Modulus of rigidity $\eta = \large\frac{F}{A \theta}$
Here $A=L^2$
and $\theta=\large\frac{x}{L}$ for small '$\theta$'
$\therefore$ restoring force $= F=- \eta A \theta$
$\qquad=-\eta L x$
or acceleration $a= \large\frac{F}{m}$
$\qquad= \frac{-\eta L}{M}$$x$ -------(1)
Since $a \alpha -x$
Since the acceleration is dependent on the displacement linearly we have simple harmonic motion. and hence the
time period is given by
$T= 2 \pi \sqrt {\large \bigg|\frac{x}{a}\bigg|}$
$\qquad= 2 \pi\sqrt {\frac{M}{\eta L}}$