# 355345056-Assignment-Solutions11111

1. A sound wave propagates a point about 50 meters below the surface of a calm sea.
The instantaneous pressure at the point is given by: p = 6x
where t is in seconds and p in Pascals.
a) What is the value of static pressure at the point?
+1000sin
(400πt),
is pressure from the height of the column of fluid above the
wave.
=6x
Pa
b) What is the value of maximum (or peak) acoustic pressure at the point?
= 1000 Pa
c) What is the root-mean-square acoustic pressure?
=
=
= 707.107 Pa
√
√
d) What is the acoustic pressure when t=0, 1.25, 2.5, 3.75, 5.00 milliseconds?
= 1000 sin (400πt)
(t = 0 mS) = 1000 sin (400πx0 ) = 1000 sin 0 = 0 Pa
(t = 1.25 mS) = 1000 sin (400πx1.25x
) = 1000 sin(0.5π) = 1000 Pa
(t = 2.5 mS) = 1000 sin (400πx2.5x
) = 1000 sin(0.5π) = 0 Pa
(t = 3.75 mS) = 1000 sin (400πx 3.75 x
) = 1000 sin(1.5π) = -1000 Pa
(t = 5.00 mS) = 1000 sin (400πx 5.00 x
) = 1000 sin(2π) = 0 Pa
e) What is the average acoustic intensity of the sound wave? (The density of the
water is 1000 kg/
and the sound speed is 1500 m/sec.)
=
=
=
=
√
=
√
= 0.3333 W/
f) What is the intensity level, L, in dB re 1 μPa?
L = 10log ( )
=
=
L= 10log (
= 6.66667x
) = 176. 99dB
2. A plane acoustic wave is propagating in a medium of density ρ=1000 kg/ . The
equation for a particle displacement in the medium due to the wave is given by:
s = (1x
seconds.
)cos(8πx −12000πt), where distances are in meters and time is in
a) What is the rms particle displacement?
=
√
=
√
=
M
b) What is the wavelength of the sound wave?
From : s =
K=
cos( x + ωt)
= 8π
=
= 0.25 M
= 0.25 M
c) What is the frequency?
s = cos( x + ωt)
ω = 2 f = 12000
=
= 6000Hz
= 6 kHz
d) What is the speed of sound in the medium?
c=
c = 0.25 x 6000 M/S
c = 1500M/S
e) What is the value of maximum (or peak) particle velocity?
s = (1x
= -(-1x
)cos(8πx −12000πt)
)sin(8πx −12000πt)
v = 3.7699 x
sin(8πx −12000πt) m/s
= 3.7699x
m/s
f) What is the value of maximum acoustic pressure?
= V(t)x
=
= 1000 x 1500
V(t) = 3.7699 x
sin(8πx −12000πt) m/s
P = (1000 x 1500 x 3.7699 x
sin(8πx −12000πt) m/s
P = (5.65485x
sin(8πx −12000πt) m/s
= 56,548. 5 Pa
g) What is the specific acoustic impedance of the medium?
=
= 1000 x 1500 = 1.5 x
rayls
h) What is the bulk modulus of the medium?
=√
=
= (1500
= 2.25x 1
x 1000
N/
i) What is the acoustic intensity of the sound wave?
=
=
=1065.9 W/
1065.9 W/
j) What is the acoustic power radiated over a 3
area?
P = Intensity X Area
P = 1065.9 W/
P =3197.73 W
3. A plane acoustic wave is propagating in a medium of density ρ and sound speed c.
The equation for pressure amplitude in the medium due to the wave is given by:
p=
cos (kx −ωt), where p0 is the maximum pressure amplitude of the sound.
cos (kx −ωt)
p=
K=
; ω= 2 f ; c =
 f=
Substituting eqn (2) in eqn (1).
p = cos ( x − 2 f t)
p=
cos ( x − 2
........ (1)
....... (2)
t)
Factoring out
p=
cos
x - t)
b) Show that maximum pressure amplitudes (compressions) can be found at the following
locations in space: x=nλ+ct where n= 0, 1, 2, 3, …
for maximum amplitude(compression) , cos (kx −ωt) = 1
substituting x =n
We have:
p=
p=
in the pressure equation , p =
cos (kx −ωt).
cos (kx −ωt)
cos
x – t)
p=
cos
p=
cos
– t)}
)}
p = cos
from the identity cos
Therefore: p =
= 1 for all n i.e n = 1 , 2, 3,...
which is always a maximum for n = 1 , 2, 3,...
c) Show that maximum pressure amplitudes (rarefactions) can be found at the following
location in space x= (n+1/2) λ+ct, where n = 0, 1, 2, 3,...
For maximum pressure amplitude (rarefaction), cos (kx −ωt) = -1
Substituting x= (n+1/2) λ+ct in the pressure equation, p =
cos
x – t)
We have:
p=
cos
p=
cos
x – t)
– t)}
p = cos
}
p = cos
+ )
From the identity, cos
+ )= -1 for all n i.e n = 0, 1, 2, 3,...
Therefore: p =
which is always a maximum (rarefaction) for n = 0, 1 , 2, 3,...
s(x,t) = 6x 1
m cos (
= (6x 1
)
) m sin (
V(x,t) = 6.03185 x
(x,t) = 6.03185 x
m/s
)
V(x,t) = 6.03185 x
= 2.0 x 1 N/
c=
= 13 x 160 = 2080 m/ s
=√
; 
=
=
=
=
= 462.278
c = 462.278 x 2080 =961, 538.24 rayls
=
x V (t) = (961, 538.24 x6.03185 x
= 5800
= 5800Pa
Pa
Pa