JPJC 2026 Wave Motion Assignment Solutions

2026/JPJC/PHYSICS/9478 1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS WAVE MOTION ASSIGNMENT SOLUTIONS 1 (a) (i) From Fig. 1.2, 0.40 m = (ii) From Fig. 1.1, 2T = 2.5 ms ➔ T = 1.25 ms, hence 3 11 800 Hz 1.25 10 f T −= = =  (iii) 1800 0.40 320 m svf  −= =  = (iv) 0.90 92 2 14.1 rad0.40 2x    =  =  = = For phase difference, w e can always subtract integer multiples of 2 π, so the answer can be given as 2  rad, i.e. 1.57 rad as well. (v) From Fig. 1.2, the amplitude at Q is approximately 0.5 mm. Since the amplitude at P is 2.0 mm (from both Fig. 1.1 and Fig. 1.2), the ratio is 4.0. (vi) Recall that 2AI . Thus, 2 PP 2 Q Q A A=II = 16 (b) From the answer to (a)(iii), speed of the wave is approximately that of sound wave in air (about 330 m s−1). Hence the wave described by Fig. 1.1 and 1.2 is a sound wave. 2 (a) (i) Comparing with the equation for the displacement of particle undergoing SHM, 0 sin 200 2 200x x t f     =  =  = Therefore f = 100 Hz (ii) 335 3.35 m100vf = = = Number of wavelengths 500.00 149.25 1493.35= =  (iii) Phase difference 0.25 2 1.57 rad2=  = = Notice that we used the formula Δ = 2x   , where x   is the number of wavelengths found in (a)(ii). We only need to use the decimal portion because complete wavelengths give an integer multiple of 2 , which can be subtracted away. (b) Recall that 2 1 rI for a spreading wave. Thus, 2 2 new old 22 old new 500 (500 300) r r== −II 2 62 new old 5 6.3 10 W m2 −− = = II

2026/JPJC/PHYSICS/9478 2 3 (a) (i) A (ii) I (iii) 2AI [For (i) and (ii), because the polarising axis of Q is parallel to P.] (b) angle  amplitude intensity 180 A I 90 0 0 60 0.5A 0.25I