Question

Consider a fixed ring shaped uniform body of linear mass density $\rho$ and radius $R$.A particle at the center of ring is displaced along the axis by a small distance,show that the particle will execute SHM under gravitation of ring & find its time period neglecting other forces

• A. $\sqrt{\frac{8\pi R^2}{Gp}}$
• B. $\sqrt{\frac{4\pi R^2}{Gp}}$
• C. $\sqrt{\frac{2\pi R^2}{Gp}}$
• D. None of these

Answer 1

The correct option is C $\sqrt{\frac{2\pi R^2}{Gp}}$

Let the displacement of the particle from the center of the ring be $x$ and $\theta$ be the angle between line joining particle to center of ring to that joining particle to any point on the ring.
Hence $cos\theta =\frac{x}{\sqrt{x^2+R^2}}$
Force acting on particle towards the ring's center$=F=\frac{GMm}{R^2+x^2}cos\theta$
$=\frac{GMmx}{(R^2+x^2{)}^{3/2}}$

for small $x$

$F=\frac{GMm}{R^3}x=kx$

$⟹k=\frac{GMm}{R^3}=m{\omega }^2$

$⟹\omega =\sqrt{\frac{GM}{R^3}}=\sqrt{\frac{G\rho 2\pi R}{R^3}}$$=\sqrt{\frac{2G\rho \pi }{R^2}}$

Hence time period $=\frac{2\pi }{\omega }$

$=\sqrt{\frac{2\pi R^2}{G\rho }}$