Natural frequency: Two objects with a spring

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:
$$\begin{eqnarray} mx_1'' &=& -kx  \\ mx_2'' &=& kx \end{eqnarray}$$
Subtract (2) from (1).
$$\begin{equation} m(x_1''-x_2'') = -2kx \end{equation}$$
Since $x'' = x_1''-x_2''$, we have
$$\begin{equation} mx'' + 2kx = 0 \quad \rightarrow \quad x'' + \frac{2k}{m} = 0 \end{equation}$$
The solution of this differential equation is
$$\begin{equation} x = A\sin\left(\sqrt{\frac{2k}{m}} t\right) + B\cos\left(\sqrt{\frac{2k}{m}} t\right) \end{equation}$$
where $\sqrt{\frac{2k}{m}}$ is the natural frequency of this spring motion.

Powered by Hirophysics.com