HCI 08 Oscillations Solutions (Discussion Questions)

1 Oscillation Tutorial Hints & Suggested Solutions to Discussion Questions D1 Amplitude of oscillation = 50 mm Period of oscillation = 2 s x = 50 sin (2π /2) t (x in mm) [Note: Taking to the right of 650 mm mark as positive direction] when x = +25 mm 25 = 50 sin (2π /2) t Time duration for which shutter remained open, t = 1/6 s = 0.17 s Key Concepts: Characteristics of S.H.M? Able to write down the displacement-time function for a S.H.M. from initial condition (i.e. when t = 0 s). Note: For SHM, displacement is NOT proportional to time. D2 Hints: Check at equilibrium x = 0, is there tension? How is tension related to the extension of spring? B. Key Concepts: Link to topic of Dynamics & Forces. Note: Tension in spring, T is NOT proportional to the displacement, x. Qn D2, D10 & D11 is about oscillation of a vertical spring mass system. D3 Hints: Sketch the displacement, x vs phase angle, ϕ graph for the scenario. Note that at time t = 0 s, Mass A is released from its maximum displacement position (2 cm below equilibrium position). Mass B is released from it maximum displacement position (1 cm below equilibrium position) only when Mass A is at displacement of 1 cm b elow equilibrium position. B. 1 2cos 3    x Mass A Mass B

2 Key Concepts: Understand phase angle and phase difference for oscillations. D4 Hints: i) Can you write down the function that relates the acceleration, a to time, t? ii) Can you write down the function that relates the displacement, x to time, t? iii) Can you write down the function that relates the velocity, v to time, t? A. From a – t graph, a = ao cos ωt We can then infer that displacement, x = - xo cos ωt Hence, velocity, v = vo sin ωt Sketch the x – t and v – t graphs on the a – t graph (same time axis), and determine that at P when a is positive, x is negative and v is positive. Key Concepts: See the relationship between a-t, x-t & v-t graphs of SHM. Know the kinematics of SHM. D5 Hints: i) How do you know that the oscillation is SHM? ii) Where is the equilibrium position of the SHM? iii) What is the amplitude of the SHM? iv) What is the displacement of SHM when the depth of water is 1.5 m? v) Can you write down the appropriate function that relates displacement, x to the time, t? 2.0 m is the equilibrium position. Amplitude, x0 = 1.0 m T = 12 hrs Depth of water, x = - 1.0 cos (2π /12) t + 2.0 1.5 = - 1.0 cos (2π /12) t + 2.0 Time duration that the boat will have to wait before entering, t = 2.0 hrs Key Concepts: Characteristics of S.H.M? Able to write down the displacement-time function for a S.H.M. from initial condition (i.e. when t = 0 s). D6 (b) (i) Key Concepts: How to identify SHM? Defining equation for a SHM. (ii) 1. ω = 2πf = 2π (13) = 82 rad s-1 2. when amax = g | -ω2xo | = g Amplitude of oscillation, xo = 9.81 / 822 = 1.47 x 10-3 m ω2x0 -ω2x0 -xo xo a x

3 (c) Hint: i) Sketch a diagram of the plate and one particle of sand on plate. What forces act on the particle of sand? Which of these forces is dependent on the relative motion between sand and plate? ii) When the plate is at maximum displacemen t above the equilibrium position, what is the direction of acceleration of the plate? What is the direction of velocity of the plate just before it reach maximum displacement? Suggested Solution: If the amplitude of the oscillati ons of the plate exceeds the value calculated in (b)(ii)2, the maximum acceleration of the plate will be higher than the acceleration of free fall, g. Thus the sand will lose contact with the flat horizontal plate on its way up (pass the equilibrium point) as the plate slows down at a rate larger than the sand. D7 (a) (i) 𝜃 = 𝜔𝑡 (ii) 𝑆𝑇 = 𝑟𝑠𝑖𝑛𝜔𝑡 (How do you know that it is a sine function?) (b) ST is the displacement x of the shadow with respect to S. Comparing it with the standard expression for the displacement of an object in SHM ( 𝑥 = 𝑥0𝑠𝑖𝑛𝜔𝑡), we can see that the shadow of the peg is moving in simple harmonic motion about the point S with amplitude 𝑥0 = 𝑟. (c) (i) amplitude of shadow's SHM is same as radius of the turntable 20 cm. (ii) Since the turntable has angular speed 𝜔 = 3.5 = 2𝜋 𝑇 , the period of the turntable is 1.8s which is also the period of the shadow's SHM. (iii) Maximum speed of shadow at S is 𝑣𝑚𝑎𝑥 = 𝜔𝑥0 = (3.5)(0.2) = 0.7𝑚/𝑠 (iv) Maximum acceleration (when shadow at rest) is given by 𝑎𝑚𝑎𝑥 = 𝜔2𝑥0 = (3.5)2(0.2) = 2.45𝑚/𝑠2 D8 Hint: i) What is the expression of the total mechanical energy of a mass in SHM? Further qn: Can you prove that the mass will oscillate in SHM? Total mechanical energy for S.H.M = max. K.E. = ½ Mv2max = ½ M (ω a)2 = ½ M (a 2π/T)2 = 2π2 Ma2 / T2 Another method: Let the effective spring constant for this system be k. From 2 k m  , we get 2kM  Total mechanical energy for SHM = 2 2 2 2 2 2 2 0 22 1 1 1 4 2 2 2 2 T MaE kx M a M a TT       D9 It is given in the question that the potential energy versus time graph is for half a period. At t = 0, the potential energy is maximum. This suggests that the particle is at the maximum displacement where the velocity is zero. Half a period later, the particle is at the other maximum displacement where velocity is again zero. During this time, the velocity is unidirectional. Hence, the velocity-time graph should give a half of sine graph.

4 D10 PEmax = KEmax = ½ m (ωxo)2= ½ m (xo 2π/T)2 From graph, when xo = 0.2 m, U = 1.0 J 1.0 = ½ (4) (0.2 x 2π/T)2 1.0 = 2.0 x 0.04 x 4π2 / T2 T2 = 2.0 x 0.04 x 4π2 T = 1.8 s Key Concepts: Understand and able to determine the potential energy vs displacement relation for a SHM. D11 (a) Hint: i) How do you know that the vertical spring-mass system will oscillate in SHM? At the equilibrium point, 00 , where is the extension of spring. resF kx Mg Mgkx Mg k x x        Any of these points or other correct points from the graph; When a mass of 150 g is hung, the extension on the spring is 10.0 cm. When a mass of 300 g is hung, the extension is 20.0 cm. When a mass of 450 g is hung, the extension is 30.0 cm 10.150 14.7 N m .0.100 gk  (b) Hint: i) What is the unstretched length, lo, of the spring? ii) What is the length of the spring when 450 g is attached? What is the extension of the spring when 450 g is attached? iii) What is the amplitude of SHM? 1 14.7152 0.910 Hz (3 s.f.)2 0.450ff     (c) Hint: i) What is the displacement of the oscillation when the length of spring is 80.0 cm?

5 ii) What is the relationship (formula) that relates velocity, v with displacement, x? ( ) 0.450 9.81 0.300 1.32435 JGPE Mg extension      2 2 2 2 21 11( ) (14.715)(0.400 0.100 ) 1.103625 J22EPE k e e      1.32435 1.103625 0.220725 JKE    2110.220725 0.990 m s2 mv v    OR Note that question is asking for v for an oscillation with amplitude 20.0 cm at displacement of 10.0 cm. 2 2 2 2 1 0 2 (0.910) 0.200 0.100 0.990m sv x x      (d) Use 12 2    kff m , deduce that f is inversely proportional to m . When m increases, f is reduced OR The effective mass of the spring-mass system increases. The resonant frequency of heavier masses is at lower values of frequencies (or at greater periods). Hence, the frequency of the oscillation would be reduced. D12 Qn D2, D