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

The time period of small oscillation is $T=2 \pi \sqrt{(} \frac{R}{g} )$.

Explanation:

• The time period of small oscillation can be derived from the equation $T=2 \pi \sqrt{\frac{\text { displacement }}{\text { acceleration }}}$ as the small block oscillates back and forth on a smooth concave surface.  
• We know that the driving force $F=m g \sin \theta$  and acceleration $a=g \sin \theta$.  For small value of $\theta, \sin \theta=\theta$.  
• Therefore, acceleration $a=g \theta$.  Let us consider the displacement from the mean position of the body to be x.  
• Hence, $\theta=\left(\frac{x}{R}\right)=>a=\frac{g}{R}$ where R is radius as the body makes the simple harmonic motion. Hence, on substituting the known values, we get $T=2 \pi \sqrt{(} \frac{R}{g} )$.