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

  $\frac{4 \pi^{2} \mathrm{f}^{2} \mathrm{~L}}{\mathrm{M}}$
• Correct answer
  D

  $2 \pi^{2} \mathrm{f}^{2} \mathrm{~L}$

Hence, the correct answer is option: 'D'.