91Ó°ÊÓ

1. Amplitude of platform oscillation, $\mathrm{x}=2.2\mathrm{cm}\mathrm{or}2.2×{10}^{-2}{\mathrm{m}}^2$.
2. Frequency of platform oscillation, $\mathrm{f}=6.60\mathrm{Hz}$.

In a simple harmonic motion, the body undergoes acceleration for amplitude. Maximum acceleration occurs when the object is at end of its path. At those points, the force acting on the object is also maximum.

Formula:

The acceleration of a body in simple harmonic motion is directly proportional to the displacement, given by

${\mathrm{a}}_{\mathrm{m}}={\mathrm{Ó\neg }}^2\mathrm{x}$ (i)

To find maximum acceleration, the angular frequency can be shown as follows:

$\begin{array}{rcl}\mathrm{Ó\neg }& =& 2\mathrm{Ï€´Ú}\\ & =& 2\mathrm{Ï€}×6.60\left(âˆ\mu \mathrm{f}=6.60\mathrm{Hz}\right)\\ & =& 41.448\mathrm{rad}/\sec \end{array}$

Now using equation (i) and the given & derived values, the maximum acceleration is as follows:

$\begin{array}{rcl}{\mathrm{a}}_{\mathrm{m}}& =& 41.{469}^2×2.2×{10}^{-2}\\ & =& 37.795\mathrm{m}/{\mathrm{s}}^2\\ & ≃& 37.8\mathrm{m}/{\mathrm{s}}^2\end{array}$

Hence, the value of maximum acceleration is $37.8\mathrm{m}/{\mathrm{s}}^2$.