CHAPTER 6: WAVES ANSWER 1. (a) (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 (e)
(f) (i) kinetic energy
(f)(ii) gravitational potential energy 4. Question 2: Kedah 07
(a) (i)
Zero error / systematic error (a) (ii) 24.0 s (b) (i) T = 1.20 s / 1.2 s (b) (ii) 0.3647 0.365 m / 36.5 cm 6.2 – 6.5 REFLECTION, REFRACTION, DIFFRACTION, INTERFERENCE
5. Question 7: Melaka Mid 08 (a)
(i)
The wavelength decreases (a)
(ii)
The depth of water decreases (a) (iii) Refraction (b) (i) 1 st : f = 5 . 1 2 2nd: = 1.33 Hz (b) (ii) 1.33 2 . 1 f v = 0.90 m (c) (i)
Relocate the jetty at the bay (c)
(ii)
The water at the bay area is calmer
The energy of the wave has spread out to wider area
2 2 1 2 142 3 4 10 ) . ( ) . (
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) (i)
Transverse wave // mechanical wave
(a) (ii)
The gap is bigger than the wavelength (b) smaller amplitude same wavelength (c) (i) made of concrete
because concrete is strong (c)
(ii)
make many holes on the wall 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
(d) λ = ax = 0.5 x10-3x 6 x 10-3
D 5
= 6 x 10-7m
11. Question 4: SBP 08
(a) Interference is the superposition of two waves originating from two coherent sources.
(b) (i)
(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 10-4x 1.5 x 10-1
3
= 6.25 x 10-7m (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
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,
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) (i) Diffraction (b) (ii) 1 Wavelength 2 Width separation 3.0 x 108= 50 x 106x // 3.0 x 108= 500 x 106x // 3.0 x 108= 5 x 109x 1 6 m 2 0.6 m // 60 cm 3 0.06 m // 6 cm (d) 1 50 MHz 2 wavelength > width separation (e) 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
λ λ
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
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
24. Qualitative problem: Teknik 07 Loudspeakers are positioned at quite a distance away. so that the distance between consecutive constructive or destructive interference is smaller.
The two main loudspeakers are not positioned opposite to each other . To prevent multiple reflections Fix softboards/ materials which are sound absorbers Reflection effects can be reduced
Use (thick) carpets or wooden floor Reduce unnecessary reflection of the floor. Assemble a higher power speaker system To produce a louder or clearer sound.
25. Making Decision: Melaka Mid 08 The wall barrier
location is built a few metre in front the entrance
so the water wave will be reflected.
Water depth is high
so bigger ship can go through
Gap between wall barriers is small
so diffraction is more obvious and the energy can be reduced Base of the wall barrier is larger so it can withstand high pressure at the bottom.
chose model V 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
sound-proof 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
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 Reason Large diameter receives more
signal The distance of
signal receiver from the centre of the parabolic disc is same as the focal length
Radar gives out parallel beam//signals focused to the receiver Use microwave wave High energy Short wavelength Easily reflected
High frequency High energy / can travel at longer distance
The position of the parabolic disc is high
The signal is not blocked //much coverage //can detect signal Strong material 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
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 Explanation Shape of the boat
is streamline To reduce the water resistance / drag Material used is strong Can withstands high water pressure Uses ultrasonic waves
Have high energy / sounds can travel at further distance Put fish in a
polisterine box containing ice
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
Made of fiber glass / less dense
material