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JEE PHYSICS MCQ: OSCILLATIONS

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The Displacement of a particle varies with time as x=12sin⁡ωt−16sin⁡3ωtx=12 \sin \omega t-16 \sin ^3 \omega t (in cm\mathrm{cm}). if its motion is S.H.M, Then its maximum accelaration is
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Correct option: B
36ω236 \omega^2
Option B (36ω236 \omega^2) is correct because the displacement equation can be simplified to x=4sin⁡(3ωt)x = 4\sin(3\omega t), and the maximum acceleration is Aω2A\omega^2 where AA is the amplitude and ω\omega is the angular frequency.

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