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Question: Answered & Verified by Expert
A man of mass ' $\mathrm{M}^{\prime}$ ' is standing on the platform. The platform is executing S.H.M. of
frequency 'f'in vertical direction. The span of oscillation is 'L'. Then the
acceleration of the platform at the top of the oscillation is
PhysicsOscillationsMHT CETMHT CET 2020 (20 Oct Shift 1)
Options:
  • A $4 \pi^{2} \mathrm{f}^{2} \mathrm{~L}$
  • B $\frac{2 \pi^{2} \mathrm{f}^{2} \mathrm{~L}}{\mathrm{M}}$
  • C $\frac{4 \pi^{2} \mathrm{f}^{2} \mathrm{~L}}{\mathrm{M}}$
  • D $2 \pi^{2} \mathrm{f}^{2} \mathrm{~L}$
Solution:
2585 Upvotes Verified Answer
The correct answer is: $2 \pi^{2} \mathrm{f}^{2} \mathrm{~L}$
$a=-\omega^{2} L=-4 \pi^{2} f^{2} L$

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