Question

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance $\left(\frac{R}{2}\right)$ from the earth's centre, where $R$ is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period:

• A. $2\pi \sqrt{\frac{R}{g}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{g}{R}}$
• C. $\frac{2\pi R}{g}$
• D. $\frac{g}{2\pi R}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{R}{g}}$

$\cos \theta =\frac{x}{d}$

If displaced from equilibrium position,

$F_{restoring}=\left(\frac{GMmd}{R^3}\right)\cos \theta$

$F_{restoring}=\frac{GMmd}{R^3}.\frac{x}{d}=\frac{GMmx}{R^3}$

$a=\frac{GMx}{R^3}$

we know that the time period given by,

$T=2\pi \sqrt{\left|\begin{array}{c}\frac{x}{a}\end{array}\right|}$

$T=2\pi \sqrt{\frac{\frac{x}{GMx}}{R^3}}$

$T=2\pi \sqrt{\frac{R}{g}}$ $(\because g=\frac{GM}{R^2})$