A small block oscillates back and forth on as smooth concave surface of radius R in figure. Find the time period of small oscillation.

Answer

`=2sqrt(r/g)`

Explanation

Given that `r=`radius
Let `n=`normal reaction
Driving force F=`mgsintheta`
Acceleration `=a=g(theta)`
As `sintheta` is very small `sinthetararrtheta`
Acceleratioin `a=g(theta)`
let x be the displacement from the mean position of the body
`:. theta=x/r`
`rarr a=g(theta)=g(x/r)`
`rarr (a/x)=(g/r)`
so teh body makes S.H.M
`T=2pisqrt(x/a)`
`=2pi sqrt(x/(gx/r))=2sqrt(r/g)`