2020

WAVES I

TEACHERS OF PHYSICS
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10/2/2020
 1. Explain the following terms and state their S.I units
(i) Wavelength
A wavelength is the distance between two points on a wave train
which are in phase.
It is the distance between two successive crests or troughs in a
transverse wave or the distance between two successive rarefactions
or compressions in a longitudinal wave.
Its SI unit is metre.

(ii) Amplitude
Is the maximum displacement on either side of the mean position of a
wave. Its SI unit is metre

(iii) Periodic time

It is the time taken by a particle to complete one oscillation. Its SI
unit is second
(iv) Frequency
Is the number of complete oscillations made by a particle in one
second. Its SI unit is hertz.

2. Differentiate between;
(i) Mechanical and electromagnetic waves. Give an example of each (4mk)
Mechanical wave require material medium for transmission e.g water
wave and sound wave while electromagnetic waves do not require
material medium for transmission e.g radiowaves, microwaves

(ii) Longitudinal and transverse (2mk)

Longitudinal wave has vibration of particles directed parallel to the
direction of the wave travel e.g sound wave while transverse wave has
vibration of the particles directed at right angles to the direction of the
wave travel e.g water waves.

3. State ONE difference between mechanical and electromagnet waves.

Mechanical wave require material medium for transmission e.g water
wave and sound wave while electromagnetic waves do not require
material medium for transmission e.g radiowaves, microwaves

4. Define the term progressive waves (1mk)

Waves that move continually away from the source

5. Name two types of progressive wave motion

Transverse wave motion
Longitudinal wave motion
6. State two differences between sound waves and electromagnetic waves (2mk)
(i) Sound wave require a material medium for transmission while
electromagnetic waves do not require a material medium for
transmission

(ii) Sound waves are longitudinal in nature while electromagnetic waves

are transverse in nature
7. Name a property of light that shows it is a transverse wave.
Light wave can be polarized.
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 8. State THREE differences between light waves and sound waves.

Light waves Sound waves

Transverse in nature Longitudinal in nature

Do not require material medium for Require material medium for

transmission (electromagnetic) transmission(mechanical)

Form crests and troughs Form rarefactions and

compressions

9. Explain the term ‘phase’ as used in waves (1 mk)

Phase occurs when particle in a wave motion happen to oscillate in the
same direction and at the same level of displacement in their oscillation.

10. Water waves are observed as they pass a fixed point at a rate of 30 crests per
minute. A particular wave crest takes 2 second to travel between two fixed
points 6m apart. Determine the frequency and the wave length of the wave.
(4mk)
30 crests per minute implies 0.5 crests per second
Frequency = 0.5herts
Velocity, v = distance/time
= 6m/2s
= 3m/s
From v = fλ
λ = v/f
= 3ms-1/0.5
λ = 6m

11. A source generates 40 waves in a second. If the wavelength is 8.5 cm.

Calculate the time taken to reach a wall 102m from the source.
Frequency(f) = 40Hz
Wavelength(λ) = 0.085m
Speed(v) = frequency (f) x wavelength(λ)
v = 40 x 0.085
v = 3.4m/s
but v = d/t
t = d/v
= 102m/3.4ms-
t = 30seconds.

12. HERO radio broadcasts on a frequency of 60 MHz and the velocity of its
signals is 3x108m/s find the wavelength of radio waves. (2mk)
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 f = 60000000 Hz
v = 3x108m/s
v = fλ
λ= v/f
= 3x108/6x106
λ=5m

13. Radio X is broadcast on wavelength 150m at a frequency of 200KHz.

Calculate the velocity of the radio waves. (2mks)
λ=150m
f= 200000H
v=fλ
= 150m x 2x105Hz
V= 3 x107m/s

14. KASS FM station broadcasts on a frequency of 250 KHz and the wavelength of
its signals is 1200m.
(i) The speed of radio waves in m/s
v=fλ
= 250000Hzx1200m
= 3.0x108m/s

(ii) The wavelength of the signal of another station that broadcasts on a

frequency of 200KHZ.
v=fλ, λ = v/f
λ= 3x108/2x104
= 1500m

15. A radio wave speed 3 x 108m/s is transmitted at a wavelength of 2 x 10-6m.

Calculate its frequency

V=3 x 108m/s
λ=2.0x10-6m
f=3 x 108m/s
2.0x10-6m
F= 1.5x1014Hz

16. Calculate the wavelength of the KBC fm radio waves transmitted at a frequency
of 95.6 Mega Hertz. (V = 3.0 x 108m/s) (2mk)
F= 9.56x107Hz
V = 3 x 108m/s
λ=v/f
=3 x 108m/s
9.56x107Hz
λ= 3.1381m

17. Determine the resultant amplitude for two waves out of phase if one wave has
an amplitude of 0.5m and the other 3.0m (2mks)
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 Amplitude,a = 0.5m
Amplitude,b = 3.0m
Since they are out of phase, we get the difference
Resultant amplitude = b-a
= 3.0 – 0.5
= 2.5.

18. One range of frequencies used in broadcasting varies from 5 x 106 Hz to 2.0 x
107 Hz. What is the longest wavelength of this range? Velocity of light air
=3x108 / s
Frequency, f1 = 5 x 106 Hz frequency, f2 = 2.0 x 107 Hz
Velocity,v = 3x108 / s Velocity,v = 3x108 / s
λ=v/f λ=v/f
= 3x108 / s = 3x108 / s
5 x 106 Hz 2.0 x 107 Hz
λ= 60m λ = 15m
the longest wavelength of this range is 60m

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 WAVE PROFILES
1. The figure below shows a wave profile

a d
4
Displaceme

b e
nt(mm)

c f
6s Time

-4

(a) Name two sets of points that are

(i) One wavelength apart (2mk)
a and d
b and e

(ii) In phase
a and d
b and e

(iii)Out of phase
a and e
b and d
c and f
b) Determine the frequency of the wave. (3mk)
3 cycles in 6s
Periodic time =6/3
= 2s
Frequency, f = ½
0.5H
2. The sketch is a displacement – time graph of a wave traveling at 320ms-1 .the
wave takes 1.2 seconds to move from point A to B.
Displacem
ent (mm)

4
A B
Time (s)

-4

Find the
1.2 s

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 (i) Amplitude (1mk)
Amplitude = 4mm
=0.004m
(ii) Frequency (3mk)
f = 1/T
3 cycles in 1.2s gives T as 1.2/3
T = 0.4s
f = 1/0.4
f= 2.5Hz
(iii) The wavelength () (2mk)
Wavelength, λ = velocity,v
Frequency,f
λ = 320m/s
2.5Hz
λ = 128m

3. The Figure below shows a wave profile of moving at a velocity of 150m/s.

Displacement (m)

0.02 0.04 0.06 0.08

-3
Time (s)
-6

Determine:
(i) Amplitude (1mk)
6m
(ii) Period. (2mk)
0.04s
(iii) Frequency. (2mk)
f=1/T
f = 1/0.04
= 25Hz
(iv) Wavelength. (2mk)
Wavelength, λ = velocity, v
Frequency, f
λ=150m/s
25Hz
λ= 6m
(V) On the same axis of the wave above, sketch a wave with half the amplitude
and double the period. (2mk)

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 4. Determine the frequency of the wave shown below. (2mk)
Displacement (m)

0.5 1.0 1.5 2.0 Time(s

Periodic time ,T = 2.0s

Frequency, f = 1/T
=½
F = 0.5Hz

State one reason why ultrasound is preferred to audible sound in echo-sounding.

They have higher frequencies hence more penetrating power

5. Figure below shows how the displacement varies for a certain wave.

Displacement (m)
0.2
0.1

0.5 1 1.5 2 2.5

0.1 Time (s)

0.2

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 Determine the frequency of the wave (3mks)
Periodic time , T = 1s
Frequency, f = 1/T
= 1/1

= 1Hz

6. The graph below is a displacement – time graph of a wave traveling at 250m/s

3
1.5
Displacement (cm)
0.2 0.4 0.6 0.8 1.0
1.5
Time (s)
3

Find the
(ii) Amplitude (1mk)
3cm = 0.03m
(iii) Frequency (2mk)
Periodic time, T = 0.4s
Frequency, f = 1/T
= 1/0.4
= 2.5Hz

(iii) The wavelength () (2mk)

Wavelength, λ = velocity, v
Frequency, f
= 250m/s
2.5Hz
= 100m

7. The figure below shows a wave profile. The distance between the 2nd and the 4th
crest is 60cm. Determine the velocity of the wave in m/s. (3mk)

0 0.2 0.4 0.6 0.8 1.0 1.2

Time (s)

The distance between 2nd and 4th crests is two wavelengths.

Wavelength, λ = 60/2
= 30cm
= 0.3m
Frequence, f = 5Hz ( 5 cycles in 1 second)

Velocity, v = wavelength x frequency

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 = 0.3m x 4H
= 1.5m/s

8. The fig. shows a wave profile. Determine the frequency of the wave.

Displacement (cm)
3

0.2 0.4 Time (s)

-3

Periodic time, T = 0.4s

Frequency, f = 1/T
= 1/0.4
= 2.5Hz
9. The figure below show the displacement time graph of a wave traveling at
400cm/s.
Displacement

2
(cm)

0.8
(s)
-2

Determine for the wave the:

(i) Amplitude (1mk)
2cm
0.2m
(ii) Period (1mk)
T = 0.8/2
= 0.4s
(iii) Frequency (2mks)
f = 1/T
= 1/0.4
= 2.5Hz
(iv) Wavelength (3mks)
Velocity,v = 400cm/s
Frequency, f = 2.5Hz
Wavelength, λ = velocity/frequency
= 4m/s
2.5H
= 1.6m

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 10. Figure below shows a wave profile for a wave whose is 4Hz

Displacement (cm)

t1 t2 t3 t4 t5 t6 t7
Time (s)

Determine the value of t 5 (s) 2 mks

Frequency, f = 4Hz
Periodic time,T = 1/f
=¼
= 0.25s
I cycle = 0.25s
1 ¼ cycle = 0.25 x 5/4
t5 = 0.3125s

Fig below shows a wave of water ripples

Displacement

2
B C
A
(cm)

Time (s)
-2

i) What is the amplitude of its wave 1 mk

Amplitude = 2cm
= 0.02m
ii) If the speed of the ripple is 30cms -1 and the distance A to B is 3cm, find the
periodic time of the wave
Wavelength = 2x3
= 6cm
Frequency =speed/wavelength
= 30/6
= 5Hz
Periodic time = 1/5
= 0.2s
11. Fig shows the displacement – time graph for a certain wave.
4
Displacement

3
2
1
(cm)

-1
-2 Time (s)
-3
-4
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 Sketch on the same axes, a wave of both frequency and amplitude double that
of the wave.
12. The graphs in figure 4 represent the same wave.
Displacement

0.5 1.5 2 2.5 3.0 30

(cm)

Distance (m)

Determine the velocity of the wave. (3 mks)

T = 0.2s
F=1/T
= 1/0.2
=5Hz
λ = 0.4m
v= fλ
v=5Hz x0.4m
=2.0m/s

13. The sketch graph in figure (a) and (b) below represent the same wave.
Determine the velocity of the wave (3mk)
Displacement

0.05 0.10 0.15 0.2 Time(s)

T = 0.2s
F = 1/T
=1/0.2
= 5Hz
λ = 2.0m
v = fλ
= 5Hz x2.0m
10m/s

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 15. Sketch a displacement-time graph of a wave of amplitude 0.5 cm and
frequency 4Hz over a time interval of 1.25s
T = 1/f
=1/4
=0.25s

16.Figure shows air molecules in front of a hollow, wooden box B set vibrating by a
tuning fork.

Tuning fork

X Y Z

i) State the reason of mounting the tuning fork on the box which is open at one
end. (2mks)

To produce a coherent source of vibration

ii) What is the name given to this kind of wave? (1mk)

Longitudinal wave

iii) What are the sections X and Y called? (2mks)

X – compressions
Y – rarefactions

(iv) A wave front takes 0.03 seconds to travel from X to Z. The velocity of sound in
air is 330m/s. Calculate:
(a) the frequency of wave produced. (2mks)
v= 330m/s
2T= 0.03s
T = 0.03/2
= 0.015s
f=1/T
=1/0.015
= 66.67Hz

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 (b) the wavelength of the wave. (2mks)
λ= v/f
= 330/66.67
= 4.95m

17. The figure below shows sound waves in air produced by a vibrating tuning
fork. R is an air molecule on the path of the waves.
Oscillation

i) Using a line, indicate on the diagram a distance d equal to one wavelength of the
wave. (1mk)
ii) In the space provided below, show with an arrow the direction of motion of the
air molecule R as the waves pass. (1mk)

iii) Explain the reason for the answer (ii) (2mk)

The point source of vibration is from the vibrating fork.

18.The figure below shows a longitudinal wave that takes 0.32s to move from point
X to Y and at a speed of 50m/s.

X Y

rarefaction compression
(a) Show on the wave a region of
(i) Rarefaction
(ii) Compression
(b) Calculate
(i) the frequency of the wave
4λ = 0.32s
λ= 0.32/4 = 0.08s
T = 0.08s
f= 1/T
= 1/0.08s
12.5Hz

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 (ii) the wavelength of the wave
λ = v/f
= 50/12.5
= 4m

19.The figure below shows a longitudinal wave that takes 0.5s to move from point
A to B and at a speed of 36m/s.
A 0.5s B

Calculate
a. The frequency of the wave
2 ½ λ takes 0.5s
1λ take 0.5/2 ½
= 1.25s
T= 1.25s
f=1/T
f = 1/1.25
= 0.8Hz

b. The wavelength of the wave

λ= v/f
= 36/0.8
= 45m

20.The figure below shows a longitudinal wave on a spring that takes 0.05s to
move from point X to Y.
X 0.05 s Y

Calculate the frequency of the wave. (3mk)

2 ½ λ takes 0.02 s
1λ take 0.02/2.5
= 0.08s
T = 0.08s
f=1/T
= 1/0.08
= 12.5Hz

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 21.The figure below shows a longitudinal wave on a spring that takes 0.16s to
move from point X to Y and at a speed of 80m/s.

X Y

Rarefaction compression
(a) Show on the wave a region of
i) Rarefaction
ii) Compression
(c) Calculate
i) the frequency of the wave
2λ take 0.16s
1λ take 0.16/2 = 0.08s
T = 0.08s
f= 1/T
= 1/0.08
= 12.5Hz

ii) the wavelength of the wave

λ= v/f
= 80m/s
12.5Hz
= 6.4m

22.The longitudinal wave below takes 1.5s to move from point M to N.

1.5 s
M N
Calculate
(i) The frequency of the wave
2½ oscillations take 1.5s
1 oscillation takes 1.5s/1.5
= 1s
T = 1s
f=1/T
= 1/1
= 1Hz

(ii) The speed of the wave if it has a wavelength of 60cm

λ =60cm =0.60m
v= fλ
= 1Hz x 0.60m
= 0.60m/s

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