1. B 2. A 3. B 4. C

Find the relationship between the amplitude of the wave and the distance from the source.

Therefore . . The amplitude is inversely proportional to the distance. Hence, when the amplitude is twice (A to

2A), the distance is halved (4d to 2d)

5. C first resonance : 0.21 = (1/4) wavelength Next resonance: (3/4) wavelength = (3/4)( 4 x 0.21) = 0.63

6. C λa = 2L ; fa = v / λa = v / 2L

λb = 4L ; fb = v / λb = v / 4L λc = 2L ; fc = v / λc = v / 2L fa : fb : fc = 2:1:2

7.

(a) The vibration is along the motion of the waves. Hence cannot define the plane of polarization (b) D = 1 ½ wavelength (c) The person detect higher number of compressions ( rarefactions) per unit time, hence frequency is higher.

8. C

The angular differences between the polarising axes are as follows.

Malus’ law for the polarised light emerging from a pair of polarisers is given as

where I0: intensity of the unpolarised incident light. I : intensity of the polarised emerging light q : the angle between the polarising axis or directions.

Thus, in decreasing order of intensity for the emergent light, the pairs are X, Z, W and Y.

9. A

At the open end of the pipe is a displacement antinode (motion of air particles is maximum) and a pressure node (changes in air pressure is minimum). At the closed end of the pipe is a displacement node (motion of air particles is minimum) and a pressure antinode (changes in air pressure is maximum).

10. B Refer to the diagram of the Electromagnetic Spectrum found in the Wave lecture notes. 11. D The shape of the wave pulse is required.

The answer is the reflection of the displacement-time graph about the vertical axis. Imagine standing at P and visualizing the wave pulse coming towards you. How does the vertical displacement change with time?

12. C Anti-nodes must be found at the open-end of the tube.

The air molecules must be either moving away or moving towards a node. (Why?) Possible answers are II and IV.

13. B 14. C UV radiation has a smaller wavelength than Radiowaves. 15. D At equilibrium points, velocity is maximum. At amplitude, velocity is zero.

P is on its way upwards, hence velocity is positive.

Q is on its way downwards, hence velocity is negative. 16.

17. 18. 19. (a) Either radio waves or microwaves.

(b) The path length of ABC and ADC are the same. The path difference of the 2 waves meeting at C is 0, resulting in constructive interference at C. Thus, the amplitudes add up.

(c) The waves reach C by 2 paths – via DABC and DC. Path length of DABC = 20 + 9 + 20

= 49 cm

Path length of DC = 9 cm

Path difference = 49 – 9

= 40 cm = 2.5 λ

Since the path difference is )2

1( n , the waves meet out of phase at C. Destructive

interference occurs, and no signal / minimum signal is detected at C .

20. B

We need to first assume that the wave is progressing from the right to the left. By drawing the wave profile at the next instant, it can then be verified that the particles are moving in the same directions as the corresponding particles shown in the question. Hence the assumption is correct and it can then be deduced that energy is being transferred from the right to the left.

21. A 22. D 23. C 24. B 25. C 26. 27. 28. A 29. B 30. B