Q89 / 125 · BITSAT 2015 · Physics

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

  • Correct answer
    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.

Solution

The displacement of the particle is given by: $$ \begin{aligned} x &=A \sin (-2 \omega t)+B \sin ^{2} \omega t \\ &=-A \sin 2 \omega t+\frac{B}{2}(1-\cos 2 \omega t) \\ &=-\left(A \sin 2 \omega t+\frac{B}{2} \cos 2 \omega t\right)+\frac{B}{2} \end{aligned} $$ This motion represents SHM with an amplitude: $\sqrt{A^{2}+\frac{B^{2}}{4}},$ and mean position $$ \frac{B}{2} $$

Hence, the correct answer is option: 'A'.

Relevant Questions