Structural Dynamics Assignment Help
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Problem
Problem 1 Earl G. Williams Problem 2.11 Problem 2 Earl G. Williams Problem 2.12
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Solution Problem 1 Using equation (2.52) from text and substituting the expression for p(x,y,0), P kx , ky
Poe it
=
(e
i 2kx
e i 2 kx )e ik x e ik y dxdy x
y
Poe it
=
(e
i ( k x 2k ) x
e i ( 2k k x ) x )e
ik y y
dxdy
Using equation (1.5) from text, P k , k
=
x y
4 2 Po e
it
( kx 2k ) (x k 2k ) y (k )
Using equation (2.50) and substituting equation (1) above,
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(1)
1
p x, y, z
4 2 [4
=
e =
Po
=
z
x
2k ) (k
x
2k )e ik x dk (k x y )e
x
x
(k
2k ) (k
2k )
)e
dk
(k
Poeit
z ik x (k x 2k ) (k x 2k ) (k iky )e]e e
it
Poeit
2
x
x
i ( kz z ky y)
ik y y
dk x dk y
dk y
eik x dk x I1 x
(2)
Where
(k
I1
=
y
i (k z zk y y )
y
Using sifting property of the delta function integral (equation (1.37) of text) and recognizing that k z
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k 2 k x 2 k y 2 ,
I1
=
e
i
2
k k x z
(3)
Thus, substituting equation (3) above into (2) above,
p x, y, z
Poeit
=
(k 2k ) x
x
e
i
(k
x
2k )e ik
x
k 2 k x 2 z
dk x
Again, using sifting property of the delta function integral (equation (1.37) of text), i 2kx i 3kz i 2kx i 3kz p x, y, z
it
e
=
Po e
=
Poe it e
3kz
Figure 1. shows the normalized pressure field 20log p
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e
e
e
cos(2kx)
Po
dB for normalized distance 0 ≤ kx ≤ 10 and 0 ≤ kz ≤ 10.
Figure 1. Normalized pressure field 20 log
p
Po
in dB
for normalized distance 0 ≤ kx ≤ 10 and 0 ≤ kz ≤ 10.
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Problem 2 Using equation (2.61) and let z = z’ = 0,
kz W (kx , ky ,0) P(k x , o ck k ,0) .
y
(1)
By recognizing the definition in equation (2.56) and substituting (1) above into the expression, P(k x , k y ) P(k x , k y ,0) o ck . P( kx , ky ) W ( kx , k z k ,0 )
y
(2)
Using equation (2.50) and observing the form of the inverse 2-D Fourier Transform expression given in equation (1.17)
Pk , k e Pk , k
p x, y, z Fx Fy
x
1
p x, y,0 Fx F
1
1 y x y
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y
z
1
ik z
(3)
Substituting (2) above into (3) .
p x, y,0
1 x
F
1
Fy
o ck W ( k x kz k ,0 )
,
y
(4)
From the 2-D Fourier transform expression provided in text for equation (2.76), ik y p x, y,0 x 1Fy 1 o Qoe ik x e ck z k F x o
y o
From equation (2.72) and (2.74), and again recognizing that z = z’ = 0,
p x, y,0 iQo o ck
Where x x 2 o
r ro
e
ik
r ro 2
y y
2
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rro
o