Exercise 7 (Sound) Suggested Solutions
1. (a) Note Z
(b) The pitches of the three notes are identical because the frequencies of the three notes are the same.
The loudness of note X is larger than that of Y, which is larger than that of Z because the amplitude of
note X is larger than that of Y, which is larger than that of Z.
2. (a) Period = 4 x 0.1 ms, so frequency =
1
1

 2500 Hz
T 4  0.1 10 3
 = v / f = 325 / 2500 = 0.13 m
(b) (i) Constructive interference since the p.d. = 2.31 – 2.05 = 0.26 m = 2 x 0.13 = 2λ.
(ii) Pitch remains unchanged, loudness increases.
(iii) There is background noise. / Waves from P and Q are of different intensity due to different path
lengths.
(iv) Wrong, because the path difference between waves sent from P and Q to any point along XY is
always zero. Only constructive interference occurs.
3. (a) Large room can weaken the reflected sound waves which would interfere with the original sound.
Silent room has less or no sound that would interfere with the sound produced by the loudspeaker.
(b) S and T should be small so that diffraction effect is more pronounced and as a result,
the diffracted waves can interfere.
(c) (i) At places where sound waves arrive in phase, constructive interference takes place,
a loud sound is heard, i.e. the amplitude is a maximum.
At places where sound waves arrive exactly out of phase, destructive interference takes place,
a soft sound is heard, i.e. the amplitude is a minimum.
(ii) P.d. at X = 2 
2.7 – 1.8 = 2 
 = 0.45 m
(d) His statement is not correct.
Although there is always constructive interference occuring along RO,
4. (a) A longitudinal wave is a wave in which the vibrations are along the direction of travel of the wave.
Any of the following: sound waves in air / ultrasonic waves / waves in a slinky spring
(b) (i) Particles A (or I) is at the center of a compression.
(ii) Particle E is at the center of a rarefaction.
(c) (i)
Particle
Displacement / cm
A
0
B
-6
C
-8
(iii) Amplitude = 8 cm, wavelength = 80 cm
(iv) 3.2 m s-1
5. (a) Wavelength = v/f = 340/200 = 1.7m
D
-6
E
0
F
6
G
8
H
6
I
0
J
-6
K
-8
6. (a) Wavelength = v/f = 6000 / 2 x 106 = 3 x 10-3 m
(b) X is the reflected pulse. Reason:
 The amplitude of the reflected pulse should be smaller than that of the transmitted pulse due to absorption
effect.
 Some energy is lost as the transmitted pulse travels along the wall.
 Only a part of the pulse will be reflected from the other surface of the wall.
(iii) From the graph, d drops to 30 mm at t = 37.5 weeks. So the pipe should be replaced after about 37 weeks.
(d)
Another pulse is recorded by the receiver.
The crack reflects the transmitted pulse to the receiver.
7.
(a)
(b)
MC
(i)
(Correct scales for both axes)
(Correct points at t = 0, 1.25 ms, 2.5 ms, 3.75 ms and 5 ms)
(Correct shape of the curve)
(ii) Amplitude = 20 cm
1
1
Frequency = =
= 200 Hz
T 5  10 3
At t = 0, A and E are the nearest particles at their equilibrium positions.
λ
 Distance between A and E =
2
 = 2  AE = 2  (4  0.2) = 1.6 m
1-5 D A D C B
6-10 D C B D D
11-15 D C C A A
16 C
1) v = f, f is unchanged.
2) For (1), fY > fX => pitch of Y > pitch of X.
For (2), same amplitude => same loudness.
For (3), Y < X.
6) Cooking food makes use of microwave.
10) Path difference at Q =
32  42  4  1m 
1
 , so if the sources are in phase, destructive interference will
2
occur. However, the two loudspeakers are vibrating exactly out of phase, so the interference result will be opposite,
i.e. constructive interference will occur at Q. Similar result applies to P.