Question:

Two objects which have equal masses, \(m\), are connected with a spring whose force constant is \(k\). The objects are placed on a frictionless surface and oscillating without external forces. Find the natural frequency of this motion.

Answer:

\(x_1\) and \(x_2\) are the displacements of each object from the natural length. The displacement, \(x\), is the one for the spring, so \(x=x_1-x_2\) for object 1 and \(-x=x_2-x_1\) for object 2. Set up the equations of motion:

\[  mx_1” = -kx  \]

\[  mx_2” = kx  \]

Subtract (2) from (1).

\[  m(x_1”-x_2”) = -2kx  \]

Since \(x” = x_1”-x_2”\), we have

[  mx” + 2kx = 0 \quad \rightarrow \quad x” + \frac{2k}{m} = 0  \]

The solution of this differential equation is

\[  x = A\sin\left(\sqrt{\frac{2k}{m}} t\right) + B\cos\left(\sqrt{\frac{2k}{m}} t\right)  \]

where \(\sqrt{\frac{2k}{m}}\) is the natural frequency of this spring motion.