An interesting, though highly impractical example of oscillation is the motion of an object dropped down a hole that extends from one side of the earth, through its center, to the other side. With the assumption (not realistic) that the earth is a sphere of uniform density, prove that the motion is simple harmonic and find the period. [Note: The gravitational force on the object as a function of the object’s distance rfrom the center of the earth was derived in Example 13.10 (Section 13.6). The motion is simple harmonic if the acceleration \({a_x}\) and the displacement from equilibrium x are related by Eq. (14.8), and the period is then\(T = \frac{{2\pi }}{\omega }\)].

\(T = \frac{{2\pi }}{\omega }\)

The time period is measured because of simple harmonic motion is a periodic oscillation.

Gauss Law states that the “gravitational force on an object at distance rfrom the center of the earth is equal to \(GM\left( r \right)m/{r^2}\)

Where,

M(r) = amount of mass at a distance less than r from the other.

With the assumption that the earth is a sphere of uniform density,

If d is the density of the earth and we assume uniform density,

\(M(r) = d\frac{4}{3}\pi {r^3}\)

If we plug this into \(GM\left( r \right)m/{r^2}\), then we will end up with F = (bunch of constants) r

Thus we have a force which proportional to distance r from center,

By Hook’s law,

F = - kx

Where k is spring constant

Now, F= - kr

Where, k= bunch of constant

We know that,

\(\begin{aligned}{c}{\rho _{earth}} = \frac{M}{V}\\ = \frac{M}{{\frac{4}{3}\pi {r^3}}}\end{aligned}\)

For this case, the period of an oscillation is,

\(\begin{aligned}{c}T = 2\pi \sqrt {\frac{{m{R_{earth}}}}{{mg}}} \\ = 2\pi \sqrt {\frac{{{R_{earth}}}}{g}} \;\;\;\;\;\;\;\;\;\;\;\left( {\frac{{{R_{earth}}}}{g} = 6.37 \times {{10}^6}\;{\rm{m}}} \right)\\ = 5065\;{\rm{s}}\end{aligned}\)

Hence the period is 5065s.