Question

A small block oscillates back and forth on a smooth concave surface of radius R (figure 12−E17). Find the time period of small oscillation.

Figure

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.
$$\begin{gathered} \therefore \mathrm{\theta }=\frac{x}{R} \\ \Rightarrow a=g\mathrm{\theta }=g\left(\frac{x}{R}\right) \\ \Rightarrow \left(\frac{a}{x}\right)=\left(\frac{g}{R}\right) \end{gathered}$$
$\Rightarrow a=x\frac{g}{R}$
As acceleration is directly proportional to the displacement. Hence, the body will execute S.H.M.

Time period$\left(T\right)$ is given by,
$T=2\mathrm{\pi }\sqrt{\frac{\mathrm{displacement}}{\mathrm{Acceleration}}}$
$=2\mathrm{\pi }\sqrt{\frac{x}{gx\mathit{/}R}}\mathit{=}2\mathrm{\pi }\sqrt{\frac{R}{g}}$