Question

The end of a diving board can oscillate with a frequency of 2.9 Hz. Assume it behaves like a simple harmonic oscillator in what follows.

Before we deal with the diving board, first consider you have a mass, ?m, on a spring with spring constant, ?, oscillating with a certain frequency, ? and amplitude, ?. Derive an equation for ...

a) the magnitude of the maximum speed of the oscillator in terms of the frequency, ? and the amplitude of the oscillation, ?. (2 mark)

b) the magnitude of the maximum acceleration of the oscillator in terms of the frequency, ? and the amplitude of the oscillation, ?. (2 marks)

c) Now back to the diving board. You place a pebble on the end of the oscillating diving board. What is the maximum amplitude that the board can oscillate with so that the pebble doesn’t lose contact with the board at the top of the oscillation? (3 marks) (Answer: 3cm)

Answer 1

The displacement of a simple harmonic oscillation of amplitude A and frequency f is given by : y = A sin ωt,

where, ω = angular frequency of oscillation = 2πf

Hence, y = A sin (2πft)

Speed of oscillation = v = dy/dt = 2πfA cos (2πft)

(a) Hence, as maximum value of cos (2πft) = 1, maximum speed is given by : v_max = 2πfA

Acceleration of oscillation = a = dv/dt = -(2πf)² A sin (2πft)

(b) Hence, as maximum value of sin (2πft) = 1, maximum acceleration is given by : a_max = (2πf)²A

(taking magnitude only)

(c) The pebble will lose contact with the diving board if the board is accelerating at a rate greater than g (=10 m/s²), acceleration due to gravity. Since the maximum acceleration is given by : a_max = (2πf)²A,

we will put , a_max = g for the maximum allowed acceleration,

and, A = A_max for the maximum allowed amplitude.

Hence, A_max = g / (2πf)² = 10 / (2 * 3.14 * 2.9)² = 0.03 m = 3 cm.