Question

A tunnel is dug through the centre of the Earth. Show that a body of mass 'm' when dropped from
rest from one end of the tunnel will execute simple harmonic motion​

Answer 1

Answer:

Time period of SHM inside the Earth is given as

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

Explanation:

As we know that gravitational force inside the Earth at a distance "x" from the center of Earth is given as

$F = \frac{GMmx}{R^3}$

now we know that this force is always towards the center of the Earth

So acceleration of the object is given as

$a = \frac{F}{m}$

$a = \frac{GM}{R^3} x$

now we have

$\omega^2 = \frac{GM}{R^3}$

so time period of the oscillation is given as

$T = \frac{2\pi}{\omega}$

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

#Learn

Topic : SHM

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