Physics Bk3B Unit 6 Answer

3B Wave Motion II Chapter 6 Wave Phenomena

6 Wave Phenomena

Practice 6.1 (p. 48)1 D

2 D

3

4 (a) The wave speed remains to be 3 cm s1.

(b) By v = f, when the frequency is

doubled and the speed remains

unchanged, the wavelength is halved.

Therefore the new wavelength is 1 cm

and the new wavefronts are as shown.

5 Wave troughs are shown on the screen as

dark lines.

Wave crests are shown on the screen as

bright lines.

6 (a) The wavelength of the wave is 2 cm.

(b) The frequency of the wave is 10 Hz.

Speed of water waves

= f = 10 2 = 20 cm s1

(c) Increasing the frequency does not

change the wave speed, so the new

speed is 20 cm s1.

By v = f, when the frequency is

doubled and the speed remains

unchanged, the wavelength is halved.

Therefore the new wavelength is 1 cm.

7 (a) Wavelength = = 1 cm

(b) Speed =

= = 2 cm s1

(c) By v = f,

frequency = = = 2 Hz

(d) Reduce the speed of the vibrator by half.


2 B

3 D

4

+ 65 + 90 = 180

= 25

= 90 = 90 = = 25

5 =

=

deep = 6 cm

The wavelength in the deep region is 6 cm.

6 By nXY = ,

speed in region X = vY nXY

New Senior Secondary Physics at Work Oxford University Press 20091


= 4 1.25

= 5 cm s1

7

8

9

10 (a)

In region A In region B

Frequency 12 Hz 12 Hz

Wavelength 2 cm 1.5 cm

Wave speed 24 cm s1 18 cm s1

(b) Region A is deeper.



11

12 (a)

(b) Wave speed in the deep region

= f = 2 3 = 6 cm s1

(c) (i) The speed of the waves in the

shallow region is 3 cm s1.

(ii) By v = f,

wavelength = = = 1.5 cm

13 (a) Region B is deeper.

(b) By = ,

wavelength in region A

= B = = 1 cm

(c)



(d) Refractive index from A to B

= = = 0.667


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)



(b) Diffraction

(c)

6 (a)

(b) No, I do not agree with the company.

This is because ocean waves diffract

into the bay, so that the water in the bay

may not be calm enough for the sports.

7 (a) This design provides an entrance for the

ships and at the same time reduces the

amount of waves entering the typhoon

shelter.

(b) I do not agree with him.

If breakwaters are built as in Figure d,

water waves would diffract through the

opening and travel into the typhoon

shelter.


2 A

3 D

4 B

5 (a) & (b)

(P can be any point on the antinodal

lines labelled by A(P).)

(Q can be any point on the nodal lines

labelled by N(Q).)

(c) 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.



(b) Path difference at point P

= YP – XP

= 3.5 – 2.5

=

= 2 cm

Path difference at point Q

= XQ – YQ

= 3.5 – 3

= 0.5

= 1 cm

(c) Constructive interference happens at

point P.

Destructive interference happens at

point Q.

7 (a) By v = f,

wavelength = = = 2 cm

(b) Path difference at point A

= QA – PA

= 66 – 60

= 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

positions where the path difference

equals , where n = 0, 1, 2...

Destructive interference takes place at

positions where the path difference

equals n, where n = 0, 1, 2...

9 (a) (i)

(ii)

(b) When t = 2 s, P, Q and R are

momentarily at rest.

Practice 6.5 (p. 88)1 C

2 A

3 D

4 B

5 B

Wavelength = = 0.4 m

Wave speed = f = 50 0.4 = 20 m s1

6 (a) Wavelength = 70 2 = 140 cm

(b) Holding the racquet at point A can

reduce the vibrations felt by the hand.

This is because the amplitude of

vibration at point A is smaller than that

at point B.

7 (a) A travelling wave carries and transmits

energy from one place to another. On

the contrary, energy in a stationary

wave is localized.



(b) Both of them do not transfer matter. 8 (a)

(b)

9 (a) They are all momentarily at rest.

(b) (i) Particles B and C are vibrating in

phase.

(ii) Particles B and C are vibrating in

antiphase with particle D.

Revision exercise 6Multiple-choice (p. 93)1 B

By = ,

vshallow = vdeep

= 12

= 8 cm s1

2 B

3 D

4 B

5 C

6 D

(2):

By v = f,



wavelength = = = 0.02 m = 2 cm Path difference at P = 6 4 = 2 cm =

Constructive interference occurs at P.

7 C

8 D

9 C

At the mid-point between X and Y, the path

difference is 0 and constructive interference

takes place.

Then consider the left side of the mid-point.

Let the path difference be .

Constructive interference takes place when

= n = 3n

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.

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

Wavelength = 2 0.60 = 1.20 m

Speed = f = 300 1.20 = 360 m s1



Conventional (p. 96)1 (12 0.5 A)

Wave

speedWavelength

Direction

of travel

Reflection no change no change change

Refraction change change

change / no

change (at

i = 0)

Diffraction no change no change change

Interference no change no change no change

2 (Correct reflected pulse drawn) (3 1A)

(a)

(b)

(c)

3 (a) Largest possible wavelength

= 2L = 2 10 = 20 m (1A)

(b) Wave speed = f (1M)

= 4 20

= 80 m s1 (1A)

(c) A stationary wave could be produced.

(1A)

Wave speed of the new stationary wave

= 80 m s1 (1A)

By v = f,

wavelength of the new stationary wave

= = = 10 m (1A)

4

(Axes with correct labels) (1A)

(Correct amplitude) (1A)

(Correct period) (1A)

(Correct shape) (1A)

5 (a)



(Decreasing wavelength) (1A)

(Correct change in amplitude) (1A)

(b) Refraction (1A)

(c) The sloped edge of the ripple tank can

reduce reflection of waves. (1A)

(d) Using spongy edge can also achieve the

purpose mentioned in (c). (1A)

6 (a)

(Shorter wavelength) (1A)

(Less bending) (1A)

(b)

(Shorter wavelength in region A than

that in Figure e) (1A)

(Less bending) (1A)

(Shorter wavelength in region B than in

region A) (1A)

7 (a) Wavelength of waves in region A

= = 0.04 m (1A)

Speed of waves in region A

= f (1M)

= 5 0.04

= 0.2 m s1 (1A)



(b) (i) Region B is deeper. (1A)

(ii) The frequency is unchanged,

which is 5 Hz. (1A)

Speed of waves in region B

= 0.2

= 0.25 m s1 (1A)

By v = f, (1M)

wavelength of waves in region B

= = = 0.05 m (1A)

(c)

(Correct wave direction) (1A)

(Longer wavelength) (1A)

(d) We can put a sheet of perspex in the

ripple tank. The water above the

perspex is shallower than elsewhere.

(1A)

8 (a) Path difference at P

= AP – BP = 2 cm = (1A)

Therefore, destructive interference

occurs. (1A)

(b) Particle P will vibrate up and down with

a larger amplitude. (1A)



By v = f, doubling the frequency

halves the wavelength, so the new

wavelength is 2 cm. (1A)

The path difference at P, which is 2 cm,

is now equal to . (1A)

Therefore, constructive interference

occurs there. (1A)

The displacementtime graph of particle

P is as shown.

(Correct labelled axes) (1A)

(Correct shape of the graph) (1A)

9 (a) For constructive interference,

largest possible wavelength

= path difference (1A)

= 2 cm (1A)

(b) For destructive interference,

path difference = (1M)

2 =

= 4 cm (1A)

The largest possible wavelength of the

waves is 4 cm.

(c) Path difference at Q

= 22 – 21 = 1 cm (1A)

Therefore, the largest possible

wavelength is 1 cm (i.e. path difference

at Q = and path difference at P = 2).

(1A)

(d) He cannot obtain a clear interference

pattern (1A)

because the two sources are incoherent.

(1A)

10 (a) The boat oscillates up and down. (1A)

(b) When waves approach the shore, their

wave speed (1A)

and wavelength decrease. (1A)

(c) (i) 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). (1A)

(ii) Speed =

= (1M)

= 208 m s1 (1A)

(iii) The depth of seabed in the ocean

varies from place to place. (1A)

Therefore, refraction occurs and

the wavefront bends. (1A)

(iv) The statement is incorrect. (1A)

When water waves travel from the

centre of earthquake to the shore,

water is not transferred. (1A)

Only energy is transferred by the

water waves. (1A)

11 (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)



(b) (i) From Figure j, there are 4 waves

over 1.4 cm and the scale used by

the figure is 1 : 25. (1A)

Wavelength of the wave

= 25

= 8.75 cm (1A)

(ii)

(Correct shape) (2 1A)

(Constant separation between

wave crests) (1A)

(c) (i) Constructive interference (1A)

(ii) At point G, destructive

interference occurs, (1A)

so the amplitude of the wave is

always zero and there is no wave

energy at that point. (1A)

(iii)

Constructive interference occurs

when the path difference is 0, ,

2... (1A)

and destructive interference occurs

when the path difference is ,

, ... (1A)

At F and H, since the path

difference is 0 and respectively,

constructive interference occurs.

Similarly, constructive interference

occurs along PQ and TU and

forms lines of big crests and

troughs.

(1A)

At G, since the path difference is

, destructive interference

occurs. Similarly, destructive

interference occurs along RS and

so a line of calm water is formed.

(1A)

12 (a) 2 waves travel in opposite directions.

(1A)

The 2 waves should have similar

amplitude. (1A)

Stationary wave forms only at certain

frequencies. (1A)

(b) The displacement of a point on the

string is perpendicular to the mean

position of the string. (1A)

(c) The amplitude of the oscillation of point

A is larger than that of point B, (1A)

and they are in antiphase. (1A)

(d) Wavelength = 1.2 m (1M)

By v = f, (1M)



frequency = = = 5.17 Hz (1A) (e)

(i) (6 loops) (1A)

(ii) (P, Q and R located appropriately,

i.e. they are not in neighbouring

loops of each other.) (1A)

13 (HKCEE 2005 Paper I Q5)

14 (a) Node (1A)

(b) Wavelength = = 0.48 m (1M)

Speed = f = 75 0.48 = 36 m s1 (1A)

(c) A stationary wave with two loops on the

string has wavelength equal to 1.2 m.

By v = f,

frequency = = = 30 Hz (1A)

Physics in articles (p. 100) (a) The minimum size that ordinary optical

microscopes can resolve is about 200 nm.

(1A)

(b) Diffraction (1A)

(c) Light diffracts around the edges of objects of

size comparable to the wavelength. (1A)

As a result, fine details close to the

wavelength look blurred. (1A)

(d) The microscopes that use X-rays have a

higher resolving power. (1A)

This is because X-rays have a much shorter

wavelength. (1A)


Physics Bk3B Unit 6 Answer

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