Assume that a tunnel is dug across the earth (radius = R) passing through its centre. Find the time a particle takes to cover the length of the tunnel if (a) it is projected into the tunnel with a speed of \[\sqrt{gR}\] (b) it is released from a height R above the tunnel (c) it is thrown vertically upward along the length of tunnel with a speed of \[\sqrt{gR}\]

#### Solution

Given:

Radius of the earth is *R*.

Let *M *be the total mass of the earth and \[\rho\] be the density.

Let mass of the part of earth having radius *x* be *M*.

\[\therefore \frac{M'}{M} = \frac{\rho \times \frac{4}{3}\pi x^3}{\rho \times \frac{4}{3}\pi R^3} = \frac{x^3}{R^3}\]

\[ \Rightarrow M' = \frac{M x^3}{R^3}\]

Force on the particle is calculated as,

\[F_x = \frac{GM'm}{x^2}\]

\[ = \frac{GMm}{R^3}x \ldots\left( 1 \right)\]

Now, acceleration \[\left( a_x \right)\] of mass *M*' at that position is given by,

\[a_x = \frac{GM}{R^3}x\]

\[ \Rightarrow \frac{a_x}{x} = \omega^2 = \frac{GM}{R^3} = \frac{g}{R} \left( \because g = \frac{GM}{R^2} \right)\]

\[\text{So, Time period of oscillation ,} T = 2\pi\sqrt{\left( \frac{R}{g} \right)}\]

(a) Velocity-displacement equation in S.H.M is written as,

\[V = \omega\sqrt{\left( A^2 - y^2 \right)} \]

where, A is the amplitude; and y is the displacement .

When the particle is at *y* = R,

The velocity of the particle is \[\sqrt{gR}\] and \[\omega = \sqrt{\frac{g}{R}}\]

On substituting these values in the velocity-displacement equation, we get:

\[\sqrt{gR} = \sqrt{\frac{g}{R}}\sqrt{A^2 - R^2} \]

\[ \Rightarrow R^2 = A^2 - R^2 \]

\[ \Rightarrow A = \sqrt{2R}\]

Let *t*_{1} and *t*_{2}_{ }be the time taken by the particle to reach the positions X and Y.

Now, phase of the particle at point X will be greater than \[\frac{\pi}{2}\] but less than \[\pi\]

Also, the phase of the particle on reaching Y will be greater than \[\pi\] but less than \[\frac{3\pi}{2}\]

Displacement-time relation is given by,*y* = *A* sin *ωt*

Substituting* y* =* R *and A =\[\sqrt{2R}\] , in the above relation , we get :

\[R = \sqrt{2}R \sin \omega t_1\]

\[\Rightarrow \omega t_1 = \frac{3\pi}{4}\]

Also,

\[R = \sqrt{2}R \sin \omega t_2\]

\[\Rightarrow \omega t_2 = \frac{5\pi}{4}\]

\[\text{So}, \omega\left( t_2 - t_1 \right) = \frac{\pi}{2}\]

\[ \Rightarrow t_2 - t_1 = \frac{\pi}{2\omega} = \frac{\pi}{2\left( \sqrt{\frac{g}{R}} \right)}\]

Time taken by the particle to travel from X to Y:

*R*

Using the principle of conservation of energy, we get:

Change in P.E. = Gain in K.E.

\[\Rightarrow \frac{GMm}{R} - \frac{GMm}{2R} = \frac{1}{2}m v^2 \]

\[ \Rightarrow v = \sqrt{\left( gR \right)}\]

As the velocity is same as that at X, the body will take the same time to travel XY.

(c) The body is projected vertically upwards from the point X with a velocity \[\sqrt{gR}\].Its velocity becomes zero as it reaches the highest point.

The velocity of the body as it reaches X again will be,

\[v = \sqrt{\left( gR \right)}\]

Hence, the body will take same time *i.e.*