Question

The displacement of an object in SHM is given by $y(t)=(8.0 \mathrm{cm}) \sin [(1.57 \mathrm{rad} / \mathrm{s}) t] .$ What is the frequency of the oscillations?

Short answer

Answer: The frequency of the oscillations is approximately \(0.25\,\mathrm{Hz}\).

Step-by-step solution

Identify the equation of motion

The displacement of the object in SHM is given by \(y(t) = (8.0 \mathrm{cm})\sin[(1.57 \mathrm{rad} / \mathrm{s})t]\). This equation has the general form \(y(t) = A\sin(\omega t)\), where A is the amplitude and ω is the angular frequency.

Extract the angular frequency

From the given equation, we can see that the angular frequency ω is equal to \(1.57 \mathrm{rad} / \mathrm{s}\).

Find the frequency

We know the relation between angular frequency (ω) and frequency (f) is given as \(\omega = 2 \pi f\). We need to solve this equation for f. $$ f = \frac{\omega}{2 \pi} $$ Now, substitute ω with the value we found earlier: $$ f = \frac{1.57 \mathrm{rad}/\mathrm{s}}{2 \pi} $$ Calculate the frequency f: $$ f \approx 0.25\,\mathrm{Hz} $$ The frequency of the oscillations is approximately \(0.25\,\mathrm{Hz}\).