answer (chapter wave)

CHAPTER 6: WAVES ANSWER

(e)

1. (a)

(f) (i) kinetic energy (f)(ii) gravitational potential energy

(b) water waves (c) Amplitude = 12 cm Wavelength = 16 cm (d) speed = 25 x 16 = 400 cm/s = 4 m/s 2. (a) Vibrations of the air particles is parallel to the direction of propagation of sound waves. (b)

(c) (i) 12 cm (c)(ii) λ = 150/3 = 50 cm (d) f = v/λ = 330/0.5 = 660 Hz 3. (a) (i) B (a)(ii) ABCCBA (b) f = 5 s/20 = 0.25 Hz (c) frequency decreases (d) damping occurs

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4. Question 2: Kedah 07 (a) Zero error / systematic error (i) (a) 24.0 s (ii) (b) T = 1.20 s / 1.2 s (i) (b) 10 (1.2) 2 (ii) 2 4(3.142)

0.3647 0.365 m / 36.5 cm 6.2 – 6.5 REFLECTION, REFRACTION, DIFFRACTION, INTERFERENCE 5. Question 7: Melaka Mid 08 (a) The wavelength decreases (i) (a) The depth of water decreases (ii) (a) Refraction (iii) (b) 2 1st : f = (i) 1.5 2nd : = 1.33 Hz (b) v 1.2   (ii) f 1.33  (c) (i) (c) (ii)

 = 0.90 m Relocate the jetty at the bay  The water at the bay area is calmer  The energy of the wave has spread out to wider area

6. Question 4: Trengganu 07 (a) v = 5 x 0.8 = 4 ms -1 (b) The wave are circular Wavelength remains unchanged (c) Diffraction (d) Waves pass through an aperture or round a small obstacle Waves spread out.

7. Question 5 (a) Diffraction (b) wavelength : unchanged Wave pattern circular (c)(i) amplitude decreases (c)(ii) energy of the wave is spread out after passing through the gap. (d)(i) less diffraction (d)(ii)

8. Question 6: Kedah 08 (a)(i) Wave diffracts / spreads to a bigger area (a)(ii) Amplitude of the waves decrease Energy decreases (a)(iii) Diffraction (a)(iv) Unchanged (b)(i) Vibrates/oscillates (b)(ii) Stationary (c) Reflection

9. Question 7: Melaka 08 (a) Transverse wave // mechanical (i) wave (a) The gap is bigger than the (ii) wavelength (b)  smaller amplitude  same wavelength (c)  made of concrete (i)  because concrete is strong (c)  make many holes on the wall (ii)  diffraction // spread of energy (d)  The water is shallow  load and unload by using small boats

10. Question 3: Perak 07 (a) electromagnetic waves / transverse waves (b) Constructive interference takes place and bright fringes are observed. Destructive interference takes place and dark fringes are observed. (c) Lights with one colour or one wavelength 10-3 x 6 x 10-3 (d) λ = ax = 0.5 x D 5 = 6 x 10-7 m

11. Question 4: SBP 08 (a) Interference is the superposition of two waves originating from two coherent sources. (b) (i)

(b)(ii) Constructive Interference

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(b)(iii) The cork will move up to to highest level (c)(i) λ = ax = 5 x 3 = 1.5 cm D 10 (c)(ii) decreases.

12. Question 3: Trengganu 09 (a) one color/wavelength (b) constructive interference / crest meet crest /trough meet trough (c) 5 x 10-4 x 1.5 x 10-1 3 = 6.25 x 10-7 m (d)(i) decrease (d)(ii) wavelength of red > green

14. Question 6: Kelantan 09 (a) Coherence source have same frequency and same phase different (b)(i) The distance between two slits in Diagram 6.2 is larger. (b)(ii) The distance between consecutive bright fringes in Diagram 6.2 is smaller (b)(iii) The distance between the double slit and the screen is equal. (c)(i) The larger the distance between the two slits,a, the smaller the distance between consecutive bright fringes, x / a inversely proportional to x (c)(i) Interference (d) When crests meet crest, constructive interference occurred – bright fringes When crest meet trough destructive interference occurred – dark fringes

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6.6 SOUND WAVES 15. Question 4: SBP 07 (a) Reflection of wave (b)  The incident ultrasonic waves is reflected by the big rock to form reflected ultrasonic waves.  The incident angle is equal to reflected angle (c)(i) d = vt = 1560 x 1.5 2 2 = 1170 m (c)(ii)  Distance between two pulses is 3 cm  Amplitude is smaller

16. Question 7: Johor 07 7(a)(i) Reflection (ii) 1 000 Hz (iii) Less (iv) Less energy (b) The building reflects soun The interval time is shorter (c)(i)

5 cm x 1 ms/cm 5 ms = 0.005

(c)(ii)

Show on the graph From graph, T = 0.005 s, d = 1.7 m

17. Question 7: Melaka 09 (a)(i) Reflection (a)(ii) Equal (b) It has high frequency / high energy (c) s = vt = 1500 x 0.05 2 2 s = 37.5 m (d)(i)  At a higher place  Easier to receive the signal (d)(ii) Microwave (d)(iii)  Increase the diameter of the device  Receive more signals 18. Question 7: MRSM 07 (a)(i) direction / angle i = r Wavelength unchanged (a)(ii) No change (b)(i) By using sonar (b)(ii)  It has high frequency and can travel further  It cannot be heard by human ears (b)(iii)  A pulse of ultrasonic sound is transmitted  The fish will reflect the pulse to the receiver (b)(iv) s = vt = 1200 x 2.4 2 2 = 14,440 m 19. Question 8: Kedah 07 (a) Electromagnetic waves/transverse waves (b) Diffraction (i) (b) 1 Wavelength (ii) 2 Width separation 3.0 x 108 = 50 x 106 x λ // 3.0 x 108 = 500 x 106 x λ // 3.0 x 108 = 5 x 109 x λ 1 6m 2 0.6 m // 60 cm 3 0.06 m // 6 cm FIZIKMOZAC 2010

(d)

(e)

1 50 MHz 2 wavelength > width separation Microwave Higher frequency//not easy to be diffraction//higher penetrating power

6.7 ELECTROMAGNETIC WAVE 20. Question 6: Melaka 09 (a) Spectrum (b)(i) Frequency – Increases Wavelength – decreases (b)(ii) same / equal (c)(i) The higher the frequency the higher the energy (c)(ii) v = fλ (d) High frequency means high energy Able to kill cancerous cell

21. Conceptual: Teknik 07 (i) Waves that propagate in the opposite direction after they encounter a reflector (ii) - the angle of reflection = the angle of incidence - wavelength remain unchanged - frequency remain unchanged - speed remain unchanged - the direction of the reflected waves changes (iii) The angle of reflection = the angle of incidence

22. Understanding: Teknik 07 o untrasound being transmitted to the sea bed o a receiver will then detect the reflected the reflected pulses o the time taken by the pulse to travel to the seabed and return to the receiver being recorded, t o the depth of the sea can be calculated using the formula, s = v/ t/2

Assemble a higher power speaker system

23. Understanding: Melaka Mid 08 (a) Amplitude (b)  the energy of the sound is transferred to glass  the glass start vibrating with the same frequencies with the sound waves  Resonance occurs , the amplitude is at its maximum value  Increasing energy may cause the glass to break

Water depth is high Gap between wall barriers is small

24. Qualitative problem: Teknik 07 Loudspeakers are so that the positioned at quite distance between a distance away. consecutive constructive or destructive interference is smaller. The two main To prevent loudspeakers are multiple not positioned reflections opposite to each other . Fix softboards/ Reflection effects materials which can be reduced are sound absorbers Use (thick) carpets Reduce or wooden floor unnecessary reflection of the floor. FIZIKMOZAC 2010

To produce a louder or clearer sound.

25. Making Decision: Melaka Mid 08 The wall barrier so the water wave will location is built be reflected. a few metre in front the entrance

Base of the wall barrier is larger chose model V

so bigger ship can go through so diffraction is more obvious and the energy can be reduced so it can withstand high pressure at the bottom. because the wall barrier location is built a few metre in front the entrance, water depth is high, gap between wall barriers is small, and base of the wall barrier is large

26. Quantitative problem: Melaka Mid 08 v = 2s = 2 x 120 t 0.16 = 240/0.16 = 1500 ms-1 λ = 1500 66000 = 0.023 m

27. Section B: Mara 08 (a) Light of one frequency / wavelength/ colour (b)  In figure 10.1(b) distance between the double slit and screen, D is larger  In figure 10.1(b), distance between two successive bright fringes, x is smaller  In figure 10.1(b) there is more fringes  In figure 10.1(b), the width of fringes is smaller // fringes are narrower.  When the distance between the double slit and screen, D increases, the distance between two successive fringes, x increases. (c)  Build a dome-shaped or circular roof  It improves the acoustic effect of sound  Reduce the number of gaps / holes/ opening like doors and windows  Reduce effect of diffraction.  Build the walls from soundproof materials  To avoid disturbance from outside  Use soft materials/fabrics / cushion for the chairs  To absorb the sound / to avoid reflection of sound  Hang curtains on the walls / put carpets on the floors  To absorb the sound / to avoid reflection of sound  Place the loud speakers far away from each other / put at the corners of the hall  To produce more constructive interference.  Increases the number of lamps or lights  To get sufficient amount of light FIZIKMOZAC 2010

28. Section B: Melaka Mid 09 (a) Sound wave is a longitudinal waves (b)(i)  The diameter of guitar string in Diagram 10.1 is smaller than diameter of the string in Diagram 10.2  The amplitude of the wave in Diagram 10.1 is the same as in Diagram 10.2  Number of oscillations in Diagram 10.1 is higher than in Diagram 10.2 (b)(ii) When the diameter of the string increases , the frequency of the wave decreases (b)(iii) The higher the frequency , the higher the pitch of the sound (c)  When a tuning fork vibrates, air molecules will vibrate.  When the tuning fork moves forwards, the air is compressed.  When the tuning fork moves backwards, the air layers are pulled apart and cause a rarefraction.  Therefore, a series of compression and rarefactions will produce sound.  The sound energy is propagated through the air around it in the form of waves. (d) Modification Large diameter The distance of signal receiver from the centre of the parabolic disc is same as the focal length Use microwave wave Short wavelength

Reason receives more signal Radar gives out parallel beam//signals focused to the receiver High energy Easily reflected

High frequency

The position of the parabolic disc is high Strong material

High energy / can travel at longer distance The signal is not blocked //much coverage //can detect signal Not easily broken

29. Section B: Kedah 09 (a)(i) A wave in which the vibration of particles in the medium is parallel to the direction of propagation of the wave. (a)(ii)  When a tuning fork vibrates, layers of air vibrate  The particles of air undergo series of compression and rarefaction  Sound energy is propagated through the air in the form of waves. (b)  The amplitude in Diagram 10.2 is higher  The peak value in Diagram 10.2 is higher  The higher the amplitude, the higher the peak value  The higher the peak value, the louder is the sound  The higher the amplitude , the louder the sound. (c)(i)  The boat use high frequency sound waves  Smaller wavelength, less diffracted  High energy waves  High penetrating power (c)(ii) list of equipment:  Transmitter  Receiver  Sound generator  CRO (c)(iii) Method of measuring the time FIZIKMOZAC 2010

 The time taken, t from the transmitter to the receiver is recorded by the CRO  The speed of the sound wave, v in water is given  The depth of the sea, d = vt 2 30. Section B: MRSM 09 (a) Constructive interference occurs when crest meet another crest or trough meet another trough. (b)  Distance between two coherent sources in Diagram 10.2 is greater.  Vertical distance between the point P and respective coherence source in both diagrams are equal  Wavelength of the propagating water wave in both diagrams are equal  Distance between two consecutive antinodes in Diagram 10.2 is smaller.  The greater the distance between two coherence source, the smaller the distance between two consecutive antinodes. (d) The son cannot hear her mother’s voice clearly because:  During lunch time (afternoon) the air near to the surface of earth is hotter.  Hot air is less dense than cool air  Sound will travel from less dense medium to denser medium  The sound will be refracted towards normal  And it will move away from the surface of the earth.

Modification Shape of the boat is streamline Material used is strong Uses ultrasonic waves Put fish in a polisterine box containing ice

Made of fiber glass / less dense material

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Explanation To reduce the water resistance / drag Can withstands high water pressure Have high energy / sounds can travel at further distance ice has larger latent heat / ice can absorb a large quantity of heat from fish as it melts / fish can be kept at a low temperature for an extended period of time The boat is lighter