A man of mass ' $\mathrm{M}^{\prime}$ ' is standing on the platform. The platform is executing S.H.M. of…

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
  1. $4 \pi^{2} \mathrm{f}^{2} \mathrm{~L}$
  2. $\frac{2 \pi^{2} \mathrm{f}^{2} \mathrm{~L}}{\mathrm{M}}$
  3. $\frac{4 \pi^{2} \mathrm{f}^{2} \mathrm{~L}}{\mathrm{M}}$
  4. $2 \pi^{2} \mathrm{f}^{2} \mathrm{~L}$

Solution

$a=-\omega^{2} L=-4 \pi^{2} f^{2} L$

Asked in: MHT CET 2020 (20 Oct Shift 1)

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