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.

Answer 1

Give that, r = radius,

Let, n = normal reaction

Driving force $F=mgsin\theta$

Acceleration $=a=g\theta$

As $sin\theta$ is very small, $sin\theta \to \theta$

$\therefore$ Acceleration, $a=g\theta$

Let 'x' be the displacement from the mean position of the body.

$\therefore \theta =\frac{x}{r}$

$\Rightarrow a=g\theta =g\left(\frac{x}{t}\right)$

$\Rightarrow \left(\frac{a}{x}\right)=\left(\frac{g}{r}\right)$

So the body makes S.H.M.

$T=2r\pi \sqrt{\frac{\text{displacement}}{\text{Acceleration}}}$

$=2\pi \sqrt{\frac{x}{gx/r}}=2\sqrt{\frac{r}{g}}$