MHT CET · Physics · Oscillations
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
- 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 \(2 \pi^{2} \mathrm{f}^{2} \mathrm{~L}\)
Answer & Solution
Correct Answer
(D) \(2 \pi^{2} \mathrm{f}^{2} \mathrm{~L}\)
Step-by-step Solution
Detailed explanation
\(a=-\omega^{2} L=-4 \pi^{2} f^{2} L\)
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