Question

The displacement of a particle is given at time $t$, by $x=A \sin (-2 \omega t)+B \sin ^{2} \omega t$ Then

• A. the motion of the particle is SHM with an amplitude of $\sqrt{A^{2}+\frac{B^{2}}{4}}$
• B. the motion of the particle is not SHM, but oscillatory with a time period of $T = \pi \omega$
• C. the motion of the particle is oscillatory with a time period of $T = \pi \, 2\omega$
• D. the motion of the particle is a periodic.

Answer 1

Answer: (A)

The displacement of the particle is given by
$x=A \sin (-2 \omega t)+B \sin ^{2} \omega t$
$=-A \sin 2 \omega t+\frac{B}{2}(1-\cos 2 \omega t)$
$=-\left(A \sin 2 \omega t+\frac{B}{2} \cos 2 \omega t\right)+\frac{B}{2}$
This motion represents $S H M$ with an amplitude
$\sqrt{A^{2}+\frac{B}{4}}$ and mean position
$\sqrt{\frac{B}{2}}$.