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Simple Harmonic Motion

MHT CET 2024 11th May Morning Shift

MCQ+1 / -02024

A particle performing S.H.M. starts from equilibrium position and its time period is 12 second. After 2 seconds its velocity is $\pi \mathrm{m} / \mathrm{s}$. Amplitude of the oscillation is $\left[\sin 30^{\circ}=\cos 60^{\circ}=0 \cdot 5, \sin 60^{\circ}=\cos 30^{\circ}=\sqrt{3} / 2\right]$

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