Problems & Solutions >Tutorial-3 Coupled Oscillator_ PHYSICS Q&A

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Tutorial-3 Coupled Oscillator Problem 1 There are two small massive beeds, each of mass 𝑀, on a taut massless string of length 8𝐿, as shown. The tension in the string is 𝑇.

1. Derive expressions for the normal mode angular frequencies of the system, for small displacements of the system in a plane. Assume that 𝑇 is suffieciently large that you may ignore gravity.

2. If at 𝑡 = 0 both masses are stationary, one is at equilibrium and the other displaced from equilibrium by a distance 𝐴, write an expression for displacement 𝑦1 (𝑡) as a function of time of the mass initially in the equilibrium position.

Solution:

1) Masses cannot be displaced in horizontal direction (x-direction). They can be moved in vertical direction only.

Tension = 𝑇,

The force equations are as follows.

Particle (bead) 1: 𝑚

𝑑 2 𝑦1

𝑇

=−

𝑑𝑡 2

𝑦1 −

3𝐿

𝑇

(𝑦1 − 𝑦2 )

(1)

(𝑦2 − 𝑦1 )

(2)

2𝐿

Particle (bead) 2: 𝑚

𝑑 2 𝑦2 𝑑𝑡 2

=−

𝑇

𝑦2 −

3𝐿

𝑇 2𝐿

Normal modes can be found by writing 𝑦1 and 𝑦2 in linear combination form. 𝑦 +𝑦2

𝑞1 = 1

2

𝑦 −𝑦2

and 𝑞2 = 1

2

After subtracting and adding the values of 𝑞1 and 𝑞2 we can rewrite equation (1) and (2) as follows written as: 𝑚

𝑑 2 𝑞1 𝑑𝑡 2 𝑑 2 𝑞2

and

𝑚

where,

𝑘=

∴

𝑑𝑡 2 𝑇

= −𝑘𝑞1

(3)

= −(𝑘+2𝑘 ′ )𝑞2

(4)

and 𝑘 ′ =

3𝐿

𝑘

𝑇

𝑚

3𝐿𝑚

𝜔1 = √ = √

𝑇 2𝐿

and 𝜔2 = √

𝑘+2𝑘 ′ 𝑚

=√

4𝑇 3𝐿𝑚

So, the expressions for the normal mode angular frequencies are 𝜔1 = √ and 𝜔2 = √

4𝑇 3𝐿𝑚

.

2) Solutions for equation (3) and (4) are 𝑦 +𝑦2

𝑞1 = 1 𝑞2 = From equation (5),

2 𝑦1 −𝑦2 2

= 𝐴1 𝑐𝑜𝑠(𝜔1 𝑡)

(5)

= 𝐴2 𝑐𝑜𝑠(𝜔2 𝑡)

(6)

𝑇 3𝐿𝑚

𝑦1 + 𝑦2 = 𝐴1 𝑐𝑜𝑠(𝜔1 𝑡) 2 𝑦1 = 𝐴 and 𝑦2 = 0

at 𝑡 = 0,

∴ 𝐴1 = 𝐴/2 𝐴2 = 𝐴/2

Similarly, So,

𝑞1 = 𝐴/2𝑐𝑜𝑠(𝜔1 𝑡) 𝑞2 = 𝐴/2𝑐𝑜𝑠(𝜔2 𝑡)

(7) (8)

Solving equation (5) and (6), we get 𝑦1 (𝑡) = where, 𝜔1 = √

𝑇 3𝐿𝑚

𝐴 [𝑐𝑜𝑠(𝜔1 𝑡) + 𝑐𝑜𝑠(𝜔2 𝑡)] 2

and 𝜔2 = √

4𝑇 3𝐿𝑚

.

Problem 2

The figure above shows a system of masses in the x-y plane. The mass of 2𝑚 is connected to an immobile wall with a spring of constant 2𝑘, while the mass of 𝑚 is connected to an immobile wall with a spring of constant 𝑘. The masses are then coupled to each other with an elastic band of length 𝐿, under tension 𝑇 = 2𝑘𝐿. The masses are constrained to move in the x direction only. At equilibrium the masses have the same x position and the springs are uncompressed. There is no friction or

gravity. The displacements from equilibrium are small enough (𝑥1 , 𝑥2 ≪ 𝐿), so that the tension in the band stays constant.

1. Find the normal mode frequencies of the system.

2. What are the ratios of the displacements from equilibrium for two different normal modes? (Hint: Basically, you have to find out the eigen functions for corresponding normal mode frequencies) Solution: 1) Equation of motions for 2𝑚 and 𝑚 : 2𝑚 and

𝑚

𝑑 2 𝑥1 𝑑𝑡 2 𝑑 2 𝑥2 𝑑𝑡 2

𝑇

= −2𝑘𝑥1 − (𝑥1 − 𝑥2 ) 𝐿

𝑇

= −𝑘𝑥2 − (𝑥2 − 𝑥1 ) 𝐿

(1) (2)

where, 𝑇 = 2𝑘𝐿 and 𝐿 is length of band.

From equation (1) and (2), 𝑚 and

𝑚

𝑑 2 𝑥1 𝑑𝑡 2

𝑑 2 𝑥2 𝑑𝑡 2

+ 2𝑘𝑥1 − 𝑘𝑥2 = 0

(3)

− 2𝑘𝑥1 + 3𝑘𝑥2 = 0

(4)

Say this coupled oscillator oscillating with a coupled frequency 𝜔 then solution to these equations are 𝑥1 = 𝐴𝑒 𝑖𝜔𝑡

and

𝑥2 = 𝐵𝑒 𝑖𝜔𝑡

Substituting these above two expressions in (3) and (4) the following matrix equation : 2 [−𝑚𝜔 + 2𝑘 −2𝑘

0 𝐴 −𝑘 ][ ] = [ ] 2 0 −𝑚𝜔 + 3𝑘 𝐵

𝜔 can be found by solving eigenvalue equation,

(5)

(−𝑚𝜔2 + 2𝑘)(−𝑚𝜔2 + 3𝑘) − 2𝑘 2 = 0 or,

𝑚2 𝜔4 − 5𝑘𝑚𝜔2 + 4𝑘 2 = 0

or,

𝜔2 =

5𝑘𝑚±√25𝑘 2 𝑚2 −16𝑘 2 𝑚2 2𝑚2

𝜔2 =

or,

𝜔1 = √

∴

𝑘 𝑚

5𝑘±3𝑘 2𝑚

=

𝑘 4𝑘

,

𝑚 𝑚 4𝑘

and 𝜔2 = √ . 𝑚

2) Eigen functions can be calculated usin (5) for 𝝎𝟏 = √ 𝑘 [ −2𝑘

𝒌 𝒎

0 −𝑘 𝐴 ][ ] = [ ] 0 2𝑘 𝐵

𝐴 − 𝐵 = 0; or, 𝐴 = 𝐵

or,

𝒙𝟏

∴ we can write.

𝒙𝟐

=𝟏

So, 𝒙𝟏 = 𝟏 and 𝒙𝟐 = 𝟏; (say) Again, for 𝝎𝟐 = √

𝟒𝒌 𝒎

2𝑘 [ −2𝑘 or, ∴

0 −𝑘 𝐴 ][ ] = [ ] 0 −𝑘 𝐵

−2𝑘𝐴 − 𝐵 = 0; or, 𝐵 = −2𝐴 𝒙𝟐 𝒙𝟏

= −𝟐

So, 𝒙𝟏 = 𝟏 and 𝒙𝟐 = −𝟐; (say)

Problem 3 Two identical spring-mass systems 𝐴 and 𝐵 are coupled by a weak middle spring having a spring constant smaller by a factor of 100 (i.e. 100𝑘𝑚𝑖𝑑𝑑𝑙𝑒 = 𝑘𝐴 = 𝑘𝐵 ).

Mass 𝐴 is pulled by a small distance and released from rest, while mass 𝐵 is released from rest at its equilibrium position, at 𝑡 = 0. Calculate the approximate number of oscillations completed by mass 𝐴 before its oscillations die down first.

Solution:

Force equations for both the masses: 𝑚1

𝑑 2 𝑥1 𝑑𝑡 2

= 𝐹1 = −𝑘1 𝑥1 − 𝑘 ′ (𝑥1 − 𝑥2 )

(1)

= 𝐹2 = −𝑘2 𝑥2 − 𝑘 ′ (𝑥2 − 𝑥1 )

(2)

and 𝑚2

𝑑 2 𝑥2 𝑑𝑡 2

From equation (1) and (2), we can write in matrix form, 𝑑 2 𝑥1 2 𝑥1 𝑚1 0 𝑘1 + 𝑘 ′ −𝑘 ′ 𝑑𝑡 = − ( ) ( ) ( ) 0 𝑚2 𝑑 2 𝑥2 −𝑘 ′ 𝑘2 + 𝑘 ′ 𝑥2 ( 𝑑𝑡 2 ) or, or,

̂ 𝑋̂̈ = −𝐾 ̂ 𝑋̂ 𝑀 ̂ −1 𝐾 ̂ 𝑋̂ 𝑋̂̈ = −𝑀

̂ , 𝑋̂̈ , 𝐾 ̂ and 𝑋̂ all are matrix.

Where, 𝑀 Now, Let us consider a normal mode solution,

(3)

𝑋̂ = 𝐴̂𝑒 𝑖𝜔𝑡

(4)

𝑋̂̈ = −𝜔2 𝑋̂

(5)

or,

Putting the value of 𝑋̂̈ from equation (3) or,

̂ −1 𝐾 ̂ 𝑋̂ = 𝜔2 𝑋̂ 𝑀

or, or,

̂𝑀 ̂ −1 𝐾 ̂ 𝑋̂ = 𝑀 ̂ 𝜔2 𝑋̂ 𝑀 ̂ 𝑋̂ = 𝑀 ̂ 𝜔2 𝑋̂ 𝐾 ̂ = (𝑚 𝑀 0

Calculated above,

0 ); 𝑚

̂ = (𝑘 + 𝜀𝑘 𝐾 −𝜀𝑘 where, 𝑘1 = 𝑘2 = 𝑘

&

(6) 𝑚1 = 𝑚2 = 𝑚

−𝜀𝑘 ); 𝑘 + 𝜀𝑘

𝑘 ′ = 𝜀𝑘 = 𝑘/100

After solving (6) and normalized, we have 𝑘

𝜔1 = √ ; 𝑚

and

1 ( ) 1 1 1 𝐴2 = ( ) 2 √ −1 𝐴1 =

𝑘

𝜔2 = √(1 + 2𝜀) ; 𝑚

1

√2

The most most general solution of coupled oscillator is, 𝑋̂(𝑡) = 𝐶1 𝐴̂1 cos(𝜔1 𝑡) + 𝐶2 𝐴̂2 cos(𝜔2 𝑡)

(7)

At 𝑡 = 0, mass A is released from a small distance = 𝑑 and mass B kept at rest. So, 𝑥 (0) 𝑑 𝑋̂(0) = ( 1 ) = ( ) 𝑥2 (0) 0 From equation (7),

or,

𝐶 1 𝐶 1 𝑑 ( ) = 12 ( ) + 22 ( ), √ √ −1 1 0 𝐶1 = 𝐶2 = 𝑑/√2

So, equation (7) becomes, 𝑋̂ (𝑡) = 𝐶1 𝐴̂1 cos(𝜔1 𝑡) + 𝐶2 𝐴̂2 cos(𝜔2 𝑡) 𝑥 (𝑡) 𝐶 1 𝐶 1 or, ( 1 ) = 12 ( ) cos(𝜔1 𝑡) + 22 ( ) cos(𝜔2 𝑡) √ √ 𝑥2 (𝑡) 1 −1 𝑥 (𝑡) 𝑑 1 𝑑 1 or, ( 1 ) = 2 ( ) cos(𝜔1 𝑡) + 2 ( ) cos(𝜔2 𝑡) 𝑥2 (𝑡) 1 −1

Comparing both side, 𝑑 𝜔2 − 𝜔1 𝜔1 + 𝜔2 ∴ 𝑥1 (𝑡) = [cos(𝜔1 𝑡) + cos(𝜔2 𝑡)] = 𝑑 𝑐𝑜𝑠 ( 𝑡) 𝑐𝑜𝑠 ( 𝑡) 2 2 2 And

2𝜋

4𝜋

2

𝜔2 −𝜔1

𝑇 = Time period of the envelop = 𝜔2−𝜔1 = 𝑇

𝑇

𝜔1 +𝜔2

4

4

2×2𝜋

∴ Number of oscillation in time = ×

1 𝜔1 + 𝜔2 = ( ) 4 𝜔2 − 𝜔1

1

4𝜋

4

𝜔2 −𝜔1

= ×

×

𝜔1 +𝜔2 4𝜋

= =

1 1 + √1 + 2𝜀 × 4 √1 + 2𝜀 − 1

1 √1 + 2/100 + 1 × 4 √1 + 2/100 − 1

=

1 √1 + 1/50 + 1 × 4 √1 + 1/50 − 1 = 50.49 ≈ 50

Problem 4 Three identical masses 𝑚 are connected in series to four identical springs each of force constant 𝑘 (each mass coupled to two strings, and the end springs to walls):

a) Calculate the eigenfrequencies of small oscillations (normal modes) of this coupled oscillator system. b) What sets of initial displacements would you give to the three masses so that the system oscillates in each of these normal modes.

Solution:

a) Equation of motions: Mass 1→

𝑚

𝑑 2 𝑥1 𝑑𝑡 2

= −𝑘𝑥1 − 𝑘(𝑥1 − 𝑥2 )

(1)

Mass 2→

𝑚

Mass 3→

𝑚

𝑑 2 𝑥2 𝑑𝑡 2 𝑑 2 𝑥3 𝑑𝑡 2

= −𝑘((𝑥2 − 𝑥1 ) − 𝑘(𝑥2 − 𝑥3 )

(2)

= −𝑘𝑥3 − 𝑘(𝑥3 − 𝑥2 )

(3)

Say, all these masses oscillating with uniform frequency 𝜔; then 𝑥1 = 𝐴𝑒 𝑖𝜔𝑡 ;

𝑥2 = 𝐵𝑒 𝑖𝜔𝑡 and 𝑥3 = 𝐶𝑒 𝑖𝜔𝑡

(4)

Substituting these we get the following matrix equation; −𝑚𝜔2 + 2𝑘 [ −𝑘 0

−𝑘 −𝑚𝜔2 + 2𝑘 −𝑘

𝐴 0 0 ] [𝐵] = [0] −𝑘 2 0 −𝑚𝜔 + 2𝑘 𝐶

Solving for 𝜔, (−𝑚𝜔2 + 2𝑘)[(−𝑚𝜔2 + 2𝑘)2 − 𝑘 2 ] + 𝑘[−𝑘(−𝑚𝜔2 + 2𝑘)] = 0 or,

(−𝑚𝜔2 + 2𝑘)[(−𝑚𝜔2 + 2𝑘)2 − 2𝑘 2 ] = 0

or, (−𝑚𝜔2 + 2𝑘) = 0, or, 𝜔2 = 2𝑘/𝑚, or, 𝝎𝟏 = √𝟐𝒌/𝒎 and

(−𝑚𝜔2 + 2𝑘)2 − 2𝑘 2 = 0

or,

𝑚2 𝜔4 − 4𝑚𝑘𝜔2 + 2𝑘 2 = 0 4𝑚𝑘±√16𝑚2 𝑘 2 −8𝑚2 𝑘 2

or,

𝜔2 =

or,

𝝎𝟐,𝟑 = √

2𝑚2 (𝟐±√𝟐)𝒌 𝒎

b) Eigen states: For 𝝎𝟏 = √𝟐𝒌/𝒎, 0 [−𝑘 0 or,

(4)

−𝑘 0 −𝑘

0 0 𝐴 −𝑘 ] [𝐵] = [0] 0 0 𝐶

𝐵 = 0 and 𝐴 = −𝐶 = 1 (say)

(5)

at 𝑡 = 0; 𝑥1 = −𝑥3 and 𝑥2 = 0 So, 𝑥1 (0) = 1, 𝑥2 (0) = 0 and 𝑥3 (0) = −1

For 𝝎𝟐 = √(𝟐 + √𝟐)𝒌/𝒎, √2𝑘 [ −𝑘 0 or,

−𝑘 √2𝑘 −𝑘

0 𝐴 0 −𝑘 ] [𝐵] = [0] 0 √2𝑘 𝐶

√2𝑘𝐴 − 𝐵𝑘 = 0 or, √2𝐴 = 𝐵; −𝐵𝑘 + √2𝑘𝐶 = 0 or, √2𝐶 = 𝐵

and

−𝐴 + √2𝐵 − 𝐶 = 0

So,

𝐴 = 𝐶 and 𝐴 = 𝐶 = 𝐵/√2

Let,

𝐴 = 𝐶 = 1, so, 𝐵 = √2

So, at 𝑡 = 0; 𝑥1 = 𝑥3 and 𝑥2 = 𝐵𝑒 𝑖𝜔𝑡 So, 𝒙𝟏 (𝟎) = 𝒙𝟑 (𝟎) = 𝟏 and 𝒙𝟐 (𝟎) = √𝟐 Similarly, for 𝝎𝟑 = √(𝟐 − √𝟐)𝒌/𝒎, So,

𝒙𝟏 (𝟎) = 𝒙𝟑 (𝟎) = 𝟏 and 𝒙𝟐 (𝟎) = −√𝟐