physics:chapter 6 waves jpn pahang

8/7/2019 Physics:Chapter 6 Waves JPN Pahang

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JPN Pahang Physics Module Form 5Teachers Guide Chapter 6:Wave

6.1 Understanding Waves

1. Motion of Waves

A wave front is a line or plane on which the vibration of every points on it are in phaseand are at the.(same/different) position from the source of the water.

2. When we use a fingertip to touch the surface of water repeatedly, ....(circular/plane)wavefronts are produced.

3. Types of wavesThere are 2 types of waves :-(a) .(b) .

4. Transverse wave(i). Transverse wave is a wave in which the vibration of particles in the medium is

at .. (parallel/perpendicular) to the direction of propagation of the wave.

A = .(crest / compression)

B = ..(rarefaction /trough)

(ii). Example of transverse wave:i. ..ii...iii..

5. Longitudinal waves(i) A longitudinal is a wave in which the vibration of particles in the medium is

.(parallel / proportional) to the direction of propagationof wave.

(ii) Example of longitudinal wave:..P = ..(crest / compression)Q = ..(rarefaction / trough)

1

A A

BB

P

P

Q Q

trough

crest

Water wave

Light waveElectromagnetic wave

sound wave

compressionrarefaction

Transverse wave Longitudinalwave

same

perpendicular

CHAPTER 6: WAVE

parallel

circular


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Amplitude, Period and Frequency of a Wave

Figure 6.20

1. Equilibrium position is .

2. One complete oscillation of the pendulum occurs when the pendulum bob moves from............

3. The period, T of a vibrating system is ...4. Frequency, f is

The S.I. unit is ..

f =T

1or f =

t

n

5. Stroboscope is use to ..Stroboscope frequency = number of slits x rotation frequency of stroboscope

or

6. Wave speedIf the wavelength of the wave is , the waves move forward a distance offx per second. Therefore, the speed of waves, v

v = f

2

A

B

C A

B

C

A-B-C-B-A

the position of the object where is no resultant force acts on the object.

Frequency, f

The number of complete oscillations

second.The S.I. unit is Hertz (Hz)

Question 2

A child on a swing makes 20 complete swings in

30 seconds. What is the frequency of theoscillation?

Question 3

In an experiment , Lim observes that a simplependulum completes 30 ascillations in 48.0seconds. What is

the period of oscillation?The frequency of oscillation?

Question 2

A child on a swing makes 20 complete swings in

30 seconds. What is the frequency of theoscillation?

Question 3

In an experiment , Lim observes that a simplependulum completes 30 ascillations in 48.0seconds. What is

the period of oscillation?The frequency of oscillation?

Frequency, fThe number of complete oscillations second.The S.I. unit is Hertz (Hz) f = 1 or f =

T

f = np

freeze waves patterns on the screen

the time taken to complete an oscillation

the number of complete oscillations

Hertz (Hz)


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Exercise 6.1

Figure 6.10

1. (a) The wavelength of the wave in the diagram above is given by letter .

(b) The amplitude of the wave in the diagram above is given by letter

2. Indicate the interval which represents one full wavelength.

Figure 6.11

Answer:

3

A

D

AE, CG, BF,

0 0


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Exercise 6.11

1. In an experiment , Lim observes that a simple pendulum completes 30 oscillations in

15.0 seconds. What is

(a) the frequency of oscillation?(b) the period of oscillation?

Answer:

(a) f = 30/15 = 2 Hz(b) T = 1/f

= 1/2= 0.5 s

2.

Calculate the frequency of the given wave above.Answer : T = 4s

f =T

1Hz

= 0.25 Hz

3. Based in the displacement-distance graph of a wave, find(a) the amplitude(b) the wavelength of the wave

Answer:(a) amplitude = 5 cm(b) Wavelength = 0.4 cm

4

5

-5

4

Dis lacement/cm

6 Time/s0

Displacement/cm

60

Distance/cm4

5

-5


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4. A transverse wave is found to have a distance of 4 cm from a trough to a crest, afrequency of 12 Hz, and a distance of 5 cm from a crest to the nearest trough. Determinethe amplitude, period, wavelength and speed of such a wave.

Answer:

5. A girl moves a long slinky spring at a frequency of 3 Hz to produce a transverse wavewith a wavelength of 0.5 m. What is the wave speed of the waves along the slinkyspring?

Answer:

6. The figure shows a wavefront pattern in a ripple tank produced by a vibrating dipper atfrequency of 5 Hz. What is the wave speed?

Answer:

f = 5 Hz = 6.0/4 = 1.5 cm

V = f= 7.5 cm s-1

5

6.0 cm

v = f x = 3 x 0.5= 1.5 m s-1

Amplitude: 4/2 = 2 cm

Wavelength: 5 x2 = 10 cm

Speed: f x = 120 cm s-1


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7. The figure shows a wavefront pattern in a ripple tank produced by a vibrating dipper atfrequency of 5 Hz. What is the wave speed?Answer:

f = 5 Hz = 4.5/3 = 1.5 cm

V = f

= 7.5 cm s-1

8. A mechanical stroboscope has 12 slits and rotates at a frequency 5 Hz. The stroboscopeis used to observe water waves. The observer notes there are 6 successive bright bands at adistance 20 cm. Calculate the speed of the water waves.

Answer:

f = n p

= 12 x 5= 60 Hz

9. The figure shows a loudspeaker produces a sound with a frequency 300 Hz. Calculate(a) the wavelength.(b) the speed of soundAnswer:

(a) 0.6 m

(b) V = f

= 300 x 0.5

6

4.5 cm

20 cm (5)

= 20/5= 4 cm

V = f= 60 x 4= 240 cm s-1


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= 150 m s-1

10. A given wave travels at a speed of 2 x 105 ms-1. If the frequency of the wave is 1000 Hz,Calculate the wavelengthAnswer:

= v/f

=1000

102 5x= 200 m

11. (a) Baseon the figure determine,(i) the amplitude(ii) the wave length

(b) What is the frequency of the sound if the speed of sound is 330 ms-1.Answer:(a) (i) 2 cm

(ii) 50 cm

(b) f = v /

= 330 / 0.5

= 660 Hz

Damping

Figure 6.12

7


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1. Damping is a ..

2. When a system is damped , the amplitude of the of oscillation .(decreases/increases)slowly until the system stops oscillating.

3. Damping is usually caused by :-(i) external frictional forces such as air resistance(ii) internal forces , where energy is lost from the system in form of heat.

Resonans

A resonance is ..

Figure 6.12: Bartons pendulum

Pendulum B and D are the same length. When pendulum B oscilates, all the pendulum startto oscilate. But pendulum B and D have the same length, so there have same naturalfrequency. So pendulum D will oscilates with (maximum/minimum) amplitude.

6.2 ANALYSING REFRLECTION OF WAVES

1. Reflection of a wave occurs when a wave strikes an obstacle such as barrier, planereflector, mirror and wall.

2. The reflection of waves obeys the law of reflection :..

3. When the reflection of a wave happened , the wavelength ,, the frequency, f and thespeed, v .but the direction of propagation of the wave changes.

8

Experiment to show a phenomenon of resonance

AB

C

D

E

process whereby oscillations die down due to a loss of energy to friction forces.

increases

the phenomenon when the oscilating system is driven at its natural frequency by

a periodic force.

maximum

(i) The angle of incidence is equal to the angle of reflection .(ii)The incident wave, the reflected wave and the normal lie in the same plane.

do not change


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Exercise 6.2Complete the diagram below.

Reflection of water waves

1.Show the pattern of the reflection of the water waves

(b)

2. Show the dark and bright pattern on the screen below.

(a)

(b)

9


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B = BrightD = Dark

10

Water waves

Ri le tank

ScreenB D B D B D B


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Reflection of Light Wave

1. When rays of light strike any surface the rays are reflected , unless the surface isblack, when they are absorbed.

2. A plane mirror is a flat smooth surface which reflects regularly most of the light

falling on it.

Figure 6.201. The characteristics of reflection of light waves:

2. The bright and dark bands of the wave pattern formed on the screen because thesurface of water acts as lenses. The crest of water waves similar with .(concave lens /convex lens) and the trough of water waves similar with (concave lens /convex lens).

Exercise 6.21

The diagram shows a single ray of light being directed at a plane mirror. What are theangles of incidence and reflection?(a) (b)

Answer:

i =r = 90 35 = 55o

i = r = 70 o

2.

11

35o

(i) The size of the image = the size of the object

(ii) The distance of the image = the distance of the object

convex lensconcave lens


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Reflection of sound waves

1. The sound waves is reflected by walls and ceilings of buildings, unborn baby or sea

bed.

Figure 6.21

2. The sound wave from the stopwatch experience a reflection after striking the smoothwall.The angle of incidence, i is .(equal to /more than/less than) the angle ofreflection, r.

Exercise 6.22

2s = v x t , t =340

2(50)= 0.29 s

2.

1. The diagram shows a student shouting in front of a school building. Calculate the timetaken by the student to hear the echo of his voice.[The speed of sound in air is 340 m s-1 ]

Answer:

12

equal to


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If an echo is heard one second after the holler and reflects off canyon walls which are adistance of 170 meters away, then what is the speed of the wave?

Answer:

6.3 ANALYSING REFRACTION OF WAVES

1. The refraction of water waves occur when there is a

1. After refraction, the wave has the same . but a different.

13

change of direction of the propagation of waves

traveling from a medium to another medium due

to a change of speed.

2s = v x t

= 2 (170) = 340 m s-1

1

Figure 6.30

frequency

speed, wavelength and direction.


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Refraction of Water Waves

Deep shallow

Figure 6.31

1. When the water wave travel from a deep area, the direction of the waves is refracted (towards/away from) the normal.

2. The angle of incidence, i of the water is . (greater / smaller) than the angle ofrefraction, r

3. When the refraction of a wave happened , the frequency, f does not change but the ................................................................................of the wave change.

Exercise 6.3

On each of the following diagram, draw the refracted wave by the perspex.

(a) (b)

(c) (d)

(e) (f)

Refraction of water wave of the sea water

14

towards

greater

wavelength ,, the speed and the direction of propagation


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Figure 6.32

1. Why are the speed and wavelength of waves in the middle of the sea almost uniform?

2. What do you think would happen to the wave speed if the depth of water is increased?..

.

3. Why do the distances between the wavefronts decrease as the waves approach the beach?

4. Why is the water in the bay stationary compared to the water at the cape?

15

The depth of water in the middle of the sea is almost uniform.

The speed

On approaching the beach, the depth of water decreases. The speed of waves decrese and the

wavelength is decreased.

The depth of water varies across the area of the bay

The energy of the water wave spreads to a wider area as compared to the region near the capeThe amplitude of the water wave near the bay is low and hence the water at the bay is

comparatively still


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Refraction of Light Waves

Figure 6.331. When a ray propagates from one medium to an optically denser medium, the ray refracts

(towards / away from) the normal.

2. The speed of light .(decreases/increases) as it propagates in the glass block,causing it to alter the direction of propagation.

Refraction of Sound Waves

Figure 6.34 (a) Figure 6.34 (b)

1. Sound waves travel faster in warm air than in .. (cold/hot) air.

2. On a hot day, the hot surface of the Earth causes the layer of air near the surface to be (warmer/cooler)

3. This causes sound waves to be . (reflected/refracted) away from the Earth.

4. On a cool night, the sound waves travel .(slower/faster)in the cooler layer of airnear the surface of the Earth than in the upper, warmer air. The waves are refracted

towards the Earth.Hence, sound can be heard over a longer distance on a cold .(night/day) comparedwith a hot day.

16

cold

warmer

refracted

slower

night

glass block

normal

towards

decreases

air


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6.4 ANALYSING DIFFRACTION OF WAVES

1. Diffraction of waves is ..........................................................................

2. The effect of diffraction is obvious only if(a) the size of the aperture or obstacle is ..small / large enough.(b) the wavelength is small /large enough.

3. Frequency, wavelength and of waves do not change4. The . of the waves are change

Exercise 6.4

Complete the diagrams below by drawing the wavefronts to show the diffraction of water waves.

From the diagram above,1. The .(narrower/wider) the gap, the more the waves spread out.2. When the gap is much wider than the wavelength of the waves, there is .

(little/more) diffraction.

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(b) Wider gap > (a) Narrow gap

(d) Narrow obstacle

(c)Wider obstacle

the spreading out of waves whenthey move through a gap or around an obstacle.

largesmall

speed

direction of propagation and the pattern

narrowerlittle


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Diffraction of light waves

Figure 6.40 Difraction of light through a single slit

Figure 6.41 (a) Figure 6.1(b)Narrow slit Wide slit

1. The diffraction of light waves occur when the light waves pass through a small slit orsmall pin hole.

2. The wider middle bright fringe shows that the light waves diffracted after passthrough a narrow slit.

3. If the slit becomes wider, diffraction pattern becomes .. (less/more) distint.

Diffraction of sound waves

Figure 6.42

1. A listener is requested to stand on the other side of the corner of the wall so that the radiois beyond his vision.

2. The listener is able to hear the sound of the radio although it is behind the wall.3. It is because the sound of the radio spreads around the corner of the wall due to diffraction

of sound.

18

Bright at the centre

less


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1. Interference is the superposition of two waves from two coherent sources meet

2. Two waves are in coherent if they are of the same .and ,

3. There are two types of interference :

Principle of superposition

1. The principle of superposition states that

2. Constructive interferenceoccurs when

3. Destructive interference occurs when

4. (a) Constructive Interference

+ =

Before superposition During superposition

Figure 6.50 (a)

+ =


Figure 6.50 (b)

19

6.5 ANALYSING INTERFERENCE WAVE

requency amplitude

(a) Constructive interference

(b) Desctructive interference

When two waves overlap, the resultant displacement is

equal to the sum of the displacements of the individual wave

the crests or troughs of both waves coincide to produce

a wave with maximum amplitude.

the crests of one wave coincide with the trough of

the other waves to produce a wave with zero amplitude


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When the vertical displacemet of two waves are in the same direction as in Figure 6.50(a) and (b), (constructive/destructive) interference occurs.

(b) Destructive Interference

+ =


Figure 6.51

If a wave with a positive displacement meets another with a negative displacement ofthe same magnitude, they cancel each other and the combined amplitude becomes

.Inteference of Water Waves

Figure 6.52: Inteference of water waves

1. At the points of constructive interference, the water moves up and down with a (large/ small) amplitude

2. The points of constructive interferences, is known as ..( nodes/antinodes)The points of destructive interferences is known as (nodes/antinodes).

20

constructive

large

antinodes

nodes

zero.


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Exercise 6.5

Figure 6.53

1. Figure 6.53 shows the interference pattern of water by two coherent sources of waterwaves.

(i) At point A, two crest meet resulting in .. interference.(ii) At point B, two through meet resulting in ..interference.(iii) At point C, a crest meet a trough resulting in interference.

2.. Complete the table below.

Before superposition Duringsuperposition

After superposition

21

constructive

constructivedestructive

A

B

C

3a

2a


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3. Categorize each labeled position as being a position where either constructive ordestructive interference occurs.

(a) Constructive interference:.(b) Destructive interference:..

.4. The diagram shows a full-scale interference pattern of a ripple tank experiment 2

seconds after vibration started. The crests of the water ripples are repsented by the darklines.(a) What is represented at P at this instant?

.

(b) What is the frequency of the vibrator?

5. Two dot vibrators whoch has a separation of 5.0 cm are placed on a water surface in

a ripple tank. The distance between two adjacent antinodes on a screen is 3.0 cm. Ifthe perpendicular distance between the dot vibrators and the screen is 10.0 cm, whatis the wavelength of the water waves?

Answer:

=0.10

0.30.5 x= 1.5 cm

22

A , B

C,D, E, F

P is between two crest. Therefore, P represent a trough

The diagram shows crests from each source. In 2 seconds, 4 crests are produced.

Thus the frequency is2

4= 2 Hz


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Inteference of Light Waves

1. Interference occur when at least two coherent light waves overlap or superpose

2. The superposition of 2 rays produces:(i) (ii)

3. When constructive occurs there will be a (bright/dark) fringeWhen destructive occurs there will be a (bright/dark) fringe

4. The wavelength of monochromatic light can be found by the formula:

a = ..x = .D = .

23

constructive interference

destructive interference.

bright

dark

is the distance between the slits

is the distance between the double-slit and the screen

is the distance between consecutive bright or dark fringes


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Exercise 6.51

1. In a Youngs double slit experiment, a light of wavelength 633 nm passes throughtwo slit which are 0.5 mm apart. Vertical fringes are observed on a screen placed4 m from the slits.

(i) Calculate the distance between two adjacent bright fringes.(ii) What will happen to the distance between two adjacent bright fringes if

a light of shorter wavelength is used?

Answer:(a) = 633 nm = 633 x 10-9

a = 0.5 mm = 0.5 x 10-3 m

D = 4 m

x =3

9

105.0

410633

x

xx

x = 5.064 x 10-3 mm

(b)

If is smaller, x will also be smaller. Therefore , the distance between twoadjacent bright fringes of light will smaller.

2. The wavelength of light can be determined with a double-slit plate. The diagramabove shows the pattern of interference fringes obtained in a Youngs double-slitexperiment. The separation distance of the two slit is 0.20 mm and the distancebetween the screen and the double-slit plate is 4.0 m.Calculate the wavelength of the light used in the experiment.

Answer:7x = 14 mmx = 2 mm = 2 x 10-3 m

D = 3.0 m

24

14 mm

a = 0.2 mm = 2 x 10-4 m

=4

102102 34 xxx

= 1 x 10-7m


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Inteference of Sound Waves

The wavelength of sound wave can be found by the formula:

a = ..x = .D = .

Exercise 6.52

1. Two loud speakers placed 2 m apart are connected to an audio signal generator that is

adjusted to produce sound waves of frequency 550 Hz. The detection of loud and softsounds as a person moves along a line is at 4.0 m from the loudspeakers.

Calculate the(a) wavelength(b) speedof the sound waves.

Answer: (a) a = 2 m

D = 4.0 m

X =0.4

2.12x = 0.6 m

(b) f = 550 Hz

V = f

= 550 x 0.6

= 330 m s-1

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distance between two loudspeakers

distance between two consecutive loud places

perpendicular distance between the speakers and

the loud


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6.6 Analysing Sound Waves

1. Sound waves are longitudinal wavces which require a medium for its propagation.

2. The loudness of the sound is depend to the

3. The pitch of the sound is depend to the ...

4. Echo is

Application of Sound Waves

1. Ultrasonic waves with frequencies above 20 kHz cannot be heard by human ear.2. The ultrasonic ruler is used to measure the distance between itself and a target.3. The depth of sea water can be calculated using the formula :

2d = v x t

Exercise 6.60

1. In an expedition to determine the depth of a freshwater lake using an ultrasonicruler, a pulse of ultrasonic sound is generated and travels to the bottom of thelake and reflected by it. The time taken by the pulse to travel to the bottom of thelake and return to the ruler is 0.35 s. If the speed of sound in freshwater is 1482

m s-1, calculate the depth of the lake.Answer:v = 1482 m s-1, t = 0.35 s

2 d = v x t

d =2

35.01482x= 259.35 m

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a phenomenon when a sound wave has been reflected off a

surface , and is heard after the original sound.

amplitude

frequency


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2. Fill in the blank.

6.7 Analysing Electromagnetic Waves

1. The electric and magnetic field vibrate ..(perpendicular/parallel) toeach other and to the direction of propagation.

2. Properties of electromagnetic waves (i)

(ii)(iii)(iv)

2. Sources and applications of electromagnetic waves in daily life

Type of em wave Source Application

Radio waves = 10-1 - 105 m

Electrical oscillating

circuit

(a) telecommunications

(b) broadcasting : tv and radio

transmission

Microwave = 10-3 10-1 m

Oscillating electrical

charge in a microwavetransmitter

(a) satellite transmissions

(b) radar(c) cooking

Infrared = 10-6 10-3 m

Hot bodies, the sun andfires

(a) night vision

(b) thermal imaging and physiotherapy(c) remote controls

Visible light = 10-7 m The sun, hot objects, light

bulbs, fluorescent tubes

(a) sight(b) photosynthesis in plants

(c) photography

Ultravioletradiation = 10-9 10-7 m

Very hot objects, the sun,

mercury vapor lamps

(a) identification of counterfeit notes

(b) production of vit-D

(c) Sentrilisation to destroy germs

X-ray = 10-11 10-9 m

x-ray tubes

(a) Radiotherapy

(b) Detection of cracks in building

structures

Gamma rays = 10-14 10-10

m

Radioactive subtances

(a) Cancer treatment

(b) Sterilisation of equipment

(c) Pest control in agriculture

Field ApplicationMedicine Sound waves of high energy are directed

to the kidney stones to destroy them in the

cavity of the kidney.Industry Sound waves of high energy are directed

to the kidney stones to destroy them in thecavity of the kidney.

27

Transverse waves

Do not require a medium to propagate and can travel in a vacuum

The waves travel at the speed of light, c = 3 x 108 ms-1

Undergo the same waves phenomenon : reflection, refraction, diffraction andinterference.

perpendicular


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Exercise 6.70

1. The diagram below shows the pattern of spectrum electromagnetic. In the boxesprovided, write the names of the parts given .

2. Table 6.70 shows electromagnetic waves P, Q, R and S and their uses

Electromagnetic waves UsesP Remote controlQ Radar system

R PhotographS Kill cancerous cells

Table 6.70Identify P, Q, R and S

Answer: P = Infra-red rays

Q = Microwaves

R = Visible lightS = Gamma rays

28

Gammaray

X-rayUltra

VioletVisiblelight

infrared microwave radiowave

wavelength

frequency

Radiowave microwave gamma ray x-ray visible ray infra-red ray ultraviolet ray


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Reinforcement Chapter 6

Part A : Objective Questions

1. Calculate the frequency of the given wave above

A. 8 Hz.B. 1/8 Hz

C. 4 Hz.D. HzE. 12 Hz

2. Which of the following is NOT a electromagnetic wave?

A. x-ray.B. Gamma rayC. water waveD. microwave.

3. Which of the following cannot travel through vacuum?

A. x-rayB. gamma rayC. sound waveD. light wave.

4. Based on the given diagram above calculate the wavelength.

A. 4 cmB. 8 cmC. 10 cmD. 15 cmE. 20 cm

29


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5. Wave length of given wave does NOT depend on.

A. velocity.B. FrequencyC. Amplitude

D. Period.

6. What is the phenomenon shown above?

A. reflection.B. Refraction.C. Interference.

7. A given wave travels at a speed of 4 x 105 ms-1. If the frequency of the wave is 1000 Hz,calculate the wavelength

A. 100 mB. 200 mC. 300 mD. 400 mE. 500 m

8. Which of the following is a common characteristic of visible light, ultraviolet rays,

infrared rays, gamma rays and X-rays?

A. Influenced by magnetic field.B. Possess the same frequency.C. Possess the same wavelength.D. Possess the same velocity in vacuum.

9. Which of the following electromagnetic waves are arranged in the order of increasingwavelength.

A. Visible light, X-rays. Radio waves. Infrared waves.

B. X-rays. Visible light. Infrared waves .Radio waves.C. Radio waves, X-rays. Visible light. Infrared waves.

10. Which of the following waves cannot travel through vacuum?A. Sound wavesB. X-raysC. Radio waves

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Part B: Structured Questions

Figure 1

1. Figure 1 shows the use of sonar equipment to measure the depth of the sea.(a) State the phenomenon of sound wave applied in sonar equipment.

.

[1 mark]

(b) Sound waves with frequency of 6.0 x 105

Hz are used to determine the depth ofthe sea. [speed of sound in sea water = 1500 m s-1]

(i) What is the wavelength of the sound wave in sea water?

=f

v= 5100.6

1500

x= 2.5 x 10-3 m

[2 marks]

(ii) If the time interval between the instant the sound wave is sent to theinstant the echo is received is 1.5 s, what is the depth of the sea

d =2vt

=2

5.11500x = 1125 m [2 marks]

2.

Figure 2

A ping pong ball is held with a string and placed near a loudspeaker as shownin Figure 2The loudspeaker emits a low frequency sound and the ping-pong ball seems tovibrate to it.

(a) Explain how the sound from the loudspeaker produces vibration on the pingpong ball.

31

Reflection


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The vibration of the loudspeaker moves the air molecules around it.The sound wave produced by the loudspeaker transfers energy to the ping

pong ball and causes it to move

(b) What will happen to the vibration of the ping pong ball if the loudness emits

(i) a louder sound? The ping pong ball vibrates with greater amplitude

(ii) a sound with a higher pitch? The ping pong ball vibrates with higher frequency.

(c) If the frequency of the sound is 50 Hz, calculate the wavelength of the soundproduced. [Assume the speed of the sound in air in 350 ms-1]

=f

v=

50

350= 7 m

2. (a) Figure 3 shows two sets of ripple tanks used to study diffraction of waves.

Figure 3.1 Figure 3.2

(i) What is meant by the diffraction of waves.Diffraction is the spreading out of waves when they move through a gap

or around an obstacle(ii) Which of the two sets will show a bigger effect of diffraction. Explain

your answer.The set shown in Figure 4.1 will show a bigger effect of diffraction. This is

because the narrower the gap, the more the waves spread out.

(iii) Redraw and complete Figure 3.1 and Figure 3.2 to show the between thetwo diffraction patterns.

(iv) State the characteristic to the diffracted waves in terms of waves in terms

of wavelength, frequency, speed and amplitude compared to the incidentwaves.(a) Wavelength remains unchanged

(b) Frequency remains unchanged

(c) Speed remains unchanged

(d) Amplitude becomes smaller compared to the incident waves

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Part C: Essay

1. Figure 3.0 shows the arrangement of the apparatus for an experiment to study a wavephenomenon.

Figure 3.0

Figure 3.1 shows the bright and dark bands of the waves pattern formed on the whitepaper when plane waves passes through the narrow and wide gaps.

Narrow gap Wide gapFigure 3.1(a) Figure 3.1(b)

(a) Name the wave phenomenon shown in figure 1.2.

Diffraction [1 mark]

(b) (i) Explain how the dark and light bands are formed on the white paper.[4 marks]

33

Figure / Crestsact as a convexlens

Figure / Troughsact as a concavelens

Figure / Every crests willconverge the light, bright

fringes are formed

Figure / Every troughs willdiverge the light, dark fringes are

formed

B D B D B D B


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(ii) Observe Figure 3.2(a) and (b). Compare the waves patterns and the wavelength ofthe waves before and after they pass through the gaps.Relate the size of the gaps, the waves patterns and the wavelengths to deduce arelevant physics concept.

[5 marks]a. In figure 3.1(a), size of the wavelength is the same as size of the gap // In figure

3.1(b), size of the gap is wider than size of the wavelength

b. In figure 3.1(a) and figure 3.1(b), size of the wavelengths are the same before andafter they pass through the gaps.

c. The waves emerging from the narrow gap in figure 3.1(a) is circular wavefrontsd. The waves emerging from the wider gap is plane wavefrontse. When a wave passes through a gap, the wave spreads.

The narrower the gap the greater the spreading of the wave

(c) Figure 1.3 shows the seashore of a fishing village. During the rainy season, wavesare big. One year the waves eroded the seashore, caused the jetty to collapse anddamaged the fishermens boats.

Figure 3.3

To prevent similar damage in the future, the fishermen suggest building retaining wallsand relocating the jetty.Make further suggestion of ways to help the fishermen solve their problems.You should use your knowledge of reflection, refraction and diffraction of waves toexplain these suggestions, to include the following aspects.(i) the design and structure of the retaining wall,(ii) the location of the new jetty(iii) the size or energy of the waves

[10 marks]Answer:

1. Build slanting barrier to reduce speed of wave2. Build the rough barrier to reduce reflection of the waves3. Build the new jetty at the bay because the water is calm at the bay

4. Build the small opening surround the bay because diffraction happens at the opening

5. Build the barrier at the bay because the wave is spread according to the shape of thebay

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physics:chapter 6 waves jpn pahang

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