Question

A cube of mass \( M \) and side \( L \) is fixed on the horizontal surface. Modulus of rigidity of the material of the cube is \( \eta \). A force is applied perpendicular to one of the side faces. When the force is removed, the cube executes small oscillations. The time period is

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

Hint:

For oscillations of rigid bodies, the time period depends on the moment of inertia and the restoring force constant, which can be related to properties like the modulus of rigidity.

Answer 1

The Correct Option is C

Step 1: Understanding the time period of oscillation.
The time period \( T \) of small oscillations of a body can be given by the formula for oscillations due to restoring force: \[ T = 2\pi \sqrt{\frac{I}{K}} \] where \( I \) is the moment of inertia and \( K \) is the restoring force constant. For this system, \( K \) is related to the modulus of rigidity \( \eta \).
Step 2: Moment of inertia and restoring force.
For a cube fixed on a surface, the moment of inertia \( I \) about an axis perpendicular to one side is given by: \[ I = \frac{1}{6} M L^2 \] The restoring force constant \( K \) is related to the modulus of rigidity \( \eta \) and the length \( L \) of the cube. The time period \( T \) is then: \[ T = 2\pi \frac{\sqrt{M}}{\eta \sqrt{L}} \] Step 3: Conclusion.
Thus, the time period of oscillation is \( 2\pi \frac{\sqrt{M}}{\eta \sqrt{L}} \), corresponding to option (C).