Question

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

• A. $\pi \sqrt{\frac{2R}{g}}$
• B. $2\pi \sqrt{\frac{R}{3g}}$
• C. $2\pi \sqrt{\frac{2R}{3g}}$
• D. $2\pi \sqrt{\frac{R}{g}}$

Answer 1

Answer: (D)

$2\pi \sqrt{\frac{R}{g}}$

Let,

R be the radius of smooth concave surface and m is mass of the block. Also, let the $\theta$ is the angle traced by the position of block on smooth concave surface with the normal at mean position while performing small oscillations . So, the displacement, say $x$ of block is,

$x=R\sin \theta$

For small angles $\sin \theta =\theta$. hence, displacement will be

$x=R\theta$ ...................(1)

The force $F$ responsible for to and fro motion of block is,

$F=mg\sin \theta$

$F=mg\theta$ .......................(2)

Therefore angular acceleration is,

$a=\frac{F}{m}$

$a=\frac{mg\theta }{m}$ ...................from(1) and (2)

$a=g\theta$ .........................(3)

Now, the angular velocity $\omega$ of the block is,

$\omega =\sqrt{\frac{a}{x}}$

$\omega =\sqrt{\frac{g\theta }{R\theta }}$

$\omega =\sqrt{\frac{g}{R}}$ ..........................(4)

Put, $\omega =\frac{2\pi }{T}$ in (4)

where T is the time period of small oscillations.

$\frac{2\pi }{T}=\sqrt{\frac{g}{R}}$

Therefore,

$T=2\pi \sqrt{\frac{R}{g}}$