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Simple Harmonic Motion question

2025 · Shift 1 · Q40
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Simple Harmonic Motion question

2025 · Shift 1 · Q40

JEE AdvancedPhysicsSimple Harmonic MotionNumerical+4 / −1
A person sitting inside an elevator performs a weighing experiment with an object of mass 50 kg . Suppose that the variation of the height yyy(in m ) of the elevator, from the ground, with time ttt(in s) is given by y=8[1+sin⁡(2πtT)]y=8\left[1+\sin \left(\frac{2 \pi t}{T}\right)\right]y=8[1+sin(T2πt​)], where T=40π sT=40 \pi \mathrm{~s}T=40π s. Taking acceleration due to gravity, g=10m/s2g=10\mathrm{m} / \mathrm{s}^2g=10m/s2, the maximum variation of the object's weight (in N ) as observed in the experiment is ‾\underline{\hspace{2cm}}​.
Numerical answer
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