Question

The displacement of a particle along the x-axis is given by $x=a{\sin }^2\omega t$. The motion of the particle corresponds to.

• A. Simple harmonic motion of frequency $3\omega /2\pi$
• B. Simple harmonic motion of frequency $\omega /2\pi$
• C. Non simple harmonic motion
• D. Simple harmonic motion of frequency $\omega /\pi$

Answer 1

Answer: (C)

Given, $X=a{\sin }^2\omega t$
or $X=a\left(\frac{1-\cos 2\omega t}{2}\right)$ $[\because \cos 2\theta =1-2{\sin }^2\theta ]$
or $X=\frac{a}{2}-\frac{a\cos 2\omega t}{2}$
Now, $V=\frac{dx}{dt}=a\omega \sin 2\omega t$
and $a=dv/dt=2a{\omega }^2\cos 2\omega t$
Here, a is not directly proportional to $(-x)$ which is a condition for SHM.