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 1 (b) The shadow moves in simple harmonic motion. 1 (c) (i) maxv = xo  = 20 (3.5) = 70 cm s-1 (ii) maxa = xo 2 = 20 (3.5)2 = 245 cm s-2 1 1 1 1 -0.0030 +0.0030 0.0444 KE x

8-3 10 (c) (i) Max. depth = 5.0 + 3.0 = 8.0 m (ii) Min. depth = 5.0 – 3.0 = 2.0 m (iii) Time between high- and low-water = ½(Period) = ½(45600) = 22,800 s (iv) If h = 5.0 m, then 5.0 = 5.0 + 3.0 sin 45600 2 t  sin 45600 2 t = 0  45600 2 t = 0 or   t = 0 s or 22,800 s (v) If h = 7.0 m, then 7.0 = 5.0 + 3.0 sin 45600 2 t 2.0 = 3.0 sin 45600 2 t sin 45600 2 t = 2.0/3.0 Caution! Set calculator to radian mode. 45600 2 t = 0.730 or 2.41 t = 5298 s or 17490 s (These are the two instances, t2 & t1)  t2 – t1 = 17490 – 5298 = 12,200 s (to 3 s.f.) 1 1 1 1 1 1 1 2 1 11 (b) Graph of sin2 t Double the frequency Values shown on t-axis 1 1 1 (c) From T = 2 ( m k ), a smaller effective mass means the period is smaller for the subsequent motion. 1 12 (a) k = F x = 4.0 3.2 x 10-2 = 125 N m-1 1 h t 7 5 t1 t2 0.05 0.10 0.15 0.20 0 t/s Ep

8-4 (b) (i) Change in GPE = mgh = 4.0 x (0.80 x 10-2) = 3.2 x 10-2 J (ii) Change in EPE = ½kxf2 - ½kxi2 = ½ (125) (4.0 x 10-2)2 - ½ (125) (3.2 x 10-2)2 = 3.6 x 10-2 J 1 1 1 1 (c) Work done = Change in EPE - Change in GPE = PE gained = 3.6 x 10-2 - 3.2 x 10-2 = 4.0 x 10-3 J 1 (d) (i) Total energy of oscillation = PE gained = 4.0 x 10-3 J Note: Total energy of the oscillation is the amount of energy exchange between KE and PE and thus not the same as the total energy ET of the system. The total energy of the system is dependent on the reference point of the GPE. Refer to the diagram below: (ii) 1. KE lost = PE gained ½ m vmax2 - 0 = 4.0 x 10-3 vmax = 81904 10042 3 ./. x.x  = 0.14 m s-1 2. vmax = xo  = vmax xo = 0.14 0.80 x 10-2 = 17.5 2f = 17.5 f = 17.5 2 = 2.8 Hz 1 1 1 1 1 1 (e) T = 2 ( m k )  f = 1 T = 1 2 ( k m ) If the spring has some mass, the system will behave as though a greater mass is attached to the spring. Hence, the frequency will decrease. 1 1 Total energy of oscillation

8-5 13 (a) (i) Loss in gravitational potential energy = (0.400)(9.81)(0.200) = 0.785 J (ii) k = F / e = (0.400 x 9.81)/0.200 = 19.62 N m-1 U = ½ ke2 = ½ (19.62 x 0.200)2 = 0.392 J 1 1 1 (b) The difference is work done against the external force needed to support the mass while lowering it gently. This force is the difference between the mass’s weight and the tension in the spring. 1 1 (c) (i) At the lowest point, the total extension of the spring is 0.400 m. Tension T = k e = 19.62 x 0.400 = 7.848 N Weight W = m g = 0.400 x 9.81 = 3.924 N F = T – mg = 7.848 – 3.924 = 3.92 N (ii) ω = mk / = 400.0/62.19 = 7.00 rad s-1 (iii) vmax = ω x0 = 7.00 x 0.200 = 1.40 m s-1 1 1 1 1 (d) Gravitational potential energy / J Elastic potential energy / J Kinetic energy / J Total energy / J Lowest point 0 1.57 0 1.57 Equilibrium point 0.785 0.392 0.392 1.57 Highest point 1.57 0 0 1.57 1m for any two correct total, 5 (e) 1m for 1 correct or 2m for 2 correct or 3m for 4 correct 14 (a)  The amplitude is decreasing with time.  The curve does not start from zero displacement at t = 0 s. 1 1 (b) (i)  = 2/T = 2/(1.5) = 4.19 rad s-1 (ii) a = - 2 x = - (4.19)2 (2.2) = - 38.6 cm s-2 1 1 1 1

8-6 (c)  Show that amplitude is decreasing.  Show that the period is the same or larger. 1 1 15 (a) The amplitude of the block becomes larger. This is because, the rate of energy transfer from the waves to the block is greater. 1 1 (b) The amplitude of the block becomes smaller. The waves now have larger wavelength and thus smaller frequency. Hence, the block does not resonate at the new frequency of the waves. 1 1 (c) The amplitude of the block becomes smaller. The mass of the block is now larger and hence its natural frequency becomes smaller (using f = (1/2) (32/m)). Hence, the block no longer resonates at the frequency of the waves. 1