Oscillations

JEE Main level, single correct, medium

4 of 194

Question 4 of 194

A particle is executing simple harmonic motion. Its amplitude is AA and time period is 55 sec. The time required by it to move from x=Ax = A to x=A2x = \dfrac{A}{\sqrt{2}} is _______ sec.

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Answer: (C) 5/85/8

The equation of simple harmonic motion for a particle starting from the extreme position x=Ax = A is given by:

x=Acos⁡(ωt)x = A \cos(\omega t)

We need to find the time tt when the particle reaches x=A2x = \dfrac{A}{\sqrt{2}}. Substituting this value into the equation:

A2=Acos⁡(ωt)\dfrac{A}{\sqrt{2}} = A \cos(\omega t)

cos⁡(ωt)=12\cos(\omega t) = \dfrac{1}{\sqrt{2}}

ωt=π4\omega t = \dfrac{\pi}{4}

Since the angular frequency ω=2πT\omega = \dfrac{2\pi}{T}, we have:

2πTt=π4\dfrac{2\pi}{T} t = \dfrac{\pi}{4}

t=T8t = \dfrac{T}{8}

Given the time period T=5T = 5 sec, the required time is:

t=58t = \dfrac{5}{8} sec

Answer: 5/85/8

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