Question

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

Answer 1

It is given that R is the radius of the concave surface.
​Let N be the normal reaction force.
Driving force, F = mg sin θ
Comparing the expression for driving force with the expression, F = ma, we get:
Acceleration, a = g sin θ
Since the value of θ is very small,
∴ sin θ → θ
∴ Acceleration, a = gθ
Let x be the displacement of the body from mean position.

\[\therefore \theta = \frac{x}{R}\] 

\[ \Rightarrow a = g\theta = g\left( \frac{x}{R} \right)\] 

\[ \Rightarrow \left( \frac{a}{x} \right) = \left( \frac{g}{R} \right)\]

\[\Rightarrow a = x\frac{g}{R}\]

As acceleration is directly proportional to the displacement. Hence, the body will execute S.H.M.

Time period \[\left( T \right)\] is given by,

\[T = 2\pi\sqrt{\frac{\text { displacement }}{\text { Acceleration }}}\]

\[= 2\pi\sqrt{\frac{x}{gx/R}} = 2\pi\sqrt{\frac{R}{g}}\]