Tutorial: Wave Motion
JURONG JUNIOR COLLEGE
J2 H1 Physics 8866 (2011)
Tutorial 8 : Wave Motion (Solution)
Learning
Outcomes
Sub-Topic
Tutorial Questions
show an understanding and use the terms displacement,
amplitude, phase difference, period, frequency, wavelength and
speed.
deduce, from the definitions of speed, frequency and
wavelength, the equation v = fλ.
recall and use the equation v = fλ.
show an understanding that energy is transferred due to a
progressive wave.
(a)
(b)
(c)
(d)
2
recall and use the relationship, intensity ∝ (amplitude) .
analyse and interpret graphical representations of transverse
and longitudinal waves.
show an understanding that polarisation is a phenomenon
associated with transverse waves.
determine the frequency of sound using a calibrated c.r.o.
determine the wavelength of sound using stationary waves.
(e)
(f)
(g)
(h)
(i)
1(a),(b), 2, 3, 10(b) & (c),
11 (a) & (b), 12(a)&(c)
11 (c)
1(c), 12(c)
Inferred in various
questions
7, 8, 9
4, 6, 10(a)
5, 11(d)
1(a)
Covered in Superposition
y / μm
1
+2
10
30
t / μs
0
20
40
−2
The displacement-time graph (y-t) above represents the simple harmonic motion of a
particle in a progressive wave travelling at a speed of 5.0 km s-1.
(a)
The frequency of vibration is _____ kHz
(b)
The amplitude of vibration is _____ μm
(c)
The wavelength is _____ mm
Working
(a)
(c)
[ 100 mm ]
Remarks
1
1
=
= 5 × 104 = 50 kHz
T 20 × 10−6
Amplitude = magnitude of max displacement = 2 μm
Wavelength, λ =
v 5 × 103
=
= 0.1 m = 100 mm
f 5 × 10 4
2
cfs@jj 200901
[ 2 μm ]
From the graph, the period, T = 20 μs
f=
(b)
[ 50 kHz ]
Page 1 of 8
Common error : Students
may think that distance of
crest to crest is wavelength.
Tutorial: Wave Motion
displacement
A
B
time
The diagram above shows two oscillations A and B.
(a)
What is the phase difference between the oscillations?
(b)
Which oscillation is leading the other one?
(a)
(b)
3
Working
t
1 π
Phase difference, φ = (2π) = (2π) = rad
T
8 4
The crest of A occurs earlier than the corresponding
crest of B, so is the trough or any other feature of the
graph, so A leads B.
[ π/4 rad ]
[ A leads B ]
Remarks
The phase difference could
also be stated as B leading
A by
7π
rad, but we normally
4
quote the phase difference
between 0 to π
Parallel water waves of wavelength 10 m strike a straight sea wall. The wavefronts
make an angle of 30° with the wall.
What is the difference in phase at any instant between the waves at two points 5 m
apart along the wall?
[ 90° ]
Working
Step 1 : Draw the diagram
wavefronts
sea wall
Direction of
Travel of wave
30°
5m
A wavefront is always perpendicular to the direction of travel of the wave.
A distance of 5 m along the sea wall is equivalent to 2.5 m along the direction of the
wave since sin 30° = 0.5
Given that the wavelength, λ = 10 m
x
2.5 π
π
= rad = 90° or rad
Phase difference, φ = (2π) = (2π)
10 2
λ
2
cfs@jj 200901
Page 2 of 8
Tutorial: Wave Motion
4
Transverse progressive sinusoidal waves of wavelength λ are passing vertically
5λ
apart and the waves are
along a horizontal rope. P and Q are points on the rope
4
traveling from P to Q.
Which of the following correctly describes Q at an instant when P is displaced
upwards but is moving downwards?
[ B ]
A
B
C
D
E
Displacement of Q
upwards
upwards
downwards
downwards
downwards
Movement of Q
downwards
upwards
upwards
downwards
stationary
Working
P
Q
Direction of wave travel
1
2
3
4
5
6
7
8
The diagram above shows 2 wavelengths of the wave in the rope moving to the right,
divided into 8 sections of
λ
each.
4
Assume particle P is in any of the first 4 sections, displaced upwards and moving
downwards, so it has to be in section 1.
5λ
away to the right, since the waves are traveling from P to Q, so it has
Particle Q is
4
to be in section 6. So it has to be displaced upwards, and moving upwards. [ B ]
5
Which of the following is true for all transverse waves?
A
B
C
D
[ B ]
They are all electromagnetic.
They can all be polarized.
They can all travel through a vacuum.
They all involve the oscillation of atoms.
Working
A
B
C
D
cfs@jj 200901
False. All electromagnetic waves are transverse, but not all transverse waves are EM.
E.g. transverse waves of rope.
TRUE.
Only EM transverse waves can travel trough vacuum. All transverse mechanical waves
need a medium.
False. EM waves involve oscillations of fields, not atoms.
Page 3 of 8
Tutorial: Wave Motion
6
The same progressive wave is represented by the following graphs.
displacement y against time
t for constant position
y
p
displacement y against position
x for constant time
y
0
q
0
t
x
Which of the following gives the speed of the wave?
p
q
1
C
D
A pq
B
q
p
pq
[C]
Working
In the first graph, with time t as the x-axis, p = T = period.
In the second graph, with position x as the x-axis,
q = λ = wavelength.
q
1
[C]
Using v = fλ = (λ) =
T
p
7
Remarks
Emphasize
difference between
displ - time graph &
displ – distance (or
position) graph
A point source of sound emits energy equally in all directions at a constant rate and a
person 8 m from the source listens. After a while, the intensity of the source is
halved.
If the person wishes the sound to seem as loud as before, how far should he be now
from the source?
[D]
B 2 2m
A 2m
C 4m
D 4 2m
Working
For sound, loudness is measured by intensity.
P
PS
At r = 8 m, the intensity at the listener, I = S 2 =
4π r
4π (8)2
where PS = power of the source
E 8 2m
- - (1)
When the intensity of the source is halved, its power is also
halved.
For the same loudness at a new distance r, the intensity at the
listener must be the same,
1
P
i.e.
I = 2 S2
- - - - - (2)
4π r
1
PS
2 PS
=
(1) = (2);
4π (8)2 4π r 2
Æ
r2 = 21 (8)2
Æ
cfs@jj 200901
r=4 2m
Page 4 of 8
[D]
Remarks
Emphasize
energy spreads
out uniformly in
3-dimensional
space. Hence,
the power is
spread across
the area of a
sphere.
Tutorial: Wave Motion
8
A sound wave is emitted from a point source. The intensity of the sound wave is
inversely proportional to the square of the distance from the source. At a distance r
from the source, the amplitude of the wave is 8X.
Determine the amplitude of the sound wave at a distance 2r from the source in terms
of X.
[ 4X ]
Working
1
r2
I ∝ A2
I∝
Also
- - - - - (1)
- - - - - (2)
k
r
k
At a distance 2r, the new amplitude, A =
2r
(4)
:
A = 4X
(3)
From (1) and (2),
9
A∝
1
Æ
r
8X =
- - - - - (3)
- - - - - (4)
A 100 W light bulb is 10% efficient (that is, 90% of its output is invisible infra-red
radiation and only 10% is visible light). A person can see the light with the naked eye
from a distance of 20 km on a dark night.
If the area of the pupil of the person’s eye is 0.5 cm2, find the power of the light that
[1.0 × 10-13 W]
the eye of the person is receiving. (Hutchings, 2nd ed., pg 163)
Working
Intensity of light that the eye can detect,
P
P
I= =
A 4π r 2
Since only 10% of the power output is received,
(10%)(100)
= 1.99 × 10-9 W m-2
Æ
I=
4π (20 × 103 )2
For area of the eye, A = 0.5 cm2,
the power of light received,
P = IA = (1.99 × 10-9)(0.5 × 10-4)
= 9.95 × 10-14 W
= 1.0 × 10-13 W
cfs@jj 200901
Page 5 of 8
Remarks
Tutorial: Wave Motion
10
(Fisherman effective guide, pg 10-9, modified)
A progressive wave has amplitude 0.80 m and wavelength 4.0 m. At a given time the
displacement is y = 0 m at x = 0 m, and y = 0.80 m at x = 1.0 m.
(a)
Sketch the displacement-position graph for 2 cycles of the wave.
(b)
Calculate:
(i)
the displacement at x = 0.5 m and 2.20 m;
[ 0.57 m, −0.25 m ]
(ii)
the phase angles at x = 0.5 m and 1.60 m;
[ 0.25π rad, 0.80π rad ]
(iii)
the phase difference between any two points which are 1.1 m apart on
the wave.
[ 0.55π rad ]
(c)
Explain the relationship of your answers in (b)(ii) and (b)(iii).
Working
(a)
y/m
0.8
8.0
4.0
0
x/m
- 0.8
(b)
Assuming at t = 0, y = 0 m at x = 0 m
the equation representing the wave is given by
y = A sin (
2π x
λ
)
Since the amplitude, A = 0.8 m; and the wavelength, λ = 4.0 m
(i)
At x = 0.5 m,
At x = 2.2 m,
(ii)
At x = 0.5 m,
At x = 1.6 m,
(iii)
2π x
)
4.0
2π (0.5)
y = 0.8 sin (
) = 0.566 m
4.0
2π (2.2)
) = −0.247 m
y = 0.8 sin (
4.0
y = 0.8 sin (
Æ
2π (0.5)
= 0.25π rad
4.0
λ
2π x 2π (1.6)
φ=
=
= 0.80π rad
4.0
λ
φ=
cfs@jj 200901
=
For any two points 1.1 m apart,
2π (1.1)
= 0.55π rad
4.0
λ
The two points in (b)(ii) are at x = 0.5 m and x = 1.6 m, they are 1.1 m
apart. Their phase difference, φ = (0.80π − 0.25π) = 0.55π rad
which is the answer to (b)(iii).
φ=
(c)
2π x
Page 6 of 8
2π x
=
Tutorial: Wave Motion
11
(a)
State the meaning of wavelength and frequency as applied to wave motion.
(b)
Deduce, from definition of speed, the equation for the speed of a wave in
terms of its wavelength and frequency.
(c)
Distinguish between longitudinal and transverse waves.
(d)
Which phenomenon associated with transverse waves is not observed with
longitudinal waves?
Working
(a)
Wavelength is the distance travelled by the wave in one period.
Frequency is the number of oscillations performed by an element in the
wave per unit time.
(b)
By definition, speed =
distance travelled
time taken
For a wave,
speed =
Since frequency, f =
(c)
λ
T
1
,
T
speed =
λ
T
= fλ
A longitudinal wave is one where the direction of vibration of the elements
in the wave is parallel to the direction of propagation of the wave.
A transverse wave is one where the direction of vibration of the elements in
the wave is perpendicular to the direction of propagation of the wave.
(d)
12
Polarisation
A ship’s siren vibrates with displacement y, where y = a sin 200πt.
This sound causes vibrations of the diaphragm of an eardrum of an observer 500 m
away. The speed of sound is 335 m s-1. Calculate
(a)
(b)
the frequency of the sound,
[ 100 Hz ]
the number of wavelength of this sound there are between the siren and the
eardrum, in 5 sig.fig.,
[ 149.25 ]
(c)
the phase difference between the motion of the siren and the eardrum, [
π
2
rad]
If the speed and distance data were reliable only to 3 s.f., little confidence could be
placed in your answer to (c). Explain this.
cfs@jj 200901
Page 7 of 8
Tutorial: Wave Motion
Working
In general, displacement
y = a sin ωt = a sin 2πft
Given that the displacement,
y = a sin 200πt
(a)
By comparison,
f = 100 Hz
(b)
The wavelength,
λ=
v 335
=
= 3.35 m
f 100
The number of wavelength, N =
(c)
500
= 149.25
3.35
For whole number of wavelengths apart, there is no phase difference, the
phase difference only arises from the fraction of a wavelength.
In this case, the fraction,
x
λ
= 0.25 =
φ=
Æ
π
2
φ
2π
rad
If the speed and distance data were reliable only to 3 s.f. the answer to (b) would
be 149, the remaining 4th and 5th s.f. is totally unreliable.
Since the value of φ is entirely based on the value of 0.25, which is the 4th and 5th
s.f., the answer in (c) would be also totally unreliable.
Extra Question not in Tutorial
To be used as a class quiz
Step-by-Step (CS Toh) – A Level Practice Question
Qn 2, Pg 130 [duration : 20 mins]
cfs@jj 200901
Page 8 of 8