 = m 2π

10 4 dyne/cm 10 = 10 2 gram 2π

gram ⋅ cm sec 2 ⋅ cm = 10 sec −1 gram 2π

or,

ν 0 ≅ 1.6 Hz τ0 =

ν0

(1)

2π sec 10

or,

τ 0 ≅ 0.63 sec b)

(2)

1 2 1 kA = × 10 4 × 32 dyne-cm 2 2

so that E = 4.5 × 10 4 erg

(3)

c) The maximum velocity is attained when the total energy of the oscillator is equal to the kinetic energy. Therefore,

1 2 mvmax = 4.5 × 10 4 erg 2 v max =

2 × 4.5 × 10 4 100

CHAPTER 3

or, vmax = 30 cm/sec

(4)

3-2. a) The statement that at a certain time t = t1 the maximum amplitude has decreased to onehalf the initial value means that

xen = A0 e − βt1 =

1 A0 2

(1)

or, 1 2

(2)

ln 2 0.69 = t1 t1

(3)

e − βt1 = so that

β= Since t1 = 10 sec ,

β = 6.9 × 10 −2 sec −1 b)

(4)

According to Eq. (3.38), the angular frequency is

ω 1 = ω 02 − β 2

(5)

where, from Problem 3-1, ω 0 = 10 sec −1 . Therefore,

ω 1 = (10 ) − ( 6.9 × 10 −2 ) 2

2  1  ≅ 10 1 − ( 6.9) × 10 −6  sec −1  2 

(6)

so that

ν1 =

10 (1 − 2.40 × 10 −5 ) sec −1 2π

(7)

which can be written as

ν1 = ν 0 (1 − δ )

(8)

δ = 2.40 × 10 −5

(9)

where That is, ν1 is only slightly different from ν 0 .

OSCILLATIONS

c) The decrement of the motion is defined to be e βτ 1 where τ 1 = 1 ν1 . Then,

e βτ 1

1.0445

3-3. The initial kinetic energy (equal to the total energy) of the oscillator is

m = 100 g and v0 = 1 cm/sec .

1 mv02 , where 2

a) Maximum displacement is achieved when the total energy is equal to the potential energy. Therefore,

1 1 mv02 = kx02 2 2

x0 =

m 10 2 1 v0 = ×1= cm 4 k 10 10

or, x0 =

1 cm 10

(1)

b) The maximum potential energy is

U max =

1 2 1 kx0 = × 10 4 × 10 −2 2 2

or, U max = 50 ergs

(2)

3-4. a) Time average:

The position and velocity for a simple harmonic oscillator are given by

x = A sin ω 0 t

(1)

x = ω 0 A cos ω 0 t

(2)

where ω 0 = k m The time average of the kinetic energy is T =

t +τ

where τ =

2π is the period of oscillation. ω0

mx dt τ ∫ 2 2

(3)

CHAPTER 3

By inserting (2) into (3), we obtain t +τ

T =

1 mA2ω 02 ∫ cos 2 ω 0 t dt 2τ t

(4)

mA2ω 02 4

(5)

or, T =

In the same way, the time average of the potential energy is

U =

t +τ

kx dt τ ∫ 2 2

t +τ

1 kA2 ∫ sin 2 ω 0 t dt 2τ t

kA2 4

(6)

and since ω 02 = k m , (6) reduces to U =

mA2ω 02 4

(7)

From (5) and (7) we see that

T = U

(8)

The result stated in (8) is reasonable to expect from the conservation of the total energy.

E = T +U

(9)

This equality is valid instantaneously, as well as in the average. On the other hand, when T and U are expressed by (1) and (2), we notice that they are described by exactly the same function, displaced by a time τ 2 :

 mA2ω 02 cos 2 ω 0 t  2   2 mA ω 0 t 2 sin ω 0 t  U=  2 T=

(10)

Therefore, the time averages of T and U must be equal. Then, by taking time average of (9), we find

T = U =

E 2

b) Space average:

The space averages of the kinetic and potential energies are

(11)

OSCILLATIONS

1 1 mx 2 dx ∫ A0 2

(12)

and

mω 02 1 1 U = ∫ kx 2 dx = x 2 dx ∫ A0 2 2A 0 A

(13)

(13) is readily integrated to give

mω 02 A2 6

(14)

To integrate (12), we notice that from (1) and (2) we can write

(

x 2 = ω 02 A2 cos 2 ω 0 t = ω 02 A2 1 − sin 2 ω 0 t

(

= ω 02 A2 − x 2

)

(15)

Then, substituting (15) into (12), we find mω 02  A2 − x 2  dx 2 A ∫0  A

mω 02  3 A3  A − 2 A  3 

(16)

mω 02 A2 6

(17)

or, T=2

From the comparison of (14) and (17), we see that T = 2U

(18)

To see that this result is reasonable, we plot T = T(x) and U = U(x):  1 x2   mω 02 A2 1 − 2   2  A   1  2 2 U = mω 0 x  2

mA2ω 02

(19)

Energy

U = U(x)

E = const. =

1 mA2ω 02 2

T = T(x) –A

And the area between T(x) and the x-axis is just twice that between U(x) and the x-axis.

CHAPTER 3

3-5. Differentiating the equation of motion for a simple harmonic oscillator,

x = A sin ω 0 t

(1)

∆x = Aω 0 cos ω 0 t ∆t

(2)

we obtain

But from (1) sin ω 0 t =

x A

(3)

Therefore, cos ω 0 t = 1 − ( x A)

(4)

and substitution into (2) yields ∆t =

∆x

(5)

ω 0 A2 − x 2

Then, the fraction of a complete period that a simple harmonic oscillator spends within a small interval ∆x at position x is given by ∆t

∆x

ω 0τ A2 − x 2

∆x

2π

A2 − x 2

(6)

∆t⁄τ

x –A3

–A2

–A1

This result implies that the harmonic oscillator spends most of its time near x = ±A, which is obviously true. On the other hand, we obtain a singularity for ∆t τ at x = ±A. This occurs because at these points x = 0, and (2) is not valid. 3-6. k

Suppose the coordinates of m1 and m2 are x1 and x2 and the length of the spring at equilibrium is . Then the equations of motion for m1 and m2 are m1 x1 = − k ( x1 − x2 + )

(1)

m2 x2 = − k ( x2 − x1 + )

(2)

OSCILLATIONS

From (2), we have x1 =

1 ( m2 x2 + kx2 − k ) k

(3)

Substituting this expression into (1), we find d2  m1 m2 x2 + ( m1 + m2 ) kx2  = 0 dt 2 

(4)

from which x2 = −

m1 + m2 kx2 m1m2

(5)

m1 + m2 k m1m2

(6)

Therefore, x2 oscillates with the frequency

ω=

We obtain the same result for x1 . If we notice that the reduced mass of the system is defined as 1

1 1 + m1 m2

(7)

we can rewrite (6) as

ω=

(8)

k µ

This means the system oscillates in the same way as a system consisting of a single mass µ. Inserting the given values, we obtain µ

66.7 g and ω

2.74 rad ⋅ s −1 .

3-7. A hb hs

Let A be the cross-sectional area of the floating body, hb its height, hs the height of its submerged part; and let ρ and ρ0 denote the mass densities of the body and the fluid, respectively. The volume of displaced fluid is therefore V = Ahs . The mass of the body is M = ρ Ahb .

CHAPTER 3

There are two forces acting on the body: that due to gravity (Mg), and that due to the fluid, pushing the body up ( − ρ0 gV = − ρ0 ghs A ). The equilibrium situation occurs when the total force vanishes: 0 = Mg − ρ0 gV = ρ gAhb − ρ0 ghs A

(1)

which gives the relation between hs and hb : hs = hb

ρ ρ0

(2)

For a small displacement about the equilibrium position ( hs → hs + x ), (1) becomes Mx = ρ Ahb x = ρ gAhb − ρ0 g ( hs + x ) A

(3)

Upon substitution of (1) into (3), we have

ρ Ahb x = − ρ0 gxA

(4)

ρ0 x=0 ρ hb

(5)

or, x+g

Thus, the motion is oscillatory, with an angular frequency

ω2 = g

g gA ρ0 = = ρhb hs V

(6)

where use has been made of (2), and in the last step we have multiplied and divided by A. The period of the oscillations is, therefore,

τ= Substituting the given values, τ

2π

= 2π

V gA

(7)

0.18 s .

3-8. y

ℓ O

m s

2a x 2a

The force responsible for the motion of the pendulum bob is the component of the gravitational force on m that acts perpendicular to the straight portion of the suspension string. This component is seen, from the figure (a) below, to be F = ma = mv = − mg cos α

(1)

OSCILLATIONS

where α is the angle between the vertical and the tangent to the cycloidal path at the position of m. The cosine of α is expressed in terms of the differentials shown in the figure (b) as cos α =

dy ds

(2)

where ds = dx 2 + dy 2

(3)

m α α

(b)

(a)

The differentials, dx and dy, can be computed from the defining equations for x(φ) and y(φ) above: dx = a (1 − cos φ ) dφ    dy = − a sin φ dφ 

(4)

Therefore, ds2 = dx 2 + dy 2 = a 2 (1 − cos φ ) + sin 2 φ  dφ 2 = 2a 2 (1 − cos φ ) dφ 2   2

= 4 a 2 sin 2

φ 2

dφ 2

(5)

so that ds = 2a sin

φ 2

dφ

(6)

Thus, dy − a sin φ dφ = ds 2a sin φ dφ 2 = − cos The velocity of the pendulum bob is

φ 2

= cos α

(7)

CHAPTER 3

ds φ dφ = 2a sin 2 dt dt

= −4 a

d  φ cos   2 dt 

(8)

v = −4 a

φ d2  cos  2  2 dt 

(9)

from which

Letting z ≡ cos

φ 2

be the new variable, and substituting (7) and (9) into (1), we have −4maz = mgz

(10)

g z=0 4a

(11)

or, z+

which is the standard equation for simple harmonic motion, z + ω 02 z = 0

(12)

If we identify

ω0 = where we have used the fact that

(13)

= 4a .

Thus, the motion is exactly isochronous, independent of the amplitude of the oscillations. This fact was discovered by Christian Huygene (1673). 3-9. The equation of motion for 0 ≤ t ≤ t0 is

mx = − k ( x − x0 ) + F = − kx + ( F + kx0 )

(1)

mx = − k ( x − x0 ) = − kx + kx0

(2)

while for t ≥ t0 , the equation is

It is convenient to define

ξ = x − x0 which transforms (1) and (2) into

mξ = − kξ + F ; mξ = − kξ ;

0 ≤ t ≤ t0

t ≥ t0

(3) (4)

OSCILLATIONS

The homogeneous solutions for both (3) and (4) are of familiar form ξ ( t ) = Ae iω t + Be − iω t , where

ω = k m . A particular solution for (3) is ξ = F k . Then the general solutions for (3) and (4) are ξ− =

F + Ae iω t + Be − iω t ; k

ξ+ = Ce iω t + De − iω t ;

0 ≤ t ≤ t0

t ≥ t0

(5) (6)

To determine the constants, we use the initial conditions: x ( t = 0 ) = x0 and x(t = 0) = 0. Thus,

ξ− ( t = 0 ) = ξ− ( t = 0 ) = 0

(7)

The conditions give two equations for A and B: F  + A+B  k  0 = iω ( A − B)  0=

(8)

Then A=B=−

F 2k

and, from (5), we have

ξ− = x − x 0 =

F (1 − cos ω t ) ; k

0 ≤ t ≤ t0

(9)

Since for any physical motion, x and x must be continuous, the values of ξ− ( t = t0 ) and

ξ− ( t = t0 ) are the initial conditions for ξ+ ( t ) which are needed to determine C and D:     F ξ+ ( t = t0 ) = ω sin ω t0 = iω Ce iω t0 − De − iω t0   k 

ξ+ ( t = t0 ) =

F 1 − cos ω t0 ) = Ce iω t0 + De − iω t0 ( k

(10)

The equations in (10) can be rewritten as: F  1 − cos ω t0 )  ( k   − iF  sin ω t0 Ce iω t0 − De − iω t0 = k 

Ce iω t0 + De − iω t0 =

(11)

Then, by adding and subtracting one from the other, we obtain

(

)

F − iω t0  1 − e iω t0  e 2k   F iω t0 1 − e − iω t0  D= e 2k  C=

(

)

(12)

CHAPTER 3

Substitution of (12) into (6) yields

(

)

(

)

ξ+ =

F  e − iω t0 − 1 e iω t + e iω t0 − 1 e − iω t   2k 

F  iω (t − t0 ) iω t − iω t − t − e + e ( 0 ) − e − iω t  e 2k 

F  cos ω ( t − t0 ) − cos ω t  k

(13)

Thus, x − x0 =

3-10.

F  cos ω ( t − t0 ) − cos ω t  ; t ≥ t0 k

(14)

The amplitude of a damped oscillator is expressed by x ( t ) = Ae − βt cos (ω 1t + δ )

(1)

Since the amplitude decreases to 1 e after n periods, we have

β nT = β n

2π

ω1

(2)

Substituting this relation into the equation connecting ω 1 and ω 0 (the frequency of undamped oscillations), ω 12 = ω 02 − β 2 , we have 1   ω1   = ω 12 1 + 2 2    2π n   4π n 

ω 02 = ω 12 + 

(3)

Therefore,

ω1  1  = 1 + 2 2  ω 0  4π n 

−1 2

(4)

so that

ω1 1 ≅ 1− 2 2 8π n ω2

3-11.

The total energy of a damped oscillator is 1 1 2 2 mx ( t ) + kx ( t ) 2 2

(1)

x ( t ) = Ae − βt cos (ω 1t − δ )

(2)

x ( t ) = Ae − βt  −β cos (ω 1t − δ ) − ω 1 sin (ω 1t − δ ) 

(3)

E (t) = where

OSCILLATIONS

ω 1 = ω 02 − β 2 ,

ω0 =

k m

Substituting (2) and (3) into (1), we have

E (t) =

A2 −2 βt  mβ 2 + k cos 2 (ω 1t − δ ) + mω 12 sin 2 (ω 1t − δ ) e  2

(

)

+ 2mβω 1 sin (ω 1t − δ ) cos (ω 1t − δ ) 

(4)

Rewriting (4), we find the expression for E(t): E (t) =

mA2 −2 βt  2 β cos 2 (ω 1t − δ ) + β ω 02 − β 2 sin 2 (ω 1t − δ ) + ω 02  e   2

Taking the derivative of (5), we find the expression for

dE : dt

dE mA2 −2 βt  2βω 02 − 4β 3 cos 2 (ω 1t − δ ) e =  dt 2 − 4β 2 ω 02 − β 2 sin 2 (ω 1t − δ ) 0 − 2βω 2  

(

(5)

)

(6)

The above formulas for E and dE dt reproduce the curves shown in Figure 3-7 of the text. To find the average rate of energy loss for a lightly damped oscillator, let us take β ω 0 . This means that the oscillator has time to complete some number of periods before its amplitude decreases considerably, i.e. the term e −2 βt does not change much in the time it takes to complete one period. The cosine and sine terms will average to nearly zero compared to the constant term in dE dt , and we obtain in this limit dE dt

− mβω 02 A2 e −2 βt

(7)

3-12. θ ℓ mg sin θ mg

The equation of motion is − m θ = mg sin θ

θ=−

sin θ

If θ is sufficiently small, we can approximate sin θ ≅ θ , and (2) becomes

(1) (2)

CHAPTER 3

θ=− θ

(3)

θ ( t ) = θ 0 cos ω 0 t

(4)

which has the oscillatory solution

where ω 0 = g and where θ 0 is the amplitude. If there is the retarding force 2m g θ , the equation of motion becomes − m θ = mg sin θ + 2m g θ

(5)

or setting sin θ ≅ θ and rewriting, we have

θ + 2ω 0θ + ω 02θ = 0

(6)

Comparing this equation with the standard equation for damped motion [Eq. (3.35)], x + 2β x + ω 02 x = 0

(7)

we identify ω 0 = β . This is just the case of critical damping, so the solution for θ ( t) is [see Eq. (3.43)]

θ ( t ) = ( A + Bt ) e −ω0t

(8)

For the initial conditions θ ( 0 ) = θ 0 and θ(0) = 0, we find

θ ( t ) = θ 0 (1 + ω 0 t ) e −ω 0t

3-13.

For the case of critical damping, β = ω 0 . Therefore, the equation of motion becomes x + 2β x + β 2 x = 0

(1)

x (t) = y (t) e −β t

(2)

If we assume a solution of the form

we have x = ye − β t − β ye − β t

  −β t −β t −β t  2 x = ye − 2β ye + β ye 

(3)

ye − β t − 2β ye − β t + β 2 ye − β t + 2β ye − β t − 2β 2 ye − β t + β 2 ye − β t = 0

(4)

y=0

(5)

y ( t ) = A + Bt

(6)

Substituting (3) into (1), we find

or,

Therefore,

OSCILLATIONS

and x ( t ) = ( A + Bt ) e − β t

(7)

which is just Eq. (3.43). 3-14.

For the case of overdamped oscillations, x(t) and x ( t ) are expressed by x ( t ) = e − β t  A1eω 2t + A2 e −ω 2t 

(

) (

(1)

)

x ( t ) e − β t  −β A1eω 2t + + A2 e −ω 2t + A1ω 2 eω 2t − A2ω 2 e −ω 2t 

(2)

where ω 2 = β 2 − ω 02 . Hyperbolic functions are defined as cosh y =

ey + e− y , 2

sinh y =

ey − e−y 2

(3)

or, e y = cosh y + sinh y   −y e = cosh y − sinh y 

(4)

Using (4) to rewrite (1) and (2), we have x ( t ) = ( cosh β t − sinh β t ) ( A1 + A2 ) cosh ω 2t + ( A1 − A2 ) sinh ω 2t 

(5)

and x ( t ) = ( cosh β t − sinh βt ) ( A1ω 2 − A1β ) ( cosh ω 2t + sinh ω 2t ) − ( A2 β + A2ω 2 ) ( cosh ω 2t − sinh ω 2t ) 

3-15.

(6)

We are asked to simply plot the following equations from Example 3.2: x ( t ) = Ae − β t cos (ω 1t − δ )

(1)

v(t) = − Ae − β t β cos (ω 1t − δ ) + ω 1 sin (ω 1t − δ ) 

(2)

with the values A = 1 cm , ω 0 = 1 rad ⋅ s −1 , β = 0.1 s −1 , and δ = π rad. The position goes through x = 0 a total of 15 times before dropping to 0.01 of its initial amplitude. An exploded (or zoomed) view of figure (b), shown here as figure (B), is the best for determining this number, as is easily shown.

CHAPTER 3

(b)

0.5 0 –0.5

x(t) (cm) v(t) (cm/s)

–1

t (s) 1

(c)

v (cm/s)

0.5

–0.5

–1

–0.5

0.5

x (cm) 0.01

x (cm)

(B)

–0.01

t (s)

3-16.

If the damping resistance b is negative, the equation of motion is x − 2β x + ω 02 x = 0

(1)

where β ≡ − b 2m > 0 because b < 0. The general solution is just Eq. (3.40) with β changed to –β: x ( t ) = e β t  A1 exp 

( β − ω t) + A exp( − β − ω t)  2

2 0

(2)

From this equation, we see that the motion is not bounded, irrespective of the relative values of β 2 and ω 02 . The three cases distinguished in Section 3.5 now become: a) If ω 02 > β 2 , the motion consists of an oscillatory solution of frequency ω 1 = ω 02 − β 2 , multiplied by an ever-increasing exponential:

OSCILLATIONS

x ( t ) = e β t  A1e iω1t + A2 e − iω1t 

(3)

x ( t ) = ( A + Bt ) e β t

(4)

x ( t ) = e β t  A1eω 2t + A2 e −ω 2t 

(5)

ω 2 = β 2 − ω 02 ≤ β

(6)

b) If ω 02 = β 2 , the solution is

which again is ever-increasing. c) If ω 02 < β 2 , the solution is:

where

This solution also increases continuously with time. The tree cases describe motions in which the particle is either always moving away from its initial position, as in cases b) or c), or it is oscillating around its initial position, but with an amplitude that grows with the time, as in a). Because b < 0, the medium in which the particle moves continually gives energy to the particle and the motion grows without bound. 3-17.

For a damped, driven oscillator, the equation of motion is x = 2β x + ω 02 x = A cos ω t

(1)

and the average kinetic energy is expressed as T =

mA2 ω2 4 ω 02 − ω 2 2 + 4ω 2 β 2

(

)

(2)

Let the frequency n octaves above ω 0 be labeled ω 1 and let the frequency n octaves below ω 0 be labeled ω 2 ; that is

ω 1 = 2n ω 0 ω 2 = 2− n ω 0

(3)

The average kinetic energy for each case is T ω1 =

22 n ω 02 mA2 4 ω 02 − 22 n ω 02 2 + (4)22 n ω 02 β 2

(4)

T ω2 =

2−2 n ω 02 mA2 4 ω 02 − 2−2 n ω 02 2 + (4)2−2 n ω 02 β 2

(5)

(

)

Multiplying the numerator and denominator of (5) by 24 n , we have

CHAPTER 3

22 n ω 02 mA2 4 ω 02 − 22 n ω 02 2 + (4)22 n ω 02 β 2

T ω2 =

(

)

Hence, we find

T ω1 = T ω 2

(6)

and the proposition is proven. 3-18. Since we are near resonance and there is only light damping, we have ω 0 ω R ω , where ω is the driving frequency. This gives Q ω 0 2β . To obtain the total energy, we use the solution to the driven oscillator, neglecting the transients:

x ( t ) = D cos (ω t − δ )

(1)

We then have

1 1 mD2 2 ω sin 2 (ω t − δ ) + ω 02 cos 2 (ω t − δ )  mx 2 + kx 2 = 2 2 2 

1 mω 02 D2 2

(2)

The energy lost over one period is

∫ ( 2mβ x ) ⋅ ( xdt ) = 2π mωβ D T

(3)

where T = 2π ω . Since ω

ω 0 , we have E energy lost over one period

ω0 4πβ

Q 2π

(4)

which proves the assertion. 3-19.

The amplitude of a damped oscillator is [Eq. (3.59)]

(1)

(ω 02 − ω 2 ) + 4ω 2β 2 2

At the resonance frequency, ω = ω R = ω 02 − β 2 , D becomes DR =

Let us find the frequency, ω = ω ′ , at which the amplitude is A 1 1 DR = = 2 2 2β ω 02 − β 2 Solving this equation for ω ′ , we find

(2)

2β ω 02 − β 2 1 DR : 2 A

(ω − ω ′ ) + 4ω ′ β 2 0

2 2

(3)

OSCILLATIONS

 β2  ω ′ = ω − 2β ± 2βω 0 1 − 2   ω0  2

2 0

(4)

For a lightly damped oscillator, β is small and the terms in β 2 can be neglected. Therefore,

ω ′ 2 ≅ ω 02 ± 2βω 0

(5)

or, 

ω ′ ≅ ω 0 1 ± 

β  ω 0 

(6)

which gives ∆ω = (ω 0 + β ) − (ω 0 − β ) = 2β

(7)

We also can approximate ω R for a lightly damped oscillator:

ω R = ω 02 − 2β 2 ≅ ω 0

(8)

Therefore, Q for a lightly damped oscillator becomes Q≅

3-20.

ω0 ω0 ≅ 2β ∆ω

(9)

From Eq. (3.66), x=

− Aω

(ω 02 − ω 2 ) + 4ω 2β 2 2

sin (ω t − δ )

(1)

Therfore, the absolute value of the velocity amplitude v is given by v0 =

Aω

(ω − ω ) + 4ω β 2 0

2 2

(2) 2

The value of ω for v0 a maximum, which is labeled ω v , is obtained from ∂ v0 =0 ∂ω ω =ω v

(3)

and the value is ω v = ω 0 . Since the Q of the oscillator is equal to 6, we can use Eqs. (3.63) and (3.64) to express β in terms of ω 0 :

β2 =

ω 02 146

We need to find two frequencies, ω 1 and ω 2 , for which v0 = vmax We find

(4) 2 , where vmax = v0 (ω = ω 0 ) .

CHAPTER 3

Aω

vmax A = = 2 2 2β

(ω 02 − ω 2 ) + 4ω 2β 2 2

(5)

Substituting for β in terms of ω 0 from (4), and by squaring and rearranging terms in (5), we obtain

(ω − ω ) − 732 (ω ω ) = 0 2 2 1,2

2 0

2 1,2

2 0

(6)

from which 2 ω 02 − ω 1,2 =±

2 1 ω 1,2ω 0 ≅ ± ω 1,2ω 0 73 6

(7)

 ω0  ± ω0  12 

(8)

Solving for ω 1 , ω 2 we obtain

ω 1,2 ≅  ±

It is sufficient for our purposes to consider ω 1 , ω 2 positive: then

ω1 ≅

ω0 12

+ ω0 ;

ω2 ≅ −

ω0 12

+ ω0

(9)

so that ∆ω = ω 1 − ω 2 =

ω0

(10)

A graph of v0 vs. ω for Q = 6 is shown. v0

A vmax = 2β A 2 2β

∆ω

ω0 ω0 12 12

1 ω0 6

ω0

3-21.

We want to plot Equation (3.43), and its derivative: x ( t ) = ( A + Bt ) e − β t

(1)

v ( t ) = [ B − β ( A + Bt ) ] e − β t

(2)

where A and B can be found in terms of the initial conditions A = x0

(3)

B = v0 + β x 0

(4)

OSCILLATIONS

The initial conditions used to produce figure (a) were ( x0 , v0 ) = ( −2, 4 ) , (1, 4) , (4,−1) , (1,−4) , (−1,−4) , and (−4, 0) , where we take all x to be in cm, all v in cm ⋅ s −1 , and β = 1 s −1 . Figure (b) is a magnified view of figure (a). The dashed line is the path that all paths go to asymptotically as t → ∞. This can be found by taking the limits. lim v(t) = −β Bte − β t

(5)

lim x(t ) = Bte − β t

(6)

t →∞

so that in this limit, v = –βx, as required. 4

(a)

3 2

v (cm/s)

1 0 –1 –2 –3 –4

–4

–2

0 x (cm)

–0.5

–0.25

0 x (cm)

0.25

0.5

(b) 0.4

v (cm/s)

0.2

–0.2

–0.4

3-22.

For overdamped motion, the position is given by Equation (3.44) x ( t ) = A1e − β1t + A2 e − β2t

(1)

100

CHAPTER 3

The time derivative of the above equation is, of course, the velocity:

v ( t ) = − A1β1 e − β1t − A2 β 2 e − β2t

(2)

x0 = A1 + A2

(3)

v0 = − A1β1 − A2 β 2

(4)

At t = 0:

The initial conditions x0 and v0 can now be used to solve for the integration constants A1 and A2 . b) When A1 = 0 , we have v0 = − β 2 x0 and v ( t ) = −β 2 x ( t ) quite easily. For A1 ≠ 0 , however, we

have v ( t ) → −β1 A1 e − β1t = − β1 x as t → ∞ since β1 < β 2 .

3-23. Firstly, we note that all the δ = π solutions are just the negative of the δ = 0 solutions. The δ = π 2 solutions don’t make it all the way up to the initial “amplitude,” A , due to the

retarding force. Higher β means more damping, as one might expect. When damping is high, less oscillation is observable. In particular, β 2 = 0.9 would be much better for a kitchen door than a smaller β, e.g. the door closing (δ = 0), or the closed door being bumped by someone who then changes his/her mind and does not go through the door ( δ = π 2 ).

101

OSCILLATIONS

β2 = 0.1, δ = 0

β2 = 0.5, δ = 0

β2 = 0.9, δ = 0

β2 = 0.1, δ = π/2

β2 = 0.5, δ = π/2

β2 = 0.9, δ = π/2

β2 = 0.1, δ = π

β2 = 0.5, δ = π

β2 = 0.9, δ = π

0.5

–0.5

–1

0.5

–0.5

–1

0.5

–1

3-24. As requested, we use Equations (3.40), (3.57), and (3.60) with the given values to evaluate the complementary and particular solutions to the driven oscillator. The amplitude of the complementary function is constant as we vary ω, but the amplitude of the particular solution becomes larger as ω goes through the resonance near 0.96 rad ⋅ s −1 , and decreases as ω is increased further. The plot closest to resonance here has ω ω 1 = 1.1 , which shows the least distortion due to transients. These figures are shown in figure (a). In figure (b), the ω ω 1 = 6

plot from figure (a) is reproduced along with a new plot with Ap = 20 m ⋅ s −2 .

102

CHAPTER 3

ω/ω1 = 1/9

–1

ω/ω1 = 1.1

ω/ω1 = 1/3

–2

–1 0

ω/ω1 = 3

ω/ω1 = 6

0.5

–0.5

xc xp x

Legend: –1

–1

t (s)

(a) 0.5

0.5 0

–0.5

Ap = 1 –1

Ap = 20

–1

(b)

3-25. This problem is nearly identical to the previous problem, with the exception that now Equation (3.43) is used instead of (3.40) as the complementary solution. The distortion due to the transient increases as ω increases, mostly because the complementary solution has a fixed amplitude whereas the amplitude due to the particular solution only decreases as ω increases. The latter fact is because there is no resonance in this case.

103

OSCILLATIONS

ω/ω1 = 1/9

ω/ω1 = 1/3

ω/ω1 = 1.1

0.5

–0.5

–1

ω/ω1 = 6

ω/ω1 = 3

ω/ω1 = 6, Ap = 6

–0.5

–1

Legend:

3-26.

–1

10 xc

–0.5

–1

The equations of motion of this system are m1 x1 = − kx1 − b1 ( x1 − x2 ) + F cos ω t    m2 x2 = − b2 x2 − b1 ( x2 − x1 ) 

(1)

The electrical analog of this system can be constructed if we substitute in (1) the following equivalent quantities: 1 ; C

m1 → L1 ;

k→

m2 → L2 ;

F → ε0 ;

b1 → R1 ; x → q b2 → R2

Then the equations of the equivalent electrical circuit are given by 1  q1 = ε 0 cos ω t  C   L2 q2 + R2 q2 + R1 ( q2 − q1 ) = 0  L1q1 + R1 ( q1 − q2 ) +

(2)

Using the mathematical device of writing exp(iω t) instead of cos ω t in (2), with the understanding that in the results only the real part is to be considered, and differentiating with respect to time, we have

104

CHAPTER 3

(

)

I  L1I1 + R1 I1 − I 2 + 1 = iωε 0 e iω t  C   L2 I 2 + R2 I 2 + R1 I 2 − I1 = 0 

( )

(

(3)

)

Then, the equivalent electrical circuit is as shown in the figure: L1 ε0 cos ωt

I2(t) R1

I1(t) C 1

R2 L2 2

The impedance of the system Z is 1 + Z1 ωC

(4)

1 1 1 = + Z1 R1 R2 + iω L2

(5)

Z = iω L1 − i where Z1 is given by

Then, Z1 =

R1  R2 ( R2 + R1 ) + ω 2 L22 + iω L2 R1 

(6)

( R1 + R2 )2 + ω 2 L22

and substituting (6) into (4), we obtain    1  2 R1  R2 ( R2 + R1 ) + ω 2 L22  + i  R1ω L2 +  ω L1 − R1 + R2 ) + ω 2 L22  (  ω C    Z= 2 2 2 ( R1 + R2 ) + ω L2

(

3-27.

)

(7)

From Eq. (3.89), F (t) =

∞ 1 a0 + ∑ ( an cos nω t + bn sin nω t ) 2 n =1

(1)

∞ 1 a0 + ∑ cn cos ( nω t − φ n ) 2 n=1

(2)

We write F (t) =

which can also be written using trigonometric relations as F (t) =

∞ 1 a0 + ∑ cn cos nω t cos φ n + sin nω t sin φ n  2 n =1

Comparing (3) with (2), we notice that if there exists a set of coefficients cn such that

(3)

105

OSCILLATIONS

cn cos φ n = an   cn sin φ n = bn 

(4)

then (2) is equivalent to (1). In fact, from (4), cn2 = an2 + bn2   b  tan φ n = n  an 

(5)

with an and bn as given by Eqs. (3.91).

3-28. Since F(t) is an odd function, F(–t) = –F(t), according to Eq. (3.91) all the coefficients an vanish identically, and the bn are given by

bn =

ω ωπ π F ( t ′ ) sin nω t ′ dt ′ π ∫−ω

π  ω 0 ω  − ∫ − π sin nω t ′ dt ′ + ∫ 0 sin nω t ′ dt ′  π  ω 

π  ω   1  ω   1 cos cos ω ω n t n t + − ′ ′  0   π  nω π   nω − 0





2 ( cos 0 − cos nπ ) nπ

 4 for n odd  =  nπ  0 for n even 

(1)

Thus, 4  ( 2n + 1) π  n = 0, 1, 2, …  b( 2 n) = 0 

b( 2 n + 1) =

(2)

Then, we have F (t) =

sin ω t +

4 4 sin 3ω t + sin 5ω t + … 3π 5π

(3)

106

CHAPTER 3

F(t) 1 –π ⁄ω π ⁄ω

–1

0.849 Terms 1 + 2

–π ⁄ω π ⁄ω –0.849

1.099 –π ⁄ω π ⁄ω

Terms 1 + 2 + 3 t

–1.099

0.918 –π ⁄ω π ⁄ω

Terms 1 + 2 + 3 + 4 t

–0.918

3-29. In order to Fourier analyze a function of arbitrary period, say τ = 2P ω instead of 2π ω , proportional change of scale is necessary. Analytically, such a change of scale can be represented by the substitution

πt P

(1)

for when t = 0, then x = 0, and when t = τ = 2P ω , then x = 2π ω . Thus, when the substitution t = Px π is made in a function F(t) of period 2P ω ′ , we obtain the function

 Px  F   = f ( x) π 

(2)

and this, as a function of x, has a period of 2π ω . Now, f(x) can, of course, be expanded according to the standard formula, Eq. (3.91):

f ( x) = where

∞ 1 a0 + ∑ ( an cos nω x + bn sin nω x ) 2 n =1

(3)

107

OSCILLATIONS

an =

 ω 2ωπ f ( x ′ ) cos nω x ′ dx ′  ∫ 0 π 

(4)

 ω bn = ∫ ω f ( x ′ ) sin nω x ′ dx ′  π 0  2π

If, in the above expressions, we make the inverse substitutions

πt P

and

dx =

π P

(5)

the expansion becomes ∞ a  π t  P π t  nωπ t   nωπ t   f   = F  ⋅  = F ( t ) = 0 + ∑  an cos  + bn sin    2 n =1  P π P   P   P  

(6)

and the coefficients in (4) become

  nωπ t ′  dt ′  0 P    ω 2ωP   nωπ t ′  bn = ∫ F ( t ′ ) sin  dt ′   0 P  P   an =

ω F t cos ( ′)  P∫ 

(7)

4π

For the case corresponding to this problem, the period of F(t) is

, so that P = 2π. Then,

substituting into (7) and replacing the integral limits 0 and τ by the limits −

τ 2

and +

τ 2

, we

obtain an =

 ω 2ωπ  nω t ′  dt ′  2π F ( t ′ ) cos  ∫  2π − ω  2   

 ω 2ωπ  nω t ′  dt ′  bn = 2π F ( t ′ ) sin  ∫  − 2π ω  2  

(8)

and substituting into (6), the expansion for F(t) is F (t) =

a0 ∞   nω t   nω t   + ∑  an cos  + bn sin    2 n =1   2   2  

(9)

Substituting F(t) into (8) yields an =

 ω 2ωπ  nω t ′  dt ′  sin ω t ′ cos  ∫  0 2π  2   

ω 2ωπ   nω t ′  dt ′  bn = sin ω t ′ sin  ∫  2π 0  2   Evaluation of the integrals gives

(10)

108

CHAPTER 3

b2 =

1 ; bn = 0 2

0   an ( n ≥ 2 ) =  −4  2  π n − 4

a0 = a1 = 0

   n even    n odd   

for n ≠ 2

(

)

(11)

and the resulting Fourier expansion is F (t) =

3-30.

ωt 4 3ω t 5ω t 7ω t 1 4 4 4 − − − +… sin ω t + cos cos cos cos 2 3π 2 5π 2 21π 2 45π 2

(12)

The output of a full-wave rectifier is a periodic function F(t) of the form

π   − sin ω t ; − ω < t ≤ 0 F (t) =   π  sin ω t ; 0<t< ω 

(1)

The coefficients in the Fourier representation are given by an =

bn =

π  ω 0 ω sin ω t cos nω t dt ω ω sin t cos n t dt − + ′ ′ ′ ′ ′ ′  ( ) π ∫ − ∫ 0 π ω  π ω sin

ω 0  π ( − sin ω t ′ ) sin nω t ′ dt ′ + ∫ 0 π ∫ − ω

   ω t ′ sin nω t ′ dt ′    

(2)

Performing the integrations, we obtain 4    π 1 − n2 ; if n even ( or 0 )  an =     0; if n odd     for all n bn = 0 

(

)

(3)

The expansion for F(t) is F (t) =

−

4 4 cos 2ω t − cos 4ω t … 3π 15π

The exact function and the sum of the first three terms of (4) are shown below.

(4)

109

OSCILLATIONS

Sum of first three terms

F(t) 1

sin ωt

–π

3-31.

–π ⁄ 2

π⁄2

ωt

We can rewrite the forcing function so that it consists of two forcing functions for t > τ :  0  F (t)  =  a (t τ ) m   a (t − τ )  a (t τ ) − τ 

t<0 0<t<τ

(1)

t >τ

During the interval 0 < t < τ, the differential equation which describes the motion is x + 2β x + ω 02 x =

(2)

The particular solution is x p = Ct + D , and substituting this into (2), we find 2β C + ω 02Ct + ω 02 D =

(3)

from which 2β C + ω 02 D = 0    a Cω 02 − = 0  τ 

(4)

Therefore, we have D=−

2β a

ω τ 4 0

(5)

ω 02τ

which gives xp =

ωτ 2 0

t−

2β a

(6)

ω 04τ

Thus, the general solution for 0 < t < τ is x ( t ) = e − β t A cos ω 1t + B sin ω 1t  + and then,

ωτ 2 0

t−

2β a

ω 04τ

(7)

110

CHAPTER 3

x ( t ) = − β e − β t A cos ω 1t + B sin ω 1t  + ω 1e − β t − A sin ω 1t + B cos ω 1t  +

ω 02τ

(8)

The initial conditions, x(0) = 0, x ( 0 ) = 0 , implies 2β a

     a  2β 2 B= − 1  ω 1ω 02τ  ω 02  A=

ω 04τ

(9)

Therefore, the response function is x (t) = For the forcing function −

 2β  a  2β − βt e − β t  2β 2 ω + cos e t 1  2 − 1 sin ω 1t + t − 2  2  2 ω0τ ω0 ω1  ω 0 ω0   a (t − τ )

(10)

in (1), we have a response similar to (10). Thus, we add these

two equations to obtain the total response function: x (t) =

 a  2β − β t e − βt  2β 2 βt − − + cos cos e t e t ω ω τ ( )  1 1  2 − 1 2 2 ω1  ω 0 ω0τ ω0 

(

)

(11)

×  sin ω 1t + − e sin ω 1 ( t − τ )  + τ  βt

When τ → 0, we can approximate e βτ as 1 + βτ, and also sin ω 1τ ≅ ω 1τ , cos ω 1τ ≅ 1 . Then, x ( t )  → τ →0

 a  2β − β t e − β t  2β 2   ω βτ ω ω τ ω cos 1 cos sin e t t t − + + + ( ) ( 1 1 1 1 )  2 − 1 2  2  ω0τ ω0 ω1  ω 0  × sin ω 1t − (1 + βτ ) ( sin ω 1t − ω 1τ cos ω 1t )  + τ 

 β 2β 3  a  −β t − β t  2βω 1 − − − + 1 cos sin ω 1t  ω e t e 1  2 2  2 ω0  ω 1 ω 1ω 0   ω0 

(12)

If we use ω 12 = ω 02 − β 2 , the coefficient of e − β t sin ω 1t becomes β ω 1 . Therefore, → x ( t )  τ →0

 a  β 1 − e − β t cos ω 1t − e − β t sin ω 1t  2  ω0  ω1 

This is just the response for a step function. 3-32. a) Response to a Step Function:

From Eq. (3.100) H ( t0 ) is defined as

(13)

111

OSCILLATIONS

 0, t < t0 H ( t0 ) =   a1 , t > t0

(1)

With initial conditions x ( t0 = 0 ) and x ( t0 = 0 ) , the general solution to Eq. (3.102) (equation of motion of a damped linear oscillator) is given by Eq. (3.105):   β e − β (t −t0 ) a  − β ( t − t0 ) ω 1 cos sin ω 1 ( t − t0 )  for t > t0  e t t − − − 1( 0) 2  ω 0  ω1    for t < t0  x (t) = 0

x (t) =

(2)

where ω 1 = ω 02 − β 2 . For the case of overdamping, ω 02 < β 2 , and consequently ω 1 = i β 2 − ω 02 is a pure imaginary

number. Hence, cos ω 1 ( t − t0 ) and sin ω 1 ( t − t0 ) are no longer oscillatory functions; instead,

they are transformed into hyperbolic functions. Thus, if we write ω 2 = β 2 − ω 02 (where ω 2 is real), cos ω 1 ( t − t0 ) = cos iω 2 ( t − t0 ) = cosh ω 2 ( t − t0 )    sin ω 1 ( t − t0 ) = sin iω 2 ( t − t0 ) = i sinh ω 2 ( t − t0 ) 

(3)

The response is given by [see Eq. (3.105)]   β e − β (t −t0 ) a  − β ( t − t0 ) ω − − − 1 cosh sinh ω 1 ( t − t0 )  for t > t0  e t t 2( 0) 2  ω 0  ω2    for t < t0  x (t) = 0

x (t) =

(4)

For simplicity, we choose t0 = 0 , and the solution becomes

x (t) =

 H ( 0)  βe−β t −β t − − ω e t 1 cosh sinh ω 2t  2 2  ω0  ω2 

(5)

This response is shown in (a) below for the case β = 5 ω 0 . b) Response to an Impulse Function (in the limit τ → 0):

From Eq. (3.101) the impulse function I ( t0 , t1 ) is defined as 0  I ( t0 , t1 ) =  a   0

t < t0 t0 < t < t1 t > t1

For t1 − t2 = τ → 0 in such a way that aτ is constant = b, the response function is given by Eq. (3.110):

(6)

112

CHAPTER 3

x (t) =

ω1

− β t −t e ( 0 ) sin ω 1 ( t − t0 ) for t > t0

(7)

Again taking the “spike” to be at t = 0 for simplicity, we have

x (t) =

ω1

e − βt sin ω 1 ( t )

for t > 0

(8)

For ω 1 = iω 2 = i β 2 − ω 02 (overdamped oscillator), the solution is

x (t) =

ω2

e − βt sinh ω 2t ; t > 0

(9)

This response is shown in (b) below for the case β = 5 ω 0 . 1

x  H ( 0) ω 02 

(a)

0.5

ω0t 1

x (b ω 2 )

(b)

0.5

5 ω0t

3-33. a) In order to find the maximum amplitude of the response function shown in Fig. 3-22, we look for t1 such that x ( t ) given by Eq. (3.105) is maximum; that is,

(

∂ x (t)

)

∂t

(1)

t = t1

From Eq. (3.106) we have

(

∂ x (t) ∂t

) = H ( 0) e ω

2 0

−β t

 β2  ω 1 +  sin ω 1t ω1  

(2)

113

OSCILLATIONS

For β = 0.2ω 0 , ω 1 = ω 02 − β 2 = 0.98ω 0 . Evidently, t1 = π ω 1 makes (2) vanish. (This is the absolute maximum, as can be seen from Fig. 3-22.) Then, substituting into Eq. (3.105), the maximum amplitude is given by βπ −  a  ω1 x ( t ) max = x ( t1 ) = 2 1 + e  ω 0  

(3)

or, x ( t1 ) ≅ 1.53

ω 02

(4)

b) In the same way we find the maximum amplitude of the response function shown in Fig. 3-24 by using x(t) given in Eq. (3.110); then,

(

∂ x (t) ∂t

) t = t1

 β − β t −t  sin ω 1 ( t − t0 )  = be ( 0 )  cos ω 1 ( t − t0 ) − ω1   t = t1

(5)

ω  1 1.37 tan −1  1  = tan −1 ( 4.9) = ω1 ω1  β  ω1

(6)

If (5) is to vanish, t1 is given by t1 − t0 =

Substituting (6) into Eq. (3.110), we obtain (for β = 0.2ω 0 ) −β

× 1.37 b x ( t ) max = x ( t1 ) = sin ( 1.37 ) e ω1 0.98ω 0

(7)

or, x ( t1 ) ≅ 0.76

aτ

ω0

(8)

The response function of an undamped (β = 0) linear oscillator for an impulse function 2π I(0,τ ), with τ = , can be obtained from Eqs. (3.105) and (3.108) if we make the following

3-34.

ω0

substitutions:

β=0; t0 = 0 ;

ω1 = ω 0

  2π  t1 = τ =  ω0 

(For convenience we have assumed that the impulse forcing function is applied at t = 0.) Hence, after substituting we have

(1)

114

CHAPTER 3

x (t) = 0

t<0

x (t) =

ω 02 

1 − cos ω 0 t 

x (t) =

cos ( w0 t − 2π ) − cos w0 t  = 0

  2π  0<t<  ω0   2π  t >τ = ω 0 

ω 02 

(2)

This response function is shown below. Since the oscillator is undamped, and since the impulse lasts exactly one period of the oscillator, the oscillator is returned to its equilibrium condition at the termination of the impulse. 2a ω 02

a ω 02

π 2π ω0 ω0

3-35.

The equation for a driven linear oscillator is x + 2β x + w20 x = f ( t )

where f(t) is the sinusoid shown in the diagram. f(t) a I

III t

Region I:

x=0

(1)

Region II:

x + 2β x + ω 02 x = a sin ω t

(2)

Region III:

x + 2β x + ω 02 x = 0

(3)

x = e − β t ( A sin ω 1t + B cos ω 1t ) + xP

(4)

xP = Da sin (ω t − δ )

(5)

The solution of (2) is

in which

where D=

(ω − ω ) + 4β 2ω 2 2 0

(6)