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 :
$T = 2 π \sqrt{\frac{R}{g}}$
$T = 2 π \sqrt{\frac{2 R}{g}}$
$T = 2 π \sqrt{\frac{R}{2 g}}$
none of these

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 :
$T = 2 π \sqrt{\frac{R}{g}}$
$T = 2 π \sqrt{\frac{2 R}{g}}$
$T = 2 π \sqrt{\frac{R}{2 g}}$
none of these

Toppr Admin · Accepted Answer

As this case is similar to a simple pendulum- Here normal reaction works in the place of tension- So-T-2-x3C0-x221A-RgAnalogus to T-2-x3C0-x221A-lg -simple pendulum-

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 :T=2π√RgT=2π√2RgT=2π√R2gnone of these

As this case is similar to a simple pendulum. Here normal reaction works in the place of tension. So,T=2π√RgAnalogus to T=2π√lg [simple pendulum]

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 :
$T = 2 π \sqrt{\frac{R}{g}}$
$T = 2 π \sqrt{\frac{2 R}{g}}$
$T = 2 π \sqrt{\frac{R}{2 g}}$
none of these

As this case is similar to a simple pendulum. Here normal reaction works in the place of tension. So,
$T = 2 π \sqrt{\frac{R}{g}}$
Analogus to $T = 2 π \sqrt{\frac{l}{g}}$ [simple pendulum]