Classical Harmonic Oscillator

Consider two masses $m_1$ and $m_2$ at positions $x_1$ and $x_2$, connnected by a spring with spring constant $k$. If the rest length of the spring is $l_0$, then the two equations governing the motion of the masses are

$$m_1 {\ddot x}_1 = k (x_2 - x_1 - l_0)$$ (1)

$$m_2 {\ddot x}_2 = -k (x_2 - x_1 - l_0).$$ (2)

It will be more convenient to work in terms of $(x_2 - x_1 - l_0)$, which is the amount by which the spring is stretched or compressed from its natural length. Denoting this quantity simply as $x$, the equations reduce to

$$m_1 {\ddot x}_1 = k x$$ (3)

$$m_2 {\ddot x}_2 = -k x.$$ (4)

Subtracting the second equation from the first gives

$$m_1 {\ddot x}_1 + m_2 {\ddot x}_2 = 0.$$ (5)

It will be convenient to introduce the total mass and center of mass coordinates, defined as

$$M = m_1 + m_2$$ (6)

$$x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2},$$ (7)

so that we now have

$$M {\ddot x}_{cm} = 0.$$ (8)

This means that the center of mass is not accelerating or decelerating, but is either at rest or moves with constant velocity; of course this simply reflects the fact that there is no external force acting on the two masses.

Now dividing (3) by $m_1$ and (4) by $m_2$ and subtracting the second equation from the first, we can obtain

$${\ddot x}_1 - {\ddot x}_2 = kx \left( \frac{1}{m_1} + \frac{1}{m_2} \right).$$ (9)

Since $x = x_2 - x_1 - l_0$, we have ${\ddot x} = {\ddot x}_2 - {\ddot x}_1$. If we also introduce the reduced mass

$$\mu = \frac{m_1 m_2}{m_1 + m_2},$$ (10)

we obtain

$${\ddot x} = - \frac{k}{\mu} x,$$ (11)

which is a second-order differential equation describing the displacement $x$ from the rest length $l_0$ as a function of time. This can be solved to yield

$$x(t) = A cos (\omega t) + B sin (\omega t),$$ (12)

where $\omega$ is defined as

$$\omega = \sqrt{ \frac{k}{\mu} }$$ (13)

and represents the frequency of oscillation (in rad s$^{-1}$) of the oscillator. One could also define a frequency $\nu$ in Hertz ($s^{-1}$) through $\omega = 2 \pi \nu$.