2017 RI H2 Physics P3 solutions

2017 Year 6 H2 Physics Prelim Paper 3 Mark Scheme © Raffles Institution 9749/03 1 (a) Simple harmonic motion is defined as the motion of a particle a bout a fixed point such that its acceleration is proportional to its displacement from the fixed point and is always directed towards the point. [B1] [B1] (b) 0.11 m [B1] (c)  = 2f = 2   0.40 = 2.51 = 2.5 rad s 1 (2 s.f.) [M1] [A1] (d) Graph in the correct quadrant, gradient at x-axis =  and gradient at y-axis = 0. Correct values at both intercepts (x 0 = 0.11 m and vmax = 0.28 m s–1). [B1] [B1] (e) Period T = 11 0.4f  = 2.5 s Since the lamp is above the seawall for more than half the peri od (1.25 s), the equilibrium position of the lamp is above the seawall. Time taken t for the lamp to go from its highest position to the top of the seawall t = 1.8 2 = 0.90 s x = x0 cos t = 0.11 cos (2.51  0.90) = 0.0699 m Time taken t for the lamp to go from its lowest position to the top of the seawall t = 2.5 1.8 0.35 s2   x = x0 cos t = 0.11  cos (2.51  0.35) = 0.0701 m [M1] [M1] [A1] velocity / m s1 displacement / m 0 0.11  0.276 or 0.28 t 1.8 s x 2.5 t 1.8 s x

2 © Raffles Institution 9749/03 Alternatively, Time taken t for the lamp to go from equilibrium position to the top of the seawall t = 1.8 1.25 2  = 0.275 s x = x0 sin t = 0.11 cos [2.51  (1.25 + 0.275)] = 0.0697 m Time taken t for the lamp to go from equilibrium position to the top of the seawall t = 1.8 1.25 2  = 0.275 s x = x0 sin t = 0.11 cos (2.51  0.275) = 0.0700 m [M1] [M1] [A1] 2 (a) A progressive wave is one in w hich energy is transferred from one point to another without the transfer of matter. [B1] A transverse wave is one in which the particles oscillate in a direction perpendicular to the direction of energy transfer. [B1] (b) 4.0 s (1 d.p.) [B1] (c)  = (t / T)  2 = (1.4 / 4.0)  2 OR (7/20)  2 = 2.20 = 2.2 rad (2 s.f.) Ignore  signs. OR  = (t / T)  2 = (2.6 / 4.0)  2 OR (13/20)  2 = 4.08 = 4.1 rad (2 s.f.) [M1] [A1] OR [M1] [A1] (d) intensity  (amplitude) 2 intensity  1/r (for 2D circular wavefront)  (amplitude)2  1/r  rQ AQ2 = rP AP2 From Fig. 2.2, A P = 2.0 mm and AQ = 1.0 mm,  rQ 1.02 = (150)  2.02  rQ = 600 mm [B1] [M1] [A0] (e)   = 2  (x / ) 2.20 = 2  (600  150) /   = 1286 mm = 1290 or 1300 mm (2 s.f.) (allow ecf from (c)) [M1] [A1] t 1.8 s x 1.25 t t 1.8 s x 1.25 t

3 © Raffles Institution 9749/03 [Turn over (f) Graph shows amplitude of wave decreasing with distance. Accept positive/negat