HCI 08 Oscillations Solutions (Self-Review Questions)

1 2022 Oscillation Tutorial Suggested Solutions to Self-Review Questions S1 C. The defining equation of SHM is a =  ω2x. Hence we can see that to obtain the a-t graph from the x-t graph, we need to ‘flip’ the graph about the horizontal axis. OR For SHM, we can deduce from the x-t graph, 𝑥 = −𝑥0𝑐𝑜𝑠𝜔𝑡 Hence 𝑎 = −𝜔2𝑥 = 𝜔2𝑥0𝑐𝑜𝑠𝜔𝑡 S2 The general solution of SHM is 𝑥 = 𝑥0𝑠𝑖𝑛𝜔𝑡. Comparing the given equation with the general form, 𝑥 = 30𝑠𝑖𝑛50𝑡, we can deduce that the SHM has an amplitude of 𝑥0 = 30 and angular frequency 𝜔 = 50 Since  = 2f  f = 50 / 2 = 7.958 = 8.0 Hz S3 (i) (a) amplitude, x0 = 20.0 cm (b) period, T = 0.4000 s (c) frequency = 1/T = 2.500 Hz (d) angular frequency,  = 2f = 2(2.500) = 5.000 = 15.7 rad s-1 (e) displacement at A, xA = -10.0 cm (ii) x = 20.0 sin (15.7 t ) (iii) At x = -10.0 cm, -10.0 = 20.0 sin 5.000 t 5.000 t = sin-1(-0.500) =  + /6 t = 0.233 s The value t = 0.233 s tallies reasonably well with the value from the graph (t = 0.238 s). (iv) (a) velocity is maximum at C, as deduced from the gradient of the displacement-time graph. (b) acceleration is maximum at B, since B is at maximum negative displacement. (c) At C, maximum velocity is positive and at B maximum acceleration is positive. (v) (a) velocity at B, vB = 0 (the gradient of the curve at B is zero) (b) velocity at C, vC = x0 = 5.000(0.200) = 3.14 m s-1 (c) acceleration at B, aB =  2(x0) = 2x0 = (5.000)2(0.200) = 49.3 m s-2 (d) acceleration at C, aC =  2(0) = 0 (vi) (a) (b) 49.3 - 49.3 t/s 0.2 t/s 0.1 0.4 3.14 - 3.14 0 0 0.3 0.2 0.1 0.4 0.3 a/ m s-2 v/ m s-1 a = -xoω2 sin ωt v = xoω cos ωt amplitude period

2 49.3  49.3  20 20 a/m s-2 x/cm x/cm v/ m s-1 3.14  3.14 20  20 (c) & (d) S4  = 2f = 2(2.0) = 4.0 = 12.6 rad s-1 vmax = x0 = 4.0(0.0080) = 0.10 m s-1, at equilibrium position amax = 2(x0) = (4.0)2(0.0080) = 1.3 m s-2 (the magnitude of acceleration at the extreme positions) a = -ω2x v = ±𝜔√(𝑥𝑜2− 𝑥2)

3 S5 (i) (a) Point B (b) Point C (ii) At C, x = 0 PE = ½ m2x2 = 0 J KE = KEmax = ½ mv2max = ½ (1.0)(3.14)2 = 4.93 J At B, x = x0 PE = KEmax = 4.93 J KE = 0 J (iii) (iv) S6 C. Total energy = ½ m2x02 = 0.5 m (2πf)2 A2 S7 A. Recognise from the F-r graph that the motion is simple harmonic. Hence the potential energy expression is a quadratic equation with minimum potential energy at the equilibrium position. S8 A The amplitude a increases until f matches the resonance (natural) frequency of the pendulum, after which the amplitude a decreases again. -------------------------------------------------------------End-------------------------------------------------------- Energy/J Time/s (a)KE (b)PE 0 0.