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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Physics
Chapter
Oscillations

BITSAT 2015 · Physics · Oscillations

The displacement of a particle is given at time t, by: x=A (-2 ω...

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

Options

  1. Athe motion of the particle is SHM with an amplitude of A2+B24\sqrt{A^{2}+\frac{B^{2}}{4}}
  2. Bthe motion of the particle is not SHM, but oscillatory with a time period of T=π/ωT=\pi / \omega
  3. Cthe motion of the particle is oscillatory with a time period of T=π/2ωT=\pi / 2 \omega
  4. Dthe motion of the particle is a periodic.

Correct answer

A. the motion of the particle is SHM with an amplitude of A2+B24\sqrt{A^{2}+\frac{B^{2}}{4}}

Step-by-step solution

The displacement of the particle is given by: x=Asin⁡(−2ωt)+Bsin⁡2ωt=−Asin⁡2ωt+B2(1−cos⁡2ωt)=−(Asin⁡2ωt+B2cos⁡2ωt)+B2\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} \cos2 \omega t\right)+\frac{B}{2} \end{aligned} This motion represents SHM with an amplitude: A2+B24,\sqrt{A^{2}+\frac{B^{2}}{4}}, and mean position B2\frac{B}{2}
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