Question

A highly rigid cubical block A of small mass M and side L is fixed rigidly on the other cubical block of same dimensions and of low modulus of rigidity n such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the side faces of A. after the force is withdrawn, block A executes small oscillations, the time period of which is given by

• A. $2\pi \sqrt{M\eta L}$
• B. $2\pi \sqrt{\frac{M\eta }{L}}$
• C. $2\pi \sqrt{\frac{ML}{\eta }}$
• D. $2\pi \sqrt{\frac{M}{\eta L}}$

Answer 1

The correct option is D $2\pi \sqrt{\frac{M}{\eta L}}$


Modulus of rigidity $\eta =\frac{F}{A\theta }$
Here, $A=L^2$
And $\theta =\frac{x}{L}$ for small $\theta$
$\therefore$ restoring force $=F=-\eta A\theta$ $=-nLx$
Or acceleration $a=\frac{F}{m}$
$=\frac{-\eta L}{M}x\dots \dots \left(1\right)$
Since $a\alpha -x$
The acceleration is dependent on the displacement linearly, we have simple harmonic motion and hence the time period is given by
$T=2\pi \sqrt{\left|\frac{x}{a}\right|}$
$=2\pi \sqrt{\frac{M}{\eta L}}$
Dimensionally we can check, only option that has dimension of time as -
${\left[\frac{M}{M{L}^{-1}{T}^{-2}L}\right]}^{1/2}=\left[T\right]$