08 Oscillations (H2) 2020 - Tutorial Solution

8-1 TUTORIAL 8: OSCILLATIONS SOLUTIONS Level 1 Solutions 1 D. Refer to the definition of SHM to appreciate the answer. 2 D. Since F is proportional to a, the F-r graph has the same shape as a-r graph. 3 D. Speed is always at the maximum at the equilibrium position. 4 C. The mass is moving away from the equilibrium position. Its acceleration is always directed towards that point. Cannot be A, because a = 0 (since x =0),  no direction to compare with v. Cannot be B, because v=0 (since v = dx/dt=0),  no direction to compare with a. Cannot be D because it is moving towards the equilibrium position. 5 A. Since bob is released at t = 0,  v = 0 when t = 0,  KE = 0 when t = 0 Cannot be C because the frequency of energy variation is doubled. 6 (a) (i) 1. amplitude = 0.15 m, 2. period = 1.0 s, 3. frequency = 1.0 Hz, 4. angular frequency = 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 ¼ of a cycle apart. Since 1 cycle  2  ¼ cycle  2/4 = ½  phase difference = 0.5 rad (iv)  = 2f = 2(1) = 2 x =  xo cos t = -0.15 cos 2t (v) (vi) (v) At P, the 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

8-2 Level 2 Solutions 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 = 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 displacements where its velocity is zero. 1 1 1 (e) (i) KEmax = ½m 2 maxv = ½ m (xo )2 = ½(0.100) (0.0030 x 100 )2 = 0.0444 J (ii) 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) /2 = t/T   = 2  t/T = 2  (0.20)/(0.60) = 2 /3 rad 1 1 1 (c) (i) Damping is the loss of energy from an oscillating system to the environment due to dissipative forces such as air resistance or friction. (ii) 1. By attaching a card to the mass such that it is perpendicular to the direction of motion of the oscillating system, light damping can be achieved. 2. By immersing the oscillating mass in a viscous fluid such as oil, the degree of damping can be increased. Different degree of damping is achieved by using fluid of different viscosity. 1 2 1 9 (a) (i)  = t (ii) ST = r sin  = r sin t 1