Oscillations

JEE Main level, single correct, medium

53 of 194

Question 53 of 194

A simple harmonic oscillator of angular frequency 2rads−12rad{s}^{-1} is acted upon by an external force F=sin⁡⁡tN.F=\sin ⁡tN.If the oscillator is at rest in its equilibrium position at t=0,t=0, its position at later times is proportional to:

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Answer: (C) sin⁡⁡t−12sin⁡⁡2t\sin ⁡t-\dfrac{1}{2}\sin ⁡2t

It is given that oscillator at rest at t=0t=0 i.e. at t=0t=0, v=0v=0. So, in option, we can check by putting v=dxdt=0v=\dfrac{dx}{dt}=0.

(1) Ifx∝sin⁡⁡t+12cos⁡⁡2tIfx\propto \sin ⁡t+\dfrac{1}{2}\cos ⁡2t

⟹v∝cos⁡⁡t+12×2(−sin⁡⁡2t)⟹v\propto \cos ⁡t+\dfrac{1}{2}\times 2(-\sin ⁡2t)

⟹att=0,v∝1−0≠0⟹att=0,v\propto 1-0\neq 0

(2) Ifx∝cos⁡⁡t−12sin⁡⁡tIfx\propto \cos ⁡t-\dfrac{1}{2}\sin ⁡t

⟹v∝−sin⁡⁡t−12cos⁡⁡t⟹v\propto -\sin ⁡t-\dfrac{1}{2}\cos ⁡t

⟹att=0,v∝−12≠0⟹att=0,v\propto -\dfrac{1}{2}\neq 0

(3) Ifx∝sin⁡⁡t−12sin2tIfx\propto \sin ⁡t-\dfrac{1}{2}sin2t

thenυ∝cos⁡⁡t−12×2cos⁡⁡2tthenυ\propto \cos ⁡t-\dfrac{1}{2}\times 2\cos ⁡2t

⟹att=0,v∝1−1=0⟹att=0,v\propto 1-1=0

(4) Ifx∝sin⁡⁡t+12sin⁡⁡2tIfx\propto \sin ⁡t+\dfrac{1}{2}\sin ⁡2t

⟹v∝cos⁡⁡t+12×2cos⁡⁡2t⟹v\propto \cos ⁡t+\dfrac{1}{2}\times 2\cos ⁡2t

⟹att=0,v∝1+1⟹att=0,v\propto 1+1

⟹v∝2≠0⟹v\propto 2\neq 0

∴inoption(3)v=0att=0∴inoption\left(3\right)v=0att=0

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