JPJC 2026 Wave Motion Tutorial Solutions

2026/JPJC/PHYSICS/9478 1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS WAVE MOTION TUTORIAL SOLUTIONS Self-Check Questions S1 A progressive wave is one where the wave profile moves along the direction of propagation and energy is transferred from the source to other points further away without matter being transferred. S2 Displacement of a particle on a wave is the distance it travelled from its equilibrium position in a specified direction. Amplitude of a wave is the magnitude of the maximum displacement of an oscillating particle in a wave from its equilibrium position. Period of a wave is the time taken for a particle of the wave to complete one oscillation. Frequency of a wave the number of oscillations made by a particle of the wave per unit time. Wavelength of a wave is the distance between two consecutive points on the wave which are in phase, such as two successive crests. Speed of a wave is the distance travelled by the wave per unit time. S3 Phase difference between two particles in a wave is a fraction of a complete oscillation by which one is ahead of the other. It is usually expressed as an angle in radians. S4 A wavefront is a line or surface joining points on a wave that are in phase. The wave travels in a direction that is perpendicular to the wavefront. S5 For a wave that travels with a constant speed, the distance travelled by the wave is the wavelength  when the time taken is one period T. Since speed is defined as the distance travelled per unit time , the velocity of the wave, v is 1 (since ) v T ff T   = == S6 (a) The intensity I of a wave is defined as the power transferred (radiated) by a wave per unit area perpendicular to the direction of wave motion. (b) Intensity is directly proportional to the square of the amplitude of the wave. (c) Intensity is inversely proportional to the square of the distance from a point source. S7 In a transverse wave, the oscillations are perpendicular to the direction of propagation of the wave , while in a longitudinal wave, the oscillations are parallel to the direction of propagation of the wave wave travel. S8 Polarisation is a phenomenon whereby the oscillations of transverse waves are restricted to a single plane. Transverse waves can be polarised because the oscillations of the particles are perpendicular to the direction of wave propagation. Longitudinal waves cannot be polarised because the oscillations of the particles are parallel to the direction of wave propagation.

2026/JPJC/PHYSICS/9478 2 Self-Practice Questions 1 From displacement-time graph, p is the period of the wave. From displacement-distance graph, q is the wavelength of the wave. Speed of wave qvf Tp= = = Answer: C 2 From Fig. 2.1,  = 1.8 m From Fig. 2.2, 3.5T = 20 ms, hence T = 5.71 ms Hence 1 3 1.8 315 320 m s 5.71 10 vf T −−= = = =   3 Option A: Incorrect. Point P is at amplitude position and is momentarily at rest. Option B : Incorrect. Point R is at amplitude position and is momentarily at rest. The energy of point R is entirely potential. Option C: Correct. Point T is at amplitude position and is experiencing maximum acceleration directed towards its equilibrium position. Option D: Incorrect. Every point in the wave is undergoing simple harmonic motion and is not always having zero displacement. Answer: C 4 Option A: Incorrect. Length PT is the wavelength of the wave because this is a displacement-distance graph. Option B: Incorrect. Particles Q and S are in anti -phase, moving in opposite directions and cannot have the same velocity. Option C: Incorrect. Particles R and T are in anti-phase, moving in opposite directions. Option D: Incorrect. The amplitude of the wave is the vertical separations PQ, QR, RS, ST. Option E: Correct. Length PT is the wavelength. Length QS is half a wavelength. Answer E 5 Just draw the wave shape a split second later. Answer: B 6 Using cf = ,  − = =  = max min 8 9 14 3.00 10 400 10 7.50 10 Hz cf 7 Since 2AI , the new intensity of the transmitted wave is proportional to ( ) 2 22cos , i.e. cosaa  . Answer: D

2026/JPJC/PHYSICS/9478 3 8 Using 2AI and intensity is rate of energy flow per unit perpendicular area, 2EA , 2 energy of transmitted light amplitude of trans mitted light energy of incident light amplitude of incident light =  2 amplitude of transmitted light 0.70amplitude of incident light = amplitude of transmitted light 0.70 0.84amplitude of incident light = = Answer: C Discussion Questions 1 The wavelength is given by 8 spacings in Fig. 3.1. For the displacement at Q to be zero, the wave shape needs to travel 1 spacing, which is a distance of 8  . Waves move at constant speed unless the medium is non-uniform. The time for a wave to travel  is the period T. Thus, answer is 8 T , i.e. 11 0.0100 s12.5 8= . 2 Since P and Q are a distance 3 2x = apart, (2 ) 3x   = = . This means that P and Q are in antiphase. If P is displaced upwards and moving downwards, Q is doing exactly the opposite: displaced downwards and moving upwards. Answer is C. 3 In 1.0 s, the pulse would have moved 250 mm. As the pulse moves towards the centre of the bowl, the circle wavefront narrows towards the centre with decreasing radius . Waves do not collide; they simple pass through one another. With reference to the figure below, the direction of wave travel will be in a straight line from one edge of the bowl to the opposite edge. This means that after the pulse passes through the centre of the bowl, the wavefront will spread out, moving towards the sides of the bowl with increasing radius . The radius of the bowl is 200 mm, hence in 1 .0 s, the radius of the pulse will be 50 mm and it will be moving outwards.

2026/JPJC/PHYSICS/9478 4 4 For a fixed source of sound that radiates energy uniformly in all directions, 2AI and 2 1 r I . Combining above relationships, 1A r , 21 12 7 2 8 3.0 1.0 105.0 6.0 10 m Ar Ar A − − = =   = 5 The loudness depends on the intensity of the wave at the receiver. If the intensity at the source is halved, and if the person continues to stand in the same place, the intensity at the receiver will also be halved. To have the same intensity at the ear as before, the person has to move. The person would have to move closer to the source in order for the loudness to increase. At 8.0 m from the source, the intensity is half the desired intensity. Since the wave spreads out, we can use 2 1 r I . So we see that r2 must be halved of the original to double the intensity at the ear so as to get the desired intensity. So ( ) 22 1 8.02r = , which gives 4 2 5.7 m (2 s.f.)r == 6 Using Area P=I , ( ) rx 4 2 6 Area 3.0 0.80 10 4 2.0 4.77 10 W P  − − = =    = I Energy received by microphone in 5.0 s −= =   = 64.77 10 5.0 24 JE Pt 7 (a) Microwaves are transverse waves. They are electromagnetic waves, which are transverse. Microwaves can be polarised by metal grilles . Since longitudinal waves cannot be polarised, microwaves must be transverse in nature. (b) Speed of microwaves 813.00 10 m s −= , since microwaves are EM-waves. Time taken for a signal to travel from the satellite to the Earth 12 4 8 distance 4.35 10 1.45 10 sspeed 3.00 10= = =   (c) If the power is radiated uniformly in all directions, it is spread out over a sphere. The fraction of the total power received by the dish will be the ratio of the area of the dish to the area of the sphere, ( ) 23 rx 212 260 22.0 2.41 10 W 4 4.35 10 P  −=  =  

2026/JPJC/PHYSICS/9478 5 (d) The answer in (c) was obtained by assuming that the microwaves are radiated uniformly in all directions. Instead of transmitting over all-space, the transmitter could have been aimed at the d