\[\begin {array} {1 1} (a)\;2 \pi \sqrt {M \eta L} & \quad \\ (b)\;2 \pi \sqrt {\frac{M \eta}{L}} & \quad \\ (c)\;2 \pi \sqrt {\frac{M }{\eta L}} & \quad \\ (d)\;4 \pi \sqrt {\frac{M}{\eta L}}\end {array}\]

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+1 vote

+1 vote

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

$T= 2 \pi \sqrt {\large \bigg|\frac{x}{a}\bigg|}$

$\qquad= 2 \pi\sqrt {\frac{M}{\eta L}}$

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