Title 08 Oscillations Tutorial Solutions.pdf ✅

St. Andrew’s Junior College H2 Physics 8-1 Oscillations: Checkpoint Question (SOLUTIONS) Graphs of S.H.M. Starting with a displacement-time graph of cos( ) 0 x = x t , draw the corresponding graphs and state the equations for the various relationships. x 0 T t v 0 T t a 0 T t KE 0 T t v 0 x - x0 x0 a 0 x - x0 x0 KE - x0 0 x0 x


St. Andrew’s Junior College H2 Physics 8-4 Level 2 Solutions 6 (a) (i) 1. amplitude = 0.15 m, 2. period = 1.0 s, 3. frequency = 1/T = 1.0 Hz, 4. angular frequency w= 2f = 6.3 rad s-1 (ii) amplitude (iii) From the diagram we can see that at t = 0.5 s, that A is at is amplitude while B is at its equilibrium position.  they are 1⁄4 of a cycle apart. Since 1 cycle → 2  1⁄4 cycle → 2/4 = 1⁄2  phase difference = 0.5 rad (Elaboration in topic: Waves) (iv)  = 2f = 2(1) = 2 x = − xo cos t = -0.15 cos 2t (v) (vi) (v) At P, where x = 0, P is at the equilibrium position. (vii) Hence speed is maximum i.e. vB = x0 = (0.1)(2) correct amplitude = 0.63 m s-1 1 1 1 1 1 1 1 1 1 1 1 1 7(a) f = 1/T = 1/0.020 = 50 Hz 1 (b)  = 2  f = 2  (50) = 100  rad s-1 1 (c) x = xo sin t { Candidates should use this eqn as the eqn of motion if the question is silent about the initial condition. Refer sect 3.1 Q1 } = 0.0030 sin (100  t) correct amplitude & correct  1 1 (d) (i) 0.9 m s-1 (ii) For each cycle, the body passes the zero displacement point twice at the same speed, but in opposite directions. (iii) For SHM, the body always changes directions at the two extreme ends where its velocity is zero. 1 1 1 (e) (i) KEmax = 1⁄2m 2 max v = 1⁄2 m (xo ) 2 = 1⁄2(0.100) (0.0030 x 100 ) 2 = 0.0444 J (ii) ( a quadratic eqn/parabolic shape) parabolic shape values at axes 1 1 1 8 (a) (i) Period = 0.6 s (ii)  = 2/T = 2/(0.6) = 10.5 rad s-1 1 1 1 (b) (i) 0.20 s (ii) Use /2 = t/T (fr Topic: Waves) {phase angle = phase difference}   = 2  t/T = 2  (0.20)/(0.60) = 2 /3 rad 1 1 1 -0.0030 +0.0030 0.0444 KE x