Consider the situation shown in figure . Show that if the blocks are displaced slightly in opposite direction and released, they will execute simple harmonic motion. Calculate the time period.

#### Solution

The centre of mass of the system should not change during simple harmonic motion.

Therefore, if the block *m* on the left hand side moves towards right by distance *x*, the block on the right hand side should also move towards left by distance *x*. The total compression of the spring is 2*x*.

If* v *is the velocity of the block. Then

Using energy method, we can write:

\[\frac{1}{2}k \left( 2x \right)^2 + \frac{1}{2}m v^2 + \frac{1}{2}m v^2 = C\]

*⇒ **mv*^{2} + 2*kx*^{2} = C

By taking the derivative of both sides with respect to *t,* we get:

\[2mv\frac{dv}{dt} + 2k \times 2x\frac{dx}{dt} = 0\]

\[\text { Putting } v = \frac{dx}{dt}; \text { and } a = \frac{dv}{dt}\text { in above expression, we get }\]

\[ ma + 2kx = 0 \]

\[ \Rightarrow - \frac{a}{x} = \frac{2k}{m} = \omega^2 \]

\[ \Rightarrow \omega = \sqrt{\frac{2k}{m}}\]

\[ \Rightarrow \text { Time period }, T = 2\pi\sqrt{\left( \frac{m}{2k} \right)}\]