Question

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.

Answer 1

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 2x.
If v is the velocity of the block. Then
Using energy method, we can write:

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

⇒ mv² + 2kx² = C
By taking the derivative of both sides with respect to t, we get:

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

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

\[  ma + 2kx = 0                                \] 

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

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

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