Physics - SHM Question with Solution | TestHub

The ball is moving on a frictionless curved surface of radius $R$. Since the surface is frictionless, the ball will slide without rotating.

The center of mass of the spherical ball of radius $r$ moves in a circular arc of radius $(R-r)$.

When the ball is displaced by a small angle $\theta$ from its mean position (the lowest point), the restoring force acting on it is the tangential component of its weight:
$F=-mg\sin \theta$

For small oscillations, $\sin \theta \approx \theta$. The linear displacement of the center of mass along the arc is $x=(R-r)\theta$, which gives $\theta =\frac{x}{R-r}$.

Substituting $\theta$ into the force equation:
$F=-mg\left(\frac{x}{R-r}\right)$

The acceleration of the ball is:
$a=\frac{F}{m}=-\left(\frac{g}{R-r}\right)x$

Comparing this with the standard equation of simple harmonic motion, $a=-{\omega }^{2}x$, we get:
${\omega }^2=\frac{g}{R-r}⟹\omega =\sqrt{\frac{g}{R-r}}$

The time period of oscillation is:
$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{R-r}{g}}$

Thus, the time period is proportional to $\sqrt{\frac{R-r}{g}}$.

(Note: Even if the ball were rolling without slipping, the time period would be $T=2\pi \sqrt{\frac{7(R-r)}{5g}}$, which is still proportional to $\sqrt{\frac{R-r}{g}}$.)

Answer: $\sqrt{\frac{\mathrm{R}-\mathrm{r}}{\mathrm{g}}}$