3B Wave Motion II
6
Chapter 6 Wave Phenomena
Wave Phenomena
Practice 6.1 (p. 48)
By v = f, when the frequency is
1
D
doubled and the speed remains
2
D
unchanged, the wavelength is halved.
3
7
4
Therefore the new wavelength is 1 cm.
5
(a) Wavelength = = 1 cm
5
distance travelled
(b) Speed =
time taken
5
=
= 2 cm s1
2.5
(c)
(a) The wave speed remains to be 3 cm s1.
By v = f,
frequency =
(b) By v = f, when the frequency is
doubled and the speed remains
v 2
= = 2 Hz
λ 1
(d) Reduce the speed of the vibrator by half.
unchanged, the wavelength is halved.
Therefore the new wavelength is 1 cm
Practice 6.2 (p. 57)
and the new wavefronts are as shown.
1
D
2
B
3
D
4
 + 65 + 90 = 180
5
 = 25
Wave troughs are shown on the screen as dark
lines.
5
Wave crests are shown on the screen as bright
lines.
6
vshallow
=
2
=
1
(a) The wavelength of the wave is 2 cm.
(b) The frequency of the wave is 10 Hz.
λ shallow
λ deep
3
deep = 6 cm
Speed of water waves
= f  = 10  2 = 20 cm s1
(c)
 = 90   = 90   =  = 25
v deep
λ deep
6
Increasing the frequency does not
change the wave speed, so the new
The wavelength in the deep region is 6 cm.
v
By nXY = X ,
vY
speed in region X = vY  nXY
speed is 20 cm s1.
= 4  1.25
= 5 cm s1
New Senior Secondary Physics at Work
1
 Oxford University Press 2009
3B Wave Motion II
Chapter 6 Wave Phenomena
7
11
12
8
(a)
(b) Wave speed in the deep region
= f = 2  3 = 6 cm s1
(c)
(i) The speed of the waves in the
shallow region is 3 cm s1.
9
(ii) By v = f,
wavelength =
13
v 3
= = 1.5 cm
f 2
(a) Region B is deeper.
λ
v
(b) By A = A ,
vB λ B
wavelength in region A
v
2
= A  B =  1.5 = 1 cm
vB
3
10
(a)
(c)
In region A
In region B
Frequency
12 Hz
12 Hz
Wavelength
2 cm
1.5 cm
Wave speed
24 cm s1
18 cm s1
(b) Region A is deeper.
New Senior Secondary Physics at Work
2
 Oxford University Press 2009
3B Wave Motion II
Chapter 6 Wave Phenomena
(d) Refractive index from A to B
v
2
= A = = 0.667
vB 3
(b) Diffraction
(c)
Practice 6.3 (p. 64)
1
D
2
C
3
A
4
(a) Diffraction of waves is the spreading of
waves around the edge into the shadow
of an obstacle without a change in speed.
(b)
5
(a)
6
New Senior Secondary Physics at Work
3
(a)
 Oxford University Press 2009
3B Wave Motion II
Chapter 6 Wave Phenomena
(b) No, I do not agree with the company.
7
(b) Path difference at point P
This is because ocean waves diffract into
= YP – XP
the bay, so that the water in the bay may
= 3.5 – 2.5
not be calm enough for the sports.
=
(a) This design provides an entrance for the
= 2 cm
ships and at the same time reduces the
Path difference at point Q
amount of waves entering the typhoon
= XQ – YQ
shelter.
= 3.5 – 3
(b) I do not agree with him.
= 0.5
If breakwaters are built as in Figure d,
= 1 cm
water waves would diffract through the
(c)
Constructive interference happens at
opening and travel into the typhoon
point P.
shelter.
Destructive interference happens at point
Q.
Practice 6.4 (p. 76)
7
(a) By v = f,
1
D
2
A
3
D
4
B
= QA – PA
5
(a) & (b)
= 66 – 60
wavelength =
v 30
=
= 2 cm
f 15
(b) Path difference at point A
= 6 cm
(c)
Path difference at point A = 6 cm = 3
Constructive interference will be
observed at point A.
8
(a) Destructive interference
(b) Constructive interference takes place at
(c)
(P can be any point on the antinodal
positions where the path difference
1

equals  n  λ , where n = 0, 1, 2...
2

lines labelled by A(P).)
Destructive interference takes place at
(Q can be any point on the nodal lines
positions where the path difference
labelled by N(Q).)
equals n, where n = 0, 1, 2...
Move the sources further apart. /
Decrease the wavelength of waves.
6
(a) Waves are arriving in phase at point P
but in antiphase at point Q.
New Senior Secondary Physics at Work
4
 Oxford University Press 2009
3B Wave Motion II
9
Chapter 6 Wave Phenomena
(a) (i)
8
(a)
(b)
(ii)
9
(a) They are all momentarily at rest.
(b) (i)
(b) When t = 2 s, P, Q and R are
Particles B and C are vibrating in
phase.
momentarily at rest.
(ii) Particles B and C are vibrating in
Practice 6.5 (p. 88)
antiphase with particle D.
1
C
2
A
Revision exercise 6
3
D
Multiple-choice (p. 93)
4
B
1
5
B
Wavelength =
By
(a) Wavelength = 70  2 = 140 cm
=
(b) Holding the racquet at point A can
This is because the amplitude of
2
B
vibration at point A is smaller than that
3
D
at point B.
4
B
5
C
6
D
(a) A travelling wave carries and transmits
energy from one place to another. On the
λ deep
1
 12
1 .5
contrary, energy in a stationary wave is
(2):
localized.
By v = f,
wavelength =
(b) Both of them do not transfer matter.
New Senior Secondary Physics at Work
=
= 8 cm s1
reduce the vibrations felt by the hand.
7
v deep
,
vshallow λ shallow
λ
vshallow = shallow  vdeep
λ deep
0 .6
= 0.4 m
1 .5
Wave speed = f = 50  0.4 = 20 m s1
6
B
5
v 0 .1
=
= 0.02 m = 2 cm
5
f
 Oxford University Press 2009
3B Wave Motion II
Chapter 6 Wave Phenomena
Path difference at P = 6  4 = 2 cm = 
Conventional (p. 96)
 Constructive interference occurs at P.
1
(12  0.5 A)
7
C
Wave
8
D
speed
9
C
Reflection
no change
Wavelength
no change
At the mid-point between X and Y, the path
Direction
of travel
change
change / no
difference is 0 and constructive interference
Refraction
change
change
takes place.
change (at
i = 0)
Then consider the left side of the mid-point.
Diffraction
no change
no change
change
Let the path difference be .
Interference no change
no change
no change
Constructive interference takes place when
2
 = n = 3n
(Correct reflected pulse drawn)
(3  1A)
(a)
Also,   XY = 17 cm
 3n =   17
n  5.67
Therefore, the number of points of
constructive interference on the left side of
the mid-point is 5.
(b)
By symmetry, there are also 5 points of
constructive interference on the right side of
the mid-point.
 total number of points of constructive
interference = 5 + 1 + 5 = 11
10
C
11
D
12
C
13
(HKCEE 2004 Paper II Q25)
14
(HKCEE 2005 Paper II Q36)
15
(HKALE 2005 Paper II Q29)
16
A
17
(HKALE 2006 Paper II Q7)
18
C
(c)
Wavelength = 2  0.60 = 1.20 m
Speed = f = 300  1.20 = 360 m s1
New Senior Secondary Physics at Work
6
 Oxford University Press 2009
3B Wave Motion II
3
Chapter 6 Wave Phenomena
(a) Largest possible wavelength
= 2L = 2  10 = 20 m
(1A)
(b) Wave speed = f
(1M)
= 80 m s
(c)
(c)
(1A)
(1A)
(1A)
The sloped edge of the ripple tank can
(1A)
(d) Using spongy edge can also achieve the
(1A)
Wave speed of the new stationary wave
= 80 m s
(Correct change in amplitude)
reduce reflection of waves.
A stationary wave could be produced.
1
(1A)
(b) Refraction
= 4  20
1
(Decreasing wavelength)
6
purpose mentioned in (c).
(1A)
(Shorter wavelength)
(1A)
(Less bending)
(1A)
(a)
(1A)
By v = f,
wavelength of the new stationary wave
80
v
= =
= 10 m
(1A)
f 4 2
4
(b)
(Shorter wavelength in region A than
5
(Axes with correct labels)
(1A)
that in Figure e)
(1A)
(Correct amplitude)
(1A)
(Less bending)
(1A)
(Correct period)
(1A)
(Shorter wavelength in region B than in
(Correct shape)
(1A)
region A)
7
(a)
(1A)
(a) Wavelength of waves in region A
0.2
=
= 0.04 m
(1A)
5
Speed of waves in region A
= f
(1M)
= 5  0.04
= 0.2 m s1
New Senior Secondary Physics at Work
7
(1A)
 Oxford University Press 2009
3B Wave Motion II
(b) (i)
Chapter 6 Wave Phenomena
Region B is deeper.
(1A)
By v = f, doubling the frequency halves
(ii) The frequency is unchanged, which
is 5 Hz.
the wavelength, so the new wavelength
(1A)
is 2 cm.
Speed of waves in region B
5
=  0.2
4
(1A)
The path difference at P, which is 2 cm,
is now equal to .
(1A)
Therefore, constructive interference
= 0.25 m s1
(1A)
By v = f,
(1M)
occurs there.
(1A)
The displacementtime graph of particle
wavelength of waves in region B
v 0.25
= =
= 0.05 m
(1A)
5
f
P is as shown.
(c)
9
(Correct labelled axes)
(1A)
(Correct shape of the graph)
(1A)
(a) For constructive interference,
largest possible wavelength
(Correct wave direction)
(1A)
(Longer wavelength)
(1A)
1
λ
2
(c)
(1A)
New Senior Secondary Physics at Work
(1A)
Path difference at Q
= 22 – 21 = 1 cm
(1A)
Therefore, the largest possible
wavelength is 1 cm (i.e. path difference
(1A)
at Q =  and path difference at P = 2).
(b) Particle P will vibrate up and down with
a larger amplitude.
(1M)
waves is 4 cm.
Therefore, destructive interference
occurs.
(1A)
The largest possible wavelength of the
(1A)
(a) Path difference at P
= AP – BP = 2 cm =
= 2 cm
 = 4 cm
ripple tank. The water above the perspex
8
(1A)
(b) For destructive interference,
1
path difference = λ
2
1
2= λ
2
(d) We can put a sheet of perspex in the
is shallower than elsewhere.
= path difference
(1A)
(1A)
8
 Oxford University Press 2009
3B Wave Motion II
Chapter 6 Wave Phenomena
(d) He cannot obtain a clear interference
pattern
(b) (i)
(1A)
over 1.4 cm and the scale used by
because the two sources are incoherent.
the figure is 1 : 25.
(1A)
10
(b) When waves approach the shore, their
(c)
(1A)
and wavelength decrease.
(1A)
(i)
(1A)
Wavelength of the wave
1.4
=
 25
4
(a) The boat oscillates up and down. (1A)
wave speed
From Figure j, there are 4 waves
= 8.75 cm
(1A)
(ii)
A tsunami is a transverse wave.
(1A)
This is because the moving
direction of water molecules
(vertical) is perpendicular to the
direction of travel of the tsunami
(horizontal).
distance travelled
(ii) Speed =
time taken
(1A)
100  1000
8  60
(1M)
(Correct shape)
= 208 m s1
(1A)
(Constant separation between wave
=
(iii) The depth of seabed in the ocean
varies from place to place.
(c)
(1A)
(i)
(2  1A)
crests)
(1A)
Constructive interference
(1A)
(ii) At point G, destructive interference
Therefore, refraction occurs and
the wavefront bends.
(1A)
occurs,
(1A)
(iv) The statement is incorrect.
(1A)
so the amplitude of the wave is
When water waves travel from the
always zero and there is no wave
centre of earthquake to the shore,
energy at that point.
water is not transferred.
(1A)
(iii)
(1A)
Only energy is transferred by the
water waves.
11
(1A)
(a) After reflection, the reflected waves
move away from the barrier at 45 to the
normal, and
(1A)
they interfere with the incident waves.
(1A)
New Senior Secondary Physics at Work
9
 Oxford University Press 2009
3B Wave Motion II
Chapter 6 Wave Phenomena
Constructive interference occurs
(e)
when the path difference is 0, ,
2...
(1A)
and destructive interference occurs
when the path difference is
1
1
1
λ , 1 λ , 2 λ ...
2
2
2
(1A)
At F and H, since the path
(i)
difference is 0 and  respectively,
Similarly, constructive interference
i.e. they are not in neighbouring
occurs along PQ and TU and forms
loops of each other.)
lines of big crests and troughs.
(1A)
(1A)
13
(HKCEE 2005 Paper I Q5)
14
(a) Node
(1A)
1 .2
(b) Wavelength =
= 0.48 m
2 .5
(1M)
At G, since the path difference is
1
λ , destructive interference
2
Speed = f = 75  0.48 = 36 m s1 (1A)
(c)
occurs. Similarly, destructive
A stationary wave with two loops on the
interference occurs along RS and so
string has wavelength equal to 1.2 m.
a line of calm water is formed.
By v = f,
frequency =
(1A)
(a) 2 waves travel in opposite directions.
(1A)
amplitude.
microscopes can resolve is about 200 nm.
Stationary wave forms only at certain
(1A)
(1A)
(b) Diffraction
(b) The displacement of a point on the string
(c)
is perpendicular to the mean position of
the string.
and they are in antiphase.
(1A)
(d) Wavelength = 1.2 m
By v = f,
frequency =
wavelength look blurred.
New Senior Secondary Physics at Work
(1A)
(1A)
(d) The microscopes that use X-rays have a
higher resolving power.
(1M)
(1A)
This is because X-rays have a much shorter
(1M)
v 6 .2
=
= 5.17 Hz
λ 1 .2
Light diffracts around the edges of objects of
As a result, fine details close to the
The amplitude of the oscillation of point
(1A)
(1A)
size comparable to the wavelength.
(1A)
A is larger than that of point B,
(1A)
(a) The minimum size that ordinary optical
(1A)
frequencies.
v 36
=
= 30 Hz
λ 1 .2
Physics in articles (p. 100)
The 2 waves should have similar
(c)
(1A)
(ii) (P, Q and R located appropriately,
constructive interference occurs.
12
(6 loops)
wavelength.
(1A)
(1A)
10
 Oxford University Press 2009