Question

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance$(R/2)$from the earth's centre, where$\text{'}R\text{'}$ 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. $\left(\frac{1}{2\pi }\right)\sqrt{\frac{g}{R}}$
• C. $\frac{2\pi R}{g}$
• D. $\frac{g}{2\pi R}$

Answer 1

The correct option is A

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

Step 1: Given data and drawing the diagram

The radius of the earth $=R$

The perpendicular distance between the earth's center and tunnel, $d=R/2$ for zero displacements.

Let the mass of particle be $m$, $x$ is the displacement of a particle doing Simple Harmonic Motion, $d$ be the maximum distance from maximum displacement $x$ to the center of the earth and $\theta$ be an angle as shown in the figure.

Acceleration of particle $=a$

Time period $=T$

Step 2: To find the acceleration, $\mathit{a}$

From the diagram, $\cos \theta =\frac{x}{d}$

If displaced from the equilibrium position,

$$\begin{gathered} F_{restoring}=\left(\frac{GMmd}{R^3}\right)\cos \theta \\ {F}_{restoring}=\left(\frac{GMmd}{R^3}\right)\frac{x}{d} \\ ma=\frac{GMmx}{R^3} \\ a=\frac{GMx}{R^3} \end{gathered}$$

(Where $G$ is universal gravitational constant, $g$ is gravity, $M$ the mass of Earth.)

Step 3: To find the time period, $\mathit{T}$

We know that the time period is given by,

$T=2\pi \sqrt{\left|\frac{x}{a}\right|}$

$$\begin{gathered} \Rightarrow T=2\pi \sqrt{\frac{x}{{\displaystyle \frac{GMx}{R^3}}}} \\ \Rightarrow T=2\pi \sqrt{\frac{R^3}{GM}} \\ \Rightarrow T=2\pi \sqrt{\frac{R}{g}} \end{gathered}$$

$($Since we know that gravity $g=\frac{GM}{R^2}$$)$

Hence, option (A) is correct.