RI Y6 Remedial Oscillations and Wave Soln
Uploaded by blahblahblah03 · 22 November 2025
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Raffles Institution Year 5-6 Physics Department 1 2025 Year 6 H2 Physics Remedial Suggested Solutions Chapter 10: Oscillations 1 C Use amax = ω2xo ; where ω2 is the gradient of graph and ω = 2π / T 2 D 3 (a) 48.00 cm = 0.4800 m (half smallest square 2 d.p. for cm) (b) k = 0 42 0 00 9 81 0 7000 0 4800natural force mg ( . . )( . ) extension x l ( . . ) −= =−− = 18.7 N m−1 (c) 122 8 38 rad s0 75 .T. ππω −= = = (d) (i) oscillations in the absence of an external force / with no loss or gain in total energy / constant amplitude (ii) 2 22 18 7 0 266 kg 8 38 ky ma kay m k m k.m. . ω ω −= =− = = = = (iii) ( )( )0 266 9 81 0 14m 14 0 cm18 7 0 140 0 4800 0 620 m ee e natural ..mg kx x . . . xx l . . . = ⇒= = = = +=+ = OR read from m1 = 0.26 (nearest half square) x = 0.615 to 0.625 m 4 (a) From 2nd graph, v0 = ωx0 = 2 T π x0 T = 0 0 2 x v π = 2 0.45 1.2 π × = 2.4 s (b) From 1st graph, Fmax = mω2x0 m = max 2 0 2 F xT π = 2 0.63 2 0.452.36 π = 0.20 kg
Raffles Institution Year 5-6 Physics Department 2 Chapter 11: Wave Motion 1 A In 2 s, the wave travels 1 cm speed = 1 cm / 2 s = 0.5 cm s−1 2 B 3 A Option A is the answer as when θ is 0°, the transmission axes are perpendicular to one another and hence no light passes through. When θ is 90°, the axes are parallel to one another and maximum intensity of light passes through. In between these two angles, the intensity transmitted increases. 4 A For Alice to hear the music as loud as before, the intensity of the sound reaching her ears should remain the same. ( ) ( ) 22 0.75 4 15 4 13 m PP x x ππ = = 5 (a) Same frequency Same period Same type Different amplitude / phase (b) Yes. They have constant phase difference and are in anti-phase (phase difference of π rad). (c) (i) 𝐼𝐼 ∝ 𝐴𝐴2 𝐼𝐼𝐵𝐵 𝐼𝐼𝐴𝐴 = 𝐴𝐴𝐵𝐵 2 𝐴𝐴𝐴𝐴 2 = 22 32 𝐼𝐼𝐵𝐵 = 4 9 I = 0.44 I (ii) 𝐼𝐼 ∝ 𝐴𝐴2 Since the two waves are in anti-phase, they will undergo destructive interference, Resultant amplitude = 3 − 2 = 1 unit. 𝐼𝐼𝑅𝑅 𝐼𝐼𝐴𝐴 = 𝐴𝐴𝑅𝑅 2 𝐴𝐴𝐴𝐴 2 = 12 32 𝐼𝐼𝑅𝑅 = 1 9 𝐼𝐼 = 0.11 𝐼𝐼
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