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9 Nature of Waves
Practice 9.1 (p. 13)
1
2
3
4
5
6
7
8
False
False
False
C
D
C
(a) C
(b) A
(c) B
(a) The frequency of a wave is the number of waves produced in 1 s.
 5 Hz means that 5 waves are produced in 1 s.
(b)
By f =
1
1 1
,=> T =  = 0.2 s
f 5
T
The period of the wave is 0.2 s.
9
10
11
(a)
Three smallest possible values are:
(b)
0.05 =
T 5T
9T
,
and
.
4 4
4
T
=> T = 0.2 s
4
The greatest possible period is 0.2 s.
By v = f,
v = 10  0.01 = 0.1 m s1
The speed of the water wave is 0.1 m s1.
distance travelled
5
= = 1.67 m s1
time taken
3
(a)
Wave speed =
(b)
The wave speed is 1.67 m s1.
By v = f,
=
v
1.67
=
= 0.835 m
2
f
The wavelength of the wave is 0.835 m.
12
(a)
Wavelength  =
12
=4m
3
The wavelength of the wave is 4 m.
(b)
Wave speed v =
distance travelled
12
=
= 6 m s1
time taken
2
The wave speed of the wave is 6 m s1.
13
(c)
By v = f, => f =
(a)
Wavelength  =
v
=
6
= 1.5 Hz
4

The frequency of the wave is 1.5 Hz.
4
= 0.16 m
25
The wavelength of the wave is 0.16 m.
distance travelled
4
=
= 0.4 m s1
time taken
10
(b)
Wave speed v =
(c)
The wave speed of the wave is 0.4 m s1.
By v = f,
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f=
v
=
2/6
0 .4
= 2.5 Hz
0.16

The frequency of the wave is 2.5 Hz.
Practice 9.2 (p. 24)
1
C
The wave will have travelled
2
3
D
Displacement of bead c = 0.6  5 = 3 cm
Beads e and m are two successive compressions in the wave.
 Separation between them is equal to the wavelength  of the wave
  = 8  5 = 40 cm
A
Period of bead c = 4  0.1 = 0.4 s
f=
4
5
6
3
λ in 0.75 s.
4
1
= 25 Hz
T
C
By v = f,
v = 25  0.4 = 1 m s1
W and Y are momentarily at rest.
X is moving upwards.
Z is moving downwards.
(a)
(i)
Amplitude A =
0.03
= 0.015 m
2
The amplitude of the wave is 0.015 m.
(ii) Wavelength  =
4
=2m
2
The wavelength of the wave is 2 m.
(iii) f =
(b)
7
(a)
1
1
=
0.2
T
= 10 Hz
2
The frequency of the wave is 10 Hz.
(iv) By v = f,
v = 10  2 = 20 m s1
The frequency of the wave is 20 m s1.
(i) Particles X and Z are in phase.
(ii) Particles W and Y are exactly out of phase.
(iii) Particle W is on a wave crest.
(iv) Particle Y is on a wave trough.
(i) Wavelength  = 6.4 cm
(ii) Since particle P undergoes the smallest number of oscillation at the
instances shown, P should have oscillated for
(b)
 period T = 1 s and frequency f = 1 Hz
(iii) By v = f = 1  6.4 = 6.4 m s1
The wave speed of the wave is 6.4 m s1.
The wave is travelling towards the right.
1
period at t = 0.25 s.
4
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8
where '0' is the position of the vibrator.
9
Revision exercise 9
Multiple-choice (p. 28)
Section A
1
C (HKCEE 1999 Paper II Q20)
2
B (HKCEE 2000 Paper II Q24)
3
D (HKCEE 2002 Paper II Q25)
Section B
4
C
5
B
From the displacement-distance graph, wavelength of the wave is 0.5 m.
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By v = f,
f=
6
=
20
= 40 Hz
0.5

C
From the displacement-distance graph, the wavelength of the wave is 2 cm.
From the displacement-time graph of particle A, the period of the wave is 0.2 s.
By v = f,
v=
7
8
9
v
1
 0.02 = 0.1 m s1
0.2
C
C
B (HKCEE 2001 Paper II Q23)
Conventional (p. 29)
Section A
1
(a) Transverse pulse
(b)
2
(a)
(1A)
distance travelled
time taken
2
=
= 2 m s1
1
Speed of the pulse =
By v = f,
f=
v
0.01
=
= 0.25 Hz
0.04

(1M)
(1A)
(1M)
(1A)
(b)
Correct wavelength
Correct amplitude
Correct waveform at t + 1 s
3
4
distance travelled
time taken
10
=
= 10 m s1
1
Pulse speed =
(a)
(1A)
(1A)
(1A)
(1M)
(1A)
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(b)
(c)
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Correct data points
Correct curve
All correct
Decreases
(i) From the graph, the wavelength is 1.6 m when frequency is 200 Hz.
(ii)
0.7
1.0
1.5
2.5
4.0
Wavelength 
(m)
Frequency f
460
320
210
130
80
(Hz)
Wave speed
322
320
315
325
320
(1A)
1
v = f (m s )
The average wave speed =
322  320  315  325  320
5
= 320.4 m s1
The average speed of sound is 320.4 m s1.
Section B
5
(a) A is moving downwards.
B and C are moving upwards.
(b) The greatest height that each particle can attain is equal to the amplitude of the
wave and
it is 1 cm.
(c) Wavelength = 4 cm
(d)
Correct waveform
Correct positions of A, B and C
6
(HKCEE 2002 Paper I Q4)
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(1A)
(1A)
(1A)
(1A)
(1M)
(1A)
(1A)
(2A)
(1A)
(1A)
(1A)
(1A)
(1A)
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(a)
(b)
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(i) Decreases
(ii) The amplitude of the waves recorded by the trace decreases with time.
5 waves are recorded during the 20 s.
 frequency = number of waves produced per second
=
(c)
5
= 0.25 Hz
20
The frequency of the waves recorded is 0.25 Hz.
(i) A transverse wave is one in which the vibrations are at right angles to the
travelling direction of the wave.
A longitudinal wave is one in which the vibrations are along the travelling
direction of the wave.
(ii) Transverse wave: water waves
Longitudinal wave: sound waves
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(1A)
(1A)
(1A)
(1M)
(1A)
(1A)
(1A)
(1A)
(1A)
Physics in articles (p. 31)
(a)
(b)
(c)
No physical matter is transferred in the propagation of waves.
Invisible signals can be interchanged by a propagating wave.
There are transverse and
longitudinal waves.
The vibrations of a transverse wave are perpendicular to the travelling direction of the
wave, while vibrations of a longitudinal wave are parallel to the travelling direction of
the wave.
(1A)
(1A)
(1A)
(1A)
(1A)