Two identical blocks (each of mass m) are connected by a spring in its natural length with force constant k. The system is kept on a smooth horizontal surface. At $t=0$, the left side block is given a sharp impulse J towards the right and the blocks begin to slide along the table. If $\omega$ is the angular frequency of the oscillations, the velocity of left block as a function of time is $\frac{2J}{\alpha m}(1+\cos \omega t)$. Find the value of $\alpha$.

$u_1=\frac{J}{m},u_2=0$

$$\begin{gathered} v_{COM}=\frac{m.\frac{J}{m}+m.\left(0\right)}{m+m} \\ {v}_{COM}=\frac{J}{2m} \end{gathered}$$

In centre of mass frame of reference the blocks execute SHM only. (and do not slide rightwards in centre of mass frame)

$$\begin{gathered} u_{1,COM}=u_1-u_{COM}=\frac{J}{m}-\frac{J}{2m}=\frac{J}{2m} \\ {u}_{2,COM}=u_2-u_{COM}=0-\frac{J}{2m}=\frac{J}{2m} \end{gathered}$$

In centre of mass frame, the blocks are executing SHM with

$\omega =\sqrt{\frac{k}{\mu }}=\sqrt{\frac{k}{m.m/(m+m)}}=\sqrt{\frac{2k}{m}}$

At initial moment spring is relaxed. Therefore,

$$\begin{gathered} x_{1,COM}=A\sin \omega t \\ and x_{2,COM}=A\sin (\omega t+\pi ) \\ {v}_{1,COM}=\frac{d{x}_{1,COM}}{dt}=\frac{J}{2m}\cos \omega t \\ {v}_{1,COM}=v_1-v_{COM} \\ \Rightarrow v_1=v_{1COM}+v_{COM} \\ or v_1=\frac{J}{2m}\cos \omega t+\frac{J}{2m}=\frac{J}{2m}(1+\cos \omega t) \\ or in ground frame \\ {v}_1=\frac{2J}{4m}(1+\cos \omega ) \\ \therefore \alpha =4 \end{gathered}$$