Kamal A.A. 1000 Solved Problems in Classical Physics: An Exercise Book
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378 8Waves
8.46
ω =
γ RT
M
k
v
p
=
ω
k
=
γ RT
M
v
g
=
dω
dk
=
γ RT
M
∴ v
g
= v
p
8.47
v
2
p
=
g
k
+
Sk
ρ
(1)
Maximize (1)
d(v
2
p
)
dk
=−
g
k
2
+
S
ρ
= 0
∴ k =
2π
λ
=
gρ
S
∴ λ
min
= 2π
S
gρ
8.3.4 Sound Waves
8.48 Let the displacement be represented by
y = A cos(kx − ωt) (1)
∂y
∂x
=−kAsin(kx −ωt)
but P =−B
∂y
∂x
= BkA sin(kx − ωt)
where B is the bulk modulus:
B = v
2
ρ
0
P =[kρ
0
v
2
A]sin(kx − ωt) (2)
P represents the change from standard pressure P
0
. The term in square bracket
represents maximum change in pressure and is called the pressure amplitude
P
max
. Then
8.3 Solutions 379
P = P
max
sin(kx − ωt) (3)
where P
max
= kρ
0
v
2
A (4)
If the displacement wave is represented by the cosine function, (1), then the
pressure wave is represented by the sine function, (3). Here the displacement
wave is 90
◦
out of phase with the pressure wave.
8.49
I =
1
2
P
2
max
/ρ
0
v
∴ P
max
=
2Iρ
0
v =
2 × 10
−12
× 1.29 × 331 = 2.92 ×10
−5
N/m
2
8.50
IL = 10 log
I
I
0
60 = 10 log
I
10
−12
log I + log 10
12
= 6logI =−6
∴ I = 10
−6
W/m
2
= 1 μ W/m
2
8.51
I =
Power
4πr
2
=
4
4π ×25
2
= 5.093 ×10
−4
W/m
2
IL = 10 log
I
I
0
= 10 log
5.093 × 10
−4
10
−12
= 10 log(5.093 × 10
8
)
= 10[log 5.093 + 8]=87 dB
8.52
A =
P
max
kρ
0
v
2
=
P
max
2πρ
0
f v
where we have substituted k =
2π
λ
and v = f λ:
∴ A =
29
2π ×1.22 × 2000 × 331
= 5.7 ×10
−6
m
8.53
I =
P
2
max
2ρ
0
v
By problem, P
max
(air) = P
max
(water)
∴
I
Water
I
Air
=
ρ
A
v
A
ρ
W
v
W
=
1.293 × 330
1000 × 1450
= 2.94 ×10
−4
380 8Waves
8.54
P = 2.4sinπ(x − 330t) = 2.4sin2π
1
2
x − 165t
P = P
max
sin 2π
x
λ
− ft
(standard expression)
On comparing the two expressions we find
(a) 2.4N/m
2
, (b) 165 Hz, (c) 2.0m, (d)v = f λ = 165×2 = 330 m/s
8.55
I = 2π
2
ρ
0
A
2
f
2
v
∴ A =
1
π f
I
2ρ
0
v
=
1
1200π
2 × 10
−6
2 × 1.293 × 330
= 1.28 ×10
−4
m
8.56
P
e
= 1 microbar = 10
−6
bar = 0.1N/m
2
I =
P
2
e
ρ
0
v
=
(0.1)
2
1.293 × 330
= 2.34 ×10
−5
W/m
2
IL = 10 log
I
I
0
= 10 log
2.34 × 10
−5
10
−12
= 10(7 +log 2.34) = 73.7dB 74 dB
8.57
IL = 10 log
I
I
0
70 = 10 log
I
10
−12
= 10 [log I + 12]
log I =−5 I = 10
−5
W/m
2
Energy density
E =
I
v
=
10
−5
331
= 3 ×10
−8
J/m
3
Effective pressure
P
e
=
Iρ
0
v =
10
−5
× 1.293 × 331 = 0.0654 N/m
2
8.58
P
max
=
2Iρ
0
v =
√
2 × 1 × 1.293 × 331 = 29.26 N/m
2
8.59
v =
γ RT
M
=
1.4 × 8.317 × 273
2.016 × 10
−3
= 1256 m/s
8.3 Solutions 381
8.60
v =
γ P
ρ
(Laplace formula)
∴
v
H
v
O
=
ρ
O
ρ
H
=
√
16 = 4
∴ v
H
= 4v
O
= 4 ×317 = 1268 m/s
8.61
IL = 10 log
I
2
I
1
= 10 log
10
0.4
= 14 dB
8.62
IL = 10 log
I
2
I
1
= 6
log
I
2
I
1
= 0.6
∴
I
2
I
1
= 3.98 or 4
8.63 The threshold of hearing intensity is taken as 10
−12
W/m
2
.Letr be the dis-
tance from the source at which the sound can be audible
I =
power
4πr
2
= 10
−12
∴ r =
power
4π ×10
−12
=
0.009
4π ×10
−12
= 2.677 ×10
4
m = 26.8km
8.64
P
max
= kρ
0
v
2
A =
2π
λ
ρ
0
v( f λ)A = 2πρ
0
v fA
∴ A =
P
max
2πρ
0
v f
=
2 × 10
−5
2π ×1.22 × 331 × 1000
= 7.9 ×10
−12
m
8.65
(a) A =
P
max
2πρ
0
f v
A
1
A
2
=
ρ
2
ρ
1
v
2
v
1
=
4
3
×
2
3
=
8
9
(b) I =
1
2
p
2
max
ρ
0
v
I
1
I
2
=
ρ
2
v
2
ρ
1
v
1
=
4
3
×
2
3
=
8
9
382 8Waves
(c) E =
I
v
E
1
E
2
=
I
1
I
2
v
2
v
1
=
8
9
×
2
3
=
16
27
8.66
(a) λ =
v
f
∴
λ
A
λ
W
=
v
A
v
W
=
331
1450
= 0.228 (∵ f
A
= f
W
)
(b) P
max
=
√
2Iρ
0
v
∴
P
A
P
W
=
ρ
A
ρ
W
v
A
v
W
=
1.293
1000
×
331
1450
= 0.0172 (∵ I
A
= I
W
)
(c) A =
2I
ρvω
2
A
A
A
W
=
ρ
W
ρ
A
v
W
v
A
=
1000
1.293
×
1450
331
= 33.88 (∵ I
A
= I
W
and f
A
= f
w
)
8.67 Characteristic impedance of a gas
Z = ρ
0
v (1)
Now v =
B
ρ
0
or ρ
0
=
B
v
2
∴ Z =
B
v
= B
M
γ RT
(2)
Thus Z ∝
1
√
T
(a) At 0
◦
C, v = 331 m/s, ρ
0
= 1.293 kg/m
3
Z = ρ
0
v = 1.293 × 331 = 428 rayl
(b) Z ∝
1
√
T
∴ Z(80
◦
C) = Z(0
◦
C) ×
273
273 + 80
= 428 ×0.879 = 376 rayl
8.3 Solutions 383
8.68
(a) I =
Power
area
=
50
π(0.25)
2
= 255 W/m
2
(b) P
max
=
√
2Iρ
0
v =
√
2 × 255 × 10
3
× 1450 = 2.72 × 10
4
N/m
2
(c) A =
2I
ρ
0
υ4π
2
f
2
=
2 × 255
1000 × 1450 × 4π
2
× (25 × 10
3
)
2
= 1.19 ×10
−7
m
(d) U
max
= Aω = 1.19 × 10
−7
× 2π ×25 × 10
3
= 0.019 m
(e) S
max
=
2π A
λ
=
2π Af
v
=
2π ×1.19 × 10
−7
× 25 × 10
3
1450
= 1.29 ×10
−5
8.69 Consider sound waves of finite amplitude. Now, the bulk modulus is constant
only for infinitesimal volume changes:
B =−V
dp
dV
(1)
where the acoustic pressure p has been replaced by the pressure change dp.
Now, Vρ = mass = constant
∴ V
dρ
dP
+ ρ
dV
dP
= 0
or − V
dP
dV
= ρ
dP
dρ
= B
where we have used (1)
∴ v =
B
ρ
=
dP
dρ
(2)
When a second wave passes through a gas the changes in volume are assumed
to be adiabatic so that
P
ρ
γ
= C = constant (3)
where γ is the ratio of specific heats of the gas at constant pressure to that at
constant volume. Differentiation of (3) gives
− γ Pρ
γ −1
dρ +ρ
γ
dP = 0
or
dP
dρ
=
γ P
ρ
(4)
384 8Waves
Using (4) in (2), we get
v =
γ P
0
ρ
0
(Laplace formula) (5)
where P
0
and ρ
0
refer to equilibrium conditions of pressure and density.
The velocity υ
0
at 0
◦
C can be found out by substituting γ = 1.4, P
0
=
1.013 × 10
5
N/m
2
and ρ
0
= 1.293 kg/m
3
. We find v
0
= 331.2m/s, in good
agreement with the experiment.
With the assumption of isothermal changes we would have obtained the
formula v =
√
P
0
/ρ
0
, (Newton’s treatment) which gives a value of 20%
lower.
8.70
(a) v|
t=20
= 1403 +5 ×20 −0.06 ×(20)
2
+0.0003 ×(20)
3
= 1481.4m/s
(b)
dv
dt
= 5 −0.12t + 0.0009t
2
dv
dt
|
t=20
= 5 −0.12 × 20 + 0.0009 × (20)
2
= 2.62 m/s/
◦
C
8.3.5 Doppler Effect
8.71
v
s
= 72 km/h = 72 ×
5
18
m/s = 20 m/s
f
0
=
v f
s
v + v
s
(direct)
f
0
=
v f
s
v − v
s
(reflected from the wall of the rock)
f
0
f
0
=
v + v
s
v − v
s
=
340 + 20
340 − 20
=
9
8
8.72
v
0
= v
s
= 90 km/h = 90 ×
5
18
= 25 m/s
v = 350 m/s
f
=
f (v + v
0
)
v − v
s
=
520(350 + 25)
350 − 25
= 600 Hz.
8.73
f
=
f (v − v
0
)
v + v
s
= 520
(330 − 25)
330 + 25
= 446.8Hz.
8.74 Let the whistle rotate clockwise, Fig. 8.7. At point A the linear velocity of
the whistle will be towards the distant listener and at B away from the listener.
8.3 Solutions 385
Fig. 8.7
Maximum frequency will be heard when the whistle will be at A and minimum
when it is at B:
v
s
= ωr = 15 ×2 = 30 m/s
f
max
=
v f
s
v − v
s
=
330 × 540
330 − 30
= 594 Hz
f
min
=
v f
s
v − v
s
=
330 × 540
330 + 30
= 495 Hz
8.75 (a) The frequency of the rod is fixed, and so also for the air and the gas. The
distance between successive heaps of cork dust is equal to the distance
between two neighbouring nodes which is
1
2
λ,Fig.8.8:
Fig. 8.8
v
air
= f λ
air
v
gas
= f λ
gas
∴ v
gas
= v
air
λ
gas
λ
air
= 330 ×
10
8
= 425 m/s
(b)
v
rod
= f λ
rod
=
v
air
λ
air
· λ
rod
=
340
2 × 0.08
(2L
rod
)
=
340
0.16
(2 × 1.2) = 5100 m/s
8.76 (i) v = f λ
∴ λ =
v
f
=
340
514
= 0.66 m
386 8Waves
(ii) λ
=
v − v
s
f
=
340 − 15
514
= 0.63 m
(iii) λ
=
v + v
s
f
=
340 + 15
514
= 0.69 m
8.3.6 Shock Wave
8.77
(a) If an object flies with a supersonic speed (speed greater than that of
sound) a shock wave is emitted, a booming sound. In the two-dimensional
drawing, Fig. 8.9, the wave fronts CB and DB represent the V -shaped
wave. In three dimensions the bunching of the wave fronts actually forms
a cone called the Mach cone. The shock wave lies on the surface of
the cone.
(b) The half-angle θ of the cone called the Mach cone is given by
sin θ =
vt
v
s
t
=
v
v
s
=
1
2
The Mach number =
v
s
v
p
= 2
(c) The Mach angle θ = 30
◦
Fig. 8.9 Shock wave
8.3.7 Reverberation
8.78 If V is the volume, S the surface area and K the absorption coefficient then
the reverberation time t
R
is given by
t
R
=
0.16V
i
k
i
S
i
(Sabine Law)
V = 10 ×18 × 4 = 720 m
3
8.3 Solutions 387
K
i
S
i
= 10 ×18 × 0.6
+ 10 × 18 × 0.02 +(18 × 4 + 10 × 4) × 2 ×0.03 + 50 × 0.5
= 143.3
t
R
=
0.16 × 720
143.3
= 0.8s
8.3.8 Echo
8.79 The drum rate, that is, frequency when the echo is inaudible, is 40/min or
2/3 per second. Therefore, the time period of drum beats t
1
=
3
2
s. Time for
the echo, t
1
=
2x
v
, where x is the initial distance from the mountain and v is
the sound velocity.
Thus
2x
v
=
3
2
(1)
On moving 90 m towards the mountain ht is x −90 m from the mountain, the
drum rate is 60/min or 1/s and again the echo is not heard.
Thus
2(x − 90)
v
= 1(2)
Solving (1) and (2) we get
x = 270 m and v = 360 m/s.
8.80
(a) If the width of the valley is d m and the rifle shot is fired at a distance x
from one of the mountains the echoes will be heard in time t
1
and t
2
s:
t
1
=
2x
v
(1)
t
2
= 2
(d − x)
v
(2)
Adding (1) and (2)
t
1
+ t
2
= 2 +4 =
2d
v
=
2d
360
Therefore d = 1080 m.
(b) Solving (1) with t
1
= 2 s and v = 360 m we find x = 360 m and therefore
d − x = 720 m. Subsequent echoes will be heard after 6, 8, 10,...s.