JPJC 2026 Oscillations Tutorial Suggested Solutions

1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS OSCILLATIONS TUTORIAL SOLUTIONS Self-Check Questions S1 displacement, x – distance an object has moved from its equilibrium position in a stated direction amplitude, x0 – the magnitude of the maximum displacement of the object from the equilibrium point period, T – the length of time taken for one complete oscillation frequency, f – number of oscillations per unit time angular frequency,  – the scalar measure of rotation rate expressed in radians per second (rad s1). S2 1 2T f   S3 Simple harmonic motion is defined as an oscillatory motion in which the acceleration is (directly) proportional to displacement from an equilibrium position and acceleration is always directed towards the equilibrium position. The defining equation is 2a x   . S4  0 sinx x t     0 0 cos cosv x t v t    , where 0 0v x  is the maximum velocity    20 0 sin cosa x t a t    , where 20 0a x  is the maximum acceleration S5 2T 2T 2T

2 S6 S7 There is continual change of energy from kinetic energy Ek to potential energy Ep and vice versa. Ek is greatest and Ep is zero as the mass passes through the equilibrium position. As the mass approaches the extreme points, its Ek decreases and Ep increases. As the mass approaches the equilibrium point, its Ek increases and Ep decreases. The frequency of energy variation is twice that of the oscillation itself. (At any instant during the simple harmonic motion, the total energy E of the system is constant and equal to the sum of Ek and Ep. Note that E is proportional to 20x and 2f .)

3 S8 Damping refers to an oscillating system being affected by the presence of viscous forces. This causes the amplitude of oscillations to decrease exponentially with time as energy from the oscillating system is used to overcome the viscous force. A swinging pendulum or vibrating tuning fork in air are examples of lightly damped oscillations. The suspension system if a car includes shock absorbers and springs. When a car goes over a bump, the springs will be displaced from their equilibrium lengths. The viscous oil in the shock absorbers provide damping to enable the springs to smoothly and quickly return to their equilibrium lengths without oscillating up and down. The springs are critically damped. This will reduce the discomfort of the passengers in the car. Without critical damping, then body of the car will oscillate up and down after going over a bump, which is undesirable. (Refer to lecture notes on the car suspension system.) S9 Forced oscillations are produced when a body is subjected to a periodic external driving force. Resonance occurs when a system responds at maximum amplitude to an external driving force. This occurs when the frequency of the driving force is equal to the natural frequency of the driven system. S10 At steady condition, the amplitude of a forced oscillation depends upon the damping of the system and the relative values of the driving frequency f to the natural frequency f0 of the system. Oscillations with the largest amplitude (i.e. resonance) occur when f is approximately equal to f0. displacement energy time time 2T 2T T

4 The sharpness of resonance is determined by the degree of damping. When damping is light, the amplitude is large but falls off rapidly when the driving frequency of the body differs slightly from the natural frequency of the body. The resonance is sharp. When damping is moderate, the amplitude at resonance decreases. The curve falls off gradually and maximum amplitude occurs at a frequency that is lower than the natural frequency of the body. When damping is heavy, the resonance response is flat. Self-Practice Questions 1 In an undamped S.H.M., total energy of the oscillating system is a constant though kinetic and potential energies are constantly being transformed from one form to the other. Since total energy and maximum energy is constant, amplitude is also constant. Period of oscillation is constant, hence angular frequency is also constant. Ans: E 2 Player should aim at the regions where the moving target is moving the slowest, i.e. at the extreme ends of the S.H.M. Target is momentarily stationary at the amplitude positions of the oscillation. Ans: B 3 Since xa 2 , from the graph, at maximum acceleration,   2 218 2 2.09 2.1 s T T         4 At point C, the body’s velocity is positive (gradient of displacement-time graph positive) and since its displacement is positive, its acceleration is negative ( 2a x   ). Ans: C RESONANCE

5 5 (a) (i) From the graph, amplitude 0 0.11 mx  (ii) From the graph, period 1.7 sT  Better method, take average, 3 complete oscillations take 5.1 s, period 1.7 sT  (iii) Frequency, 1 1 0.588 0.59 s1.7f T    (iv) Angular frequency, 12 2 3.696 3.7 rad s1.7T      (b)

6 (i) Maximum positive velocity occurs at t = 0 s, 1.7 s, 3.4 s, 5.1 s. (ii) Maximum positive acceleration occurs at t = 1.3 s, 3.0 s, 4.7 s. (iii) Maximum negative acceleration occurs at t = 0.4 s, 2.1 s, 3.8 s, 5.5 s. (iv) At x = 0, we have maximum kinetic energy. It occurs at t = 0 s, 0,85 s, 1.70 s, 2.55 s, 3.40 s, 4.25 s, 5.10 s, 5.95 s. (v) At 0x x , we have maximum potential energy. It occurs at t = 0.40 s, 1.30 s, 2.10 s, 3.00 s, 3.80 s, 4.70 s, 5.50 s. 6 Trolley is performing S.H.M. hence its acceleration 2a x   . At maximum displacement 0x , max max max max 2 max 0 1max 0 10 10 rad s2.0 0.050 F ma Fa m Fx m F mx          Ans: D 7 From t = 0 to 4Tt  (a quarter through its oscillation), the speed of the mass increases from zero to a maximum at the equilibrium position and acceleration/restoring force decreases from a maximum to zero. The amplitude is 100 30 35 cm2  . At 2Tt  , the mass is at its maximum displacement from the equilibrium position. It is momentarily at rest. So speed and kinetic energy is zero. Ans: A 8 Compare  0.050sin 8x t  with  0 sinx x t  , (a) amplitude, 0 0.050 mx  (b) frequency, 8 4.0 Hz2 2f      (c) period, 1 1 0.25 s4.0T f   (d) velocity of the particle as its passes through its equilibrium position    0 0 1 8 0.050 1.256 1.26 m s v x       

7 (e) maximum acceleration of P during its motion     0 0 2 8 8 0.050 31.58 31.6 m s a v        

8 Discussion Questions 1 Example of free oscillation - a simple pendulum bob oscillating freely without any external force acting on it. A free oscillation cannot have a constant amplitude in practice - energy is lost in the oscillating system to overcome the resistive forces, such as air resistance. 2 Consider an object oscillating about the equilibrium position. Take rightwards as positive. (a) As the object moves from 0xx  to 0xx  , it passes through the equilibrium position with maximum velocity of 0xv  . On its return journey from 0xx  to 0xx  , it passes through the equilibrium position with minimum velocity of 0xv  . (b) Since 2 20v x x  , at both 0xx  and 0xx  , the object will have zero velocity as it comes to a rest momentarily at these positions. 3 Since 0 sinx x t  , we have )22sin(5025 t 17.0t s The shutter remained open for 0.17 s. 4 (a) Simple harmonic motion is an oscillatory motion in which the acceleration is (directly) proportional to displacement from an equilibrium position and acceleration is always directed towards the equilibrium position. (b) (i) 25.0 3.0sin 45600th       Maximum depth of water occurs when 2sin 145600t    . Hence, 0.30.5 h 0.8h m The maximum depth of water is 8.0 m.