Question

A small block oscillates back and forth on a smooth concave surface of radius $R$. The time period of small oscillation of the block is :

• A. $T=2\pi \sqrt{\frac{R}{g}}$
• B. $T=2\pi \sqrt{\frac{2R}{g}}$
• C. $T=2\pi \sqrt{\frac{R}{2g}}$
• D. none of these

Answer 1

The correct option is A $T=2\pi \sqrt{\frac{R}{g}}$

As this case is similar to a simple pendulum. Here normal reaction works in the place of tension. So,

$T=2\pi \sqrt{\frac{R}{g}}$

Analogus to $T=2\pi \sqrt{\frac{l}{g}}$ [simple pendulum]