DHS 13 Wave Motion (Tutorial Solutions)

Topic 11 : Wave Motion Tutorial Solutions (2025) 1. (i) Displacement: The distance in a specific direction from the equilibrium position of the simple harmonic motion. (ii) Amplitude: Maximum displacement from the equilibrium point in either direction. (iii) Wavelength: The shortest distance between two points on a progressive w a v e which are vibrating in phase. (iv) Frequency: Number of oscillations per unit time of a particle in a wave. (v) Period: Time taken for a particle in the wave to undergo one complete oscillation. (vi) Phase difference: A measure of how much one wave is out of step with another. It is measured in either degrees or radians. (b) One wavelength  is equal to the distance travelled by the wave energy during one complete oscillation of the source. The distance travelled by the wave energy during n complete oscillation of the source = n Speed is the distance travelled pe r unit time by the wave energy i . e . v = t n Frequency is numbe r of oscillations per unit time i.e. f = t n Therefore speed v = f (c) Progressive wave: A progressive wave transfers energy from one point to another by means of oscillations. Intensity of a wave: Wave energy incident per unit time per unit area normal to the direction of energy transfer.

(d) Transverse wave: Waves in which the oscillation of the particles in the wave is perpendicular to the direction of transfer of energy of the wave. Longitudinal wave: Oscillations of the particles in the wave are along the direction of transfer of energy of the wave. 2. (a)  = T 1 = 100 Hz, T = 0.01 s (b)  = T 1 = 200 Hz, T = 0.005 s 0.005 0.0075 0.01 0.01 y/m Time / s 0 0.0025 - 0.01 0.005 (a) (b) - 0.005

(c) For wave 1, v = f λ λ = 330/100 = 3.3 m For wave 2, v = f λ λ = 330/200 = 1.65 m (d) Sound wave. Horizontal displacement-distance graph. 3. v = f λ  vf  Minimum f = Hz14 9 8 103.410700 100.3    (red) Maximum f = Hz14 9 8 105.710400 100.3    (violet) 4. Answer: 50 mm outwards Diameter of bowl = 400 mm Radius of bowl = 200 mm Since the wave travel 250 mm in 1 second, it will travel 200 mm inward, then the remaining 50 mm outwards. 5. P gives the period T of the motion and Q the wavelength λ. v = f λ => P Q Tv   1.65 2.475 3.30 0.01 y/m Distance/ m 0 0.825 - 0.01 0.005 (a) (b) -0.005

6. Draw a graph moments after. 7. (a) False. Speed at P is a minimum (zero) (b) False. This graph is not that of a stationary wave. It will follow an oscillatory motion. (c) False. Energy of R is entirely potential energy. (d) False. Energy of S is entirely kinetic energy. 8. (a) Ans: C I 2 1 r By similar triangle, r is 8 times, I will be 64 1 I 2A Since I is 64 1 , then A is 8 1 (b) Ans: D I rh P A P  2 where h is the depth of pond I r 1 I 2A A 2 r 1 A r 1 Note: For a pond, it is a 2D distribution of wave energy, thus the usual 3D distribution I 2 1 r cannot be used. h r

9. Energy is spread over a longer wavefront. Intensity drops. Since Intensity is proportional to square of amplitude, the amplitude drops too. 10. (0.25I, f, 0.50) as the frequency is constant as the wave travels from one medium to another. 11. 4 0 125 m21 0 0 x x..      12. Let Δx be the path difference between the two points 5 m apart along the wall.    x 2 rad  5.010 5.2 2  13. (a) (i) Period, T = 1.25 ms = 1.25 x 103 s  = T 1 = 800 Hz Wavelength, λ = 0.40 m Velocity, v = f λ = 800 x 0.40 = 320 ms1 (ii)    x 2 rad . . 2 924 9 40 90 2   rad) 2 and rad 0between to(round  rad2 (iii) 45.0 2  mm mm QatAmplitude PatAmplitude Δx 5 m mxx 5.2530sin  30°

(iv)   1642 2 2  Q p A A QatIntensity PatIntensity (b) With the speed being 320ms 1, it is probably a sound wave (c) (i) A microphone and a suitably adjusted Cathode Ray Oscilloscope (CRO) placed at position P could detect the 800 Hz 320 ms 1 sound waves from an appropriate source and produce a display similar to that graph. (ii) The same method cannot be used directly to obtain the 2 nd graph. However, we can use several microphones and place them at several specified positions, at various distances x from the source. These are connected the CRO and provide the displacement y of the point it is placed at, at the time t = 0 s. By manually plotting displacement y against distance x and joining these points in a curve, we can get the graph as seen. 14. (a) (b) (c) 2  rad out of phase d p x x Patm + ΔP Patm – ΔP Patm

15. For the intensity of the emergent light to be a minimum (zero), the polarizing axis of P3 should be perpendicular to P2. Hence, θ = 30° + 90° = 120° or θ = 30° + 270° = 300° 16. (a) (i) Monochromatic: only one wavelength (color) (ii) Plane-polarized: If the oscillations in a wave are confined to one direction only, in a plane normal to the direction of transfer of energy of the wave, the wave is said to be polarised in that direction. (b) s = r θ s = (6.0)(1.2 x 10 -3) = 7.2 x 10-3 m (c) Rotate a Polaroid between the beam and a receiver through 180°. If there is an angle where the receiver detects no laser beam, the beam is plane polarized. 17. (a) λ = f c = 7 8 100.9 100.3   = 3.33 m d = 2  = 1.67 m (b) rate = distance internodal detector of speed = 67.1 10 = 6 nodes per second 18. Let I1, I2 and I3 be the intensity after passing through polarisers 1, 2, and 3 respectively. I2 = I1cos245o I3 = I2cos245o = I1cos445o  0