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JEE PHYSICS MCQ: OSCILLATIONS

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A particle performing S.H.M. starts from equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is

\pi \mathrm{~m} / \mathrm{s}

. Amplitude of oscillation is

\left(\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right)

View answer and explanation

Correct option: D

8 \sqrt{2} \mathrm{~m}

Option D (

8 \sqrt{2} \mathrm{~m}

) is correct because the amplitude of oscillation is calculated using the given velocity, time period, and initial conditions of the SHM.

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