SIMPLE HARMONIC MOTION Challenging MCQ questions by The Physics Cafe Compiled and selected by The Physics Cafe
1 Fig..1 shows a device for measuring the frequency of vibrations of an engine. The rigid metal probe and case is held against the engine so that the engine vibrations induce forced vibrations of the strip in the direction shown. Fig..1 The length L of the strip which is free to vibrate can be varied until a maximum amplitude is observed. The frequency corresponding to L can then be read from the scale. (a) (i) State what will be observed as the length L of the strip that is free to move (b) is gradually increased from a small value..... Explain why this happens. [] In one test, the end of the strip oscillated with simple harmonic motion of amplitude 5.0 mm at a frequency of 1 Hz. (i) record of frequency length of strip free to vibrate L movable strip Define simple harmonic motion. probe plate with markings to enable amplitude measurements engine direction of oscillation ThePhysicsCafe [] [1] P a g e
Determine for the end of the strip, 1. its maximum speed. its maximum acceleration maximum speed = m s -1 (c) maximum acceleration = m s - [3] The free end of the vibrating strip is an antinode and the end near the probe is a node. When the natural frequency of the strip is 1 Hz, the length L of the strip that is free to vibrate is 16 cm. This has been plotted for you in Fig... Sketch a graph on Fig.., showing how the length L varies with natural frequency of vibration of the strip. [] length free to vibrate/ cm Fig.. frequency/ Hz 3 P a g e
Ans (a) (i) The amplitude increases to a maximum and then decreases. B1 The strip has a natural frequency which depends on its length. Maximum amplitude when the frequency of engine vibration equals to the natural frequency of the length of strip free to vibrate. (b) (i) The magnitude of the acceleration is proportional to the displacement from its equilibrium position and is directed opposite to the displacement/ directed towards the equilibrium position (c) 1 1. maximum speed = B1 B1 xo fxo 1 0.005 0.38 m s C1 A1. maximum acceleration = x fx o o B1 B1 4 1 0.005 8 m s A1 v The length L is equivalent to ¼ wavelength L Lf constant = 16 1 4f Graph shows inverse relationship. Curve must pass through at least 3 points with Lf = 16 x 1, such as (6, 3), (16, 1) and (4, 8). ThePhysicsCafe M1 A1 4 P a g e
A mass of mass m was suspended from the bottom of a vertical spring, and set into oscillation with a period T. Measurements were made, and the resultant force F acting on the mass, and its velocity v were obtained. The graphs below were then plotted to show how F and v vary with the displacement x of the mass from its equilibrium position. F/ N v/ m s -1 0.63 1. -0.45 0 0.45 x/m -0.45 0 0.45 x/m -0.63 Calculate: (a) the period of the oscillation, and -1. (b) the mass m. [4] 5 P a g e
Ans (a) From nd graph, v0 = x0 = x0 T (b) x 0 T = v0 0.45 = 1. =.4 s From 1 st graph, Fmax = m x0 Fmax m = x0 T 0.63 = 0.45.36 = 0.0 kg ThePhysicsCafe 6 P a g e
3 A toy car s wheel is set up as a pendulum by hanging it vertically from a fixed support. The wheel oscillates about the fixed support as illustrated in Fig. 8.1. Fig. 8.1 The variation with displacement x of the acceleration a of the wheel is shown in Fig. 8.. a 0 x (a) (i) Fig. 8. Use Fig. 8. to explain why the motion of the wheel is simple harmonic.........[3] State a condition that must be satisfied for the oscillation of the wheel to be simple harmonic.......[1] (b) The wheel in Fig. 8.1 is displaced and released at time t = 0. The oscillations of the wheel have amplitude 14.7 cm and angular frequency.40 rad s -1. (i) Define angular frequency... 7 P a g e
[1] State an expression for the displacement x of the wheel in terms of time t. (iii)..[1] Use your expression in (b) to sketch the variation of the kinetic energy of the wheel EK with time t for one complete oscillation in Fig 8.3. [] Fig. 8.3 (c) The angular frequency value stated in (b) is calculated from the period of the simple harmonic motion. An accurate value for the period is found by timing a large number of oscillations. (i) E k/j 0 Explain why a large number of oscillations would help to achieve a more accurate value for the period.....[1] Calculate the number of oscillations that are measured in a total time of 83.8 s. number of oscillations = [] (iii) A stopwatch with display provides timing to 1/100 th of a second over a range of 9 hours, 59 minutes and 59.99 seconds was used to record the duration of the oscillations in (c). Explain why the duration of the oscillations in (c) is recorded to 0.1 seconds. ThePhysicsCafe t/s 8 P a g e
[1] (d) In order to time the oscillations of the wheel in Fig. 8.1, a stopwatch is started when the centre of the wheel passes a marker. This marker is.7 cm from the equilibrium position as shown in Fig.8.4. 14.7 cm.7 cm marker (i) Fig. 8.4 The wheel has a mass of 0.165 kg. Calculate the restoring force acting on the wheel as it passes the marker. restoring force = N [] Calculate the speed of the wheel at the instant it passes the marker. speed m s -1 [] 9 P a g e
(e) In an experiment to demonstrate resonance, the wheel in Fig. 8.1 is made to oscillate by an external periodic driving force of frequency f. Fig. 8.5 shows the variation with frequency f of the amplitude of the forced oscillations of the wheel. (i) amplitude Fig. 8.5 Describe how the amplitude of oscillation depends on the forcing frequency........[] Sketch in Fig. 8.5 the shape of the graph if the forced oscillations of the wheel were repeated in a vacuum. Label your sketch as A. [] (AJC, 014) ThePhysicsCafe f 10 P a g e
Ans ai SHM is the motion of a body such that its acceleration is proportional to its displacement from a fixed point and is always directed towards that point. The oscillations are simple harmonic because: 1. The straight line passes through the origin indicating that the acceleration a of the plate is directly proportional to the displacement x from its equilibrium position. The negative gradient of the graph indicates that its acceleration a is opposite in direction to its displacement x, and is directed towards the equilibrium position. Note: acceleration and displacement not inversely proportional ii bi The angle of oscillation must be small. Angular frequency of a body undergoing simple harmonic motion is a constant of a given oscillator and is related to its natural frequency f by f ii x 0.147 cos(.40t) OR x 0.147cos(.40t) OR x 14.7cos(.40t) OR x 14.7cos(.40t) iii - Correct shape of cos function - cycles within one period with Ek max at T/4, 3T/4 and Ek = 0 at 0, T/, T. ci ii iii To reduce fractional error of time measurement. T.6180 s.40 83.8 number of oscillations = 3 (round to lower integer).6180 As the average human reaction time is between 0. s and 0.4 s, the timing for the oscillations is recorded to the same precision as the human reaction time. 11 P a g e
di ii ei ii F ma m x 1 st method v 0.165.40 0.070 0.057 N(3sf). 40 0. 147 0. x0 x 070 = 0.347 m s -1 (3sf) nd method x x o cos t.7 14.7 cos(.40t ) t 0.57753 s v x 0 sin t.40(14.7 10 ) sin(.40t ) 0.347 ms The amplitude of oscillation increases when f increases to natural frequency f0. When f is larger than f0, the amplitude decreases. Amplitude is maximum when the frequency is equalled to f0. (Learning point: f0 = 1/T or / =.40/ = 0.38 Hz) ThePhysicsCafe - greater values of amplitude for all values of f - the peak is shaper and at the right of f0-1 1 P a g e