I need a small clarification regarding the following system:

In the centre of momentum frame, $2m$ has a speed of $u/3$ where the centre of mass itself moves in the same direction with a speed $2u/3$. As this is the natural length of the spring, this is the maximum speed block A can have. As this can't be greater than centre of mass's speed, block A will always be in motion. But the answer as per my book says, A's velocity "may be zero at certain instants of time".

Do you agree with me that the block A's velocity can never be zero?

And the book's answer is incorrect?

Suppose we work in the COM frame.

In this frame the two blocks are both performing SHM about their equilibrum positions.

Yes sir. And both are in the mean position at the same time.

As they both have the same time periods this situation remains the same.

Let's call the natural length of the spring $3a$, so the large block oscillates about the position $x = +a$ with some amplitude $A$ and the small block oscillates about the point $x = -2a$ with some amplitude $2A$.

The value of $A$ will depend on the initial conditions.

@GuruVishnu OK so far?

Ok sir.

So for the big block we have $x(t) = +a + A\sin(\omega t)$ and the velocity of the big block is therefore $v = dx/dt = A\omega\cos(\omega t)$

Ok sir. As per this, at $t=0$, A has its maximum speed, am I right?

The initial conditions are that at $t=0$ the small block is stationary, the large block moves with speed $u$ and the spring is at its natural length?

Yes sir.

If that's the case on the return journey (from the right to the left), A will have the same speed ($u/3$) but in the opposite direction. So shifting to the ground frame from the COM frame ($2u/3$ towards right), we get its speed to be $u/3$ towards right.

OK, so in the COM frame the speed of the large block is $v(t) = \tfrac13 u\cos\omega t$

Yes sir.

So add back $\tfrac23u$ to get back to the ground frame and we get $v(t) = u(\tfrac23 + \tfrac13\cos\omega t)$

So you're correct that in the ground frame the speed of the large block can never be zero.

Perfect! Then you agree with me, right? Actually, I didn't use the sinusoidal equation for SHM here, just analysed it qualitatively.

@JohnRennie Thank you very much for the clarification sir! :-)

Sometimes, it takes a lot of effort to prove book is wrong. Because most of the time it's me missing something.

I would be cautious about qualitative arguments as it's so easy to miss something.

Ok sir. I'll have that it mind :-)