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A highly rigid cubical block A of small mass M and side L is fixed rigidly to another cubical block B of the same dimensions and modulus of rigidity $\eta$ Such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on horizontal surface. A small force is applied perpendicular to one of the side face of A. After the force is withdrawn block A executes small oscillations the time period of which is given by

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

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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
$T= 2 \pi \sqrt {\large \bigg|\frac{x}{a}\bigg|}$
$\qquad= 2 \pi\sqrt {\frac{M}{\eta L}}$
answered Oct 8, 2013 by meena.p
edited Mar 21, 2014 by balaji.thirumalai

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