Simple harmonic motion of a floating pier

Question

A 5000 kg floating pier is moved up and down by the changing tide. If the period of this motion is 12 hours and the amplitude is 2.5 m and we treat the motion as being simple harmonic then:

a) what is the frequency of the motion?

b) What is the maximum vertical velocity of the pier?

c) what is the maximum vertical acceleration of the pier (hint: use circular motion as an analogue)

Answer 1

The generic equation governing the displacement of simple harmonic motion is

x = A cos(ωt)

Where A is the amplitude, ω = 2πf, is angular frequency and t is time.

The frequency, f = 1/T [with T the period]

a)

We know the period, T = 12h = 43200s ==> f = 1/(43200s) = 2.3x10^-5Hz

From this, the angular frequency ω = 2πf = ω = 2π(2.3x10^-5)rad/s = 14.5x10^-5rad/s

b) differentiating displacement with respect to time gives the velocity, v i.e

v = -Aωsin(ωt) ==>v_max = -Aω = -(2.5m)(14.5x10^-5rad/s) = 36.4x10^-5m/s [ignore the negative sign since we are interested in the magnitude]

c) differentiate (with respect to time) the velocity equation in b) to get the acceleration, a, i.e

a = -Aω²cos(ωt) ==> a_max = -Aω² = -(2.5m)(14.5x10^-5rad/s)² = 5.3x10^-8m/s² [once again, ignore the negative sign since we are interested in the magnitude]