The gravitational force on a particle of mass \(m\) inside the earth at a distance \(r\) from the center \((r<\text { the radius of the earth } R)\) is \(F=-m g r / R\) (Chapter \(6,\) Section 8 Problem 21). Show that a particle placed in an evacuated tube through the center of the earth would execute simple harmonic motion. Find the period of this motion.

Gravitational force is a fundamental force that attracts two bodies with mass toward one another.
Inside the Earth, at a distance \(r\) from its center, the gravitational force acting on a particle of mass \(m\) is given by:
\[ F = - \frac{m g r}{R} \]
This force is directed toward the center of the Earth, indicating it acts as a restoring force.

This is essential because the restoring force is a key component in simple harmonic motion (SHM).
Understanding the nature and direction of the gravitational force here helps us realize why the particle can exhibit SHM inside the Earth. It always pulls the particle back toward equilibrium, similar to how a spring pulls back after being stretched or compressed.

A restoring force is a force that always acts to bring an object back to its equilibrium position.
In many physical systems, this force is proportional to the displacement of the object from equilibrium.
Mathematically, this force can be represented as:
\[ F = - k x \]
where \(k\) is a constant and \(x\) is the displacement.

In the context of our exercise, the gravitational force inside the Earth acts as a restoring force with:
\[ F = - \frac{m g r}{R} \]
Here, \(k\) is equal to \( \frac{mg}{R} \) and \(x\) is the distance \(r\) from the center.
This proportional relationship ensures that the force is always directed toward the center of the Earth, making the motion of the particle similar to that of a mass on a spring.

Newton's Second Law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
This can be written as:
\[ F = ma \]
In our exercise, the restoring force inside the Earth is given by:
\[ F = -\frac{m g r}{R} \]
Inserting this force into Newton's second law gives us:
\[ ma = - \frac{m g r}{R} \]
Since acceleration \(a\) can be expressed as \(\frac{d^2 r}{dt^2}\), we arrive at:
\[ m \frac{d^2 r}{dt^2} = -\frac{m g r}{R} \]
This equation will help us identify the simple harmonic motion.

Differential equations describe the relationship between a function and its derivatives.
They are vital in modeling continuous systems.
In our example, we derived the differential equation from Newton's second law:
\[ m \frac{d^2 r}{dt^2} = - \frac{m g r}{R} \]
We can simplify this by canceling the mass \(m\) from both sides:
\[ \frac{d^2 r}{dt^2} = - \frac{g r}{R} \]
This equation is a standard form of the simple harmonic motion differential equation:
\[ \frac{d^2 r}{dt^2} = - \omega^2 r \]
where \(\omega\) is the angular frequency.

Angular frequency \(\omega\) is an important concept in simple harmonic motion (SHM).
It describes how quickly the oscillations occur.
From the simplified differential equation:
\[ \frac{d^2 r}{dt^2} = -\frac{g r}{R} \]
we can see that:
\[ \omega^2 = \frac{g}{R} \]
Thus, the angular frequency \(\omega\) is:
\[ \omega = \sqrt{\frac{g}{R}} \]
Now, to find the period \(T\) of the simple harmonic motion, we use the relation:
\[ T = \frac{2\pi}{\omega} \]
Substituting \(\omega\) gives us the period as:
\[ T = 2\pi \sqrt{\frac{R}{g}} \]
This formulates how long it takes for the particle to complete one full oscillation within the Earth.