Question

A simple harmonic motion is represented by $x\left(t\right)={\sin }^2\omega t-2{\cos }^2\omega t$. The angular frequency of oscillation is given by?

Answer 1

Step 1: Given data

The equation of the simple harmonic motion is $x\left(t\right)={\sin }^2\omega t-2{\cos }^2\omega t$.

Step 2: Simple harmonic motion

1. Simple harmonic motion is a motion where the acceleration is proportional to the displacement and the force on the body is always directed toward a fixed point.
2. A simple harmonic motion is defined by the form, $y=A\cos \omega t$, where, y is the displacement of the body, A is the amplitude, $\omega$ is the angular frequency and t is the time.

Step 3: Finding the angular frequency

Here we have,$x\left(t\right)={\sin }^2\omega t-2{\cos }^2\omega t$

So,

$$\begin{gathered} x\left(t\right)=\left(1-{\cos }^2\omega t\right)-2{\cos }^2\omega t\left(Since,{\sin }^2\omega t=1-{\cos }^2\omega t\right) \\ orx\left(t\right)=1-3{\cos }^2\omega t \\ orx\left(t\right)=1-3\frac{1+\cos 2\omega t}{2} \\ orx\left(t\right)=1-\frac{3}{2}-\frac{3}{2}\cos 2\omega t..................\left(1\right) \end{gathered}$$

Equation (1) is a periodic motion of angular frequency $2\omega$.

Therefore, the angular frequency of the harmonic oscillation is $2\omega$.