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CHAPTER 11 Sound Waves
Question 11.1
A sound note with frequency 512 Hz is sounded together with
another sound note offrequency 510 Hz. How many beats can be heard
in a time interval of 3 seconds?
Solution 11.1
Beat frequencyF = (512 510) Hz= 2 Hz
3 2 = 6 beats6 beats are heard every 3 seconds.
Question 11.2
When a sound note with frequencyfis sounded together with
another sound note with afrequency of 480 Hz, five beats are heard
every second. However, only two beats are heardwhen a sound note of
frequency 483 Hz is used. What is the value off?
Solution 11.2
Fbeats frequency = difference in frequency between the two sound
notesTherefore possible equations to calculate Fare F=f 480 or F=
480 f
However by increasing the frequency of 480 Hz to 483 Hz, the
beats frequency is reduced.Thus the correct equation must be
F = f 480 5 = f 480
f = 480 + 5= 485 Hz
Question 11.3
In an experiment with a resonant tube, the shortest length which
resonants with afrequency of 512 Hz is 14.0 cm. The next resonant
length observed was 46.0 cm.Calculate(a) the velocity of sound in
air and(b) the end correction.
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2c
1c
Question 11.4
An unknown tuning fork when sounded with a fork of known
frequency of 250 Hzproduces 6 beats in exactly 2 seconds both
before and after a small piece of plasticine is
attached to its prongs. Find the frequency before and after the
plasticine is used.
Solution 11.4
Let the frequency of the fork before and after the plasticine is
used be f1
and f2
respectively. With the plasticine, the prongs of the fork
vibrate slower, hence we expect
f1
>f2, thus
6 6f
1 250 = and 250 f
2=
2 2
f1
= 253 Hz and f2
= 247 Hz
Question 11.5
Two organ pipes with both ends open sounded together to produce
9 beats in 2 seconds.If one of the pipes is 60.0 cm long and the
other pipe is 61.0 cm long, find the velocity of
sound in air.
Solution 11.3
1 v(a) = c+
1but =
4 f
1 V = c+
14 f
1 V = c+ 0.14 1
4 (512)1 V
= c+ 24 f
3 V = c+ 0.46 24 (512)
1 V() = 0.32 2 12 512
V = 327.68 m s1 or 328 m s1
(b) Substituting V = 327.68 m s1 into 1
327.68 = c+ 0.144(512)
0.16 = c+ 0.14c = 0.02 m or 2 cm
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12
Question 11.7
A stationary ambulance emits sound waves of frequency 400 Hz
from its siren. Calculatethe maximum and minimum frequency heard by
the driver of a car which is moving at aconstant speed of 30 m s1
as he passes the ambulance. Velocity of sound in air is340 m
s1.
Question 11.6
A train is moving with a constant speed of 30 m s1 and sounding
its whistle which emitssound waves of 330 Hz. What is the minimum
and maximum frequencies of the soundheard by a boy standing by the
side of the track when the train passes him? The velocityof sound
in air is 340 m s1.
Solution 11.6
When the train is approaching the boy, the apparent wavelength
is given by
V U V V = f = = ()f
f V U
340f = ()330
340 30
When the train is moving away from the boy, the apparent
frequency heard is given by
340f = ()330
340 + 30
= 303 Hz
Solution 11.5
1 = 2
= 2
V VFor fundamental note f = =
2
V Vf1
= and f2
= 2(0.60) 2(0.61)
9f
1f
2=
2
V V = 4.5
2(0.60) 2(0.61)
1 1V( ) = 4.5
1.20 1.22
V = 329.4 m s1
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=12
Solution 11.7
When approaching the ambulance, the relative velocity of the
sound wave observed bythe driver is given by
Vr
= V + Uwhere V = velocity of sound
U = velocity of car
Vr
V + U f = =
V/fV+ U
f = ()fV
340 + 30= ()400 Hz
340
= 435 Hz
While moving away from the ambulance, the relative velocity of
the sound isV
r= V U or V
r= 340 30
V U Apparent frequency f = ()fV
340 30f = ()400 Hz
340
= 365 Hz
Question 11.8
A 140 cm long wire is stretched between 2 points by a force of
160 N. If the mass of thewire is 50 g, find the frequency of
(a) the fundamental mode of vibration and(b) the first harmonic
of the situation.
Solution 11.8
1(a) =
2
= 2 = 2(140 cm)= 280 cm
FVelocity of wave in string V =
V 50 103but f = = kg m1
1.40
1 F = 3.57 102 kg m1 f
0=
1 160=
2.80 3.57 102
= 23.9 Hz
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