Question:

Two blocks of masses $m_1=1~\mathrm{kg}$ and $m_2=2~\mathrm{kg}$ are connected by a spring of spring constant $k=24~\mathrm{N/m}$ and placed on a frictionless horizontal surface. The block $m_1$ is imparted an initial velocity $v_0=12~\mathrm{cm/s}$ to the right. The amplitude of the oscillation is _____.

My approach:

Since there is no horizontal external force, the momentum of the system (spring+blocks) must be conserved. I observed the system from the centre of mass's reference frame. It initially travels with a speed of $4~\mathrm{cm/s}$ relative to ground and hence in it's frame, the block $m_1$ moves towards it with a speed of $8~\mathrm{cm/s}$ and $m_2$ moves towards right with the same speed as of COM i.e., $4~\mathrm{cm/s}$.

I also found the value of spring constants of the spring in between the blocks and the centre of mass.

Using energy conservation in this frame, I got the value of amplitudes of both blocks relative to the centre of mass. But none of them were equal to the options (1,2,3 and 4)

So, I wanted to ask how would we define the amplitude when the system doesn't oscillate in one frame due to it's horizontal motion?

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Whenever you have a system like always consider working in the centre of mass frame. In this case it makes the question very easy.

Yes sir. I'd assume the centre of mass point as a fixed wall and sit on it and observe the system :-)

My doubt was, how do we define amplitude when both the blocks are oscillating and translating in the horizontal direction?

In the centre of mass frame the velocity of m1 is twice the velocity of m2, so in this frame m1 has velocity 8cm/s and m2 has velocity 4cm/sc

Calculate the total KE of both and equate this to the PE of the spring.

Yes sir. And this velocity relation ($v_1=2v_2$) is always true in order for the momentum to be conserved within the centre of mass frame.

I get $A_1=\frac{4}{3}$ cm and $A_2=\frac{2}{3}$ cm.

Does that match an answer?

The correct answer is $A_1+A_2=2$ cm. No problem. But when $m_1$ approaches minimum value, $m_2$ approaches maximum as their velocities are always in the same direction. So shouldn't this be the difference instead $A_1-A_2=\frac{2}{3}$?

Why do we need to consider the sum of amplitudes instead of the difference? The latter is the one we would be observing from the centre of mass frame, right sir?

Short version: Why are we concerned about $A_1+A_2$ instead of $|A_1-A_2|$?

@GuruVishnu we've been a bit cavalier about the signs of the velocities.

The velocities are always equal and opposite, so actually $v_1 = -2v_2$. Yes?

@JohnRennie Yes sir. I think that's where this discrepancy arises. Taking this into consideration, I hope the sum of amplitudes would make sense as both blocks either move towards each other or away from each other.