What's wrong with this mass on a spring solution?

The error occurs in the problem in a tacit assumption made in the first paragraph. Although the lighter mass' amplitude is 0.5 cm from its equilibrium position, it is wrongly assumed that that the spring is stretched/compressed 0.5 cm from its nominal length at this time.

In fact, the spring is stretched/compressed 0.5 cm plus the displacement of the larger mass. This affects the energy calculations. Because no net forces are acting from outside the system, the center of mass is not accelerating - or vibrating - and the total momentum of the system is constant. If we take the frame of reference in which the overall velocity of the center of mass is zero, we can treat the system as if the spring were pinned at the center of mass, which will make each spring part into two separate springs.

If the spring constant of the overall spring is k, then the force exerted by a fraction f of the spring when stretched by length x will be the same as if the overall spring were stretched by x/f, making the effective spring constant of the part between the center of mass and the lighter mass k/f. If m is the mass of the lighter weight, then the heavier mass will be 2m. Because the momentum of the system is constant, the momentum of the two masses always equal and opposite, so

m*v1 + (2m)*v2 = 0

and the velocity of the heavier mass v2 is -1/2 that of the lighter one.

Because the displacement is merely a time integral of the velocity, we can get the displacement of the heavier mass from its equilibrium position at any time by multiplying the light mass' displacement by the constant -½. The heavier mass' amplitude is therefore 0.25 cm and its total displacement is 0.5 cm.

We have, by a completely different method, solved the original problem. The problem can of course correctly be solved by using energy if we note that the value of f for the lighter mass will be

2m / (2m + m) or 2/3. The correct potential energy at maximum displacement will therefore be

½(k/f)x²

= ½k/(⅔)x²

= ¾kx²

Because the velocity of the heavier mass will always be half that of the lighter one, the magnitude of its displacement from its own equilibrium position will also be half that of the lighter mass. The heavier mass' effective spring constant will also be twice that of the lighter mass, giving a maximum potential energy for the heavier mass of

(3/2)k(½x)²

= ⅜kx²

The conclusion that the maximum spring (potential) energy of the heavier mass is half that of the lighter one is therefore correct, but the conclusion that the effective spring constant will be the same for the two masses is not. Therefore the proportionality assumed in equation (2) is also incorrect.

You're overthinking this.

The center of mass doesn't move. It doesn't matter how the masses interact. Via a spring or gravity or electrostatic repulsion or nuclear forces or whatever.

If you move the light mass a certain distance, the heavy mass must move to compensate:

dheavy = dlight * mlight / mheavy

Easy as that.

If you want to know where you went wrong, it was statement (1). You can't calculate the energy of the small mass' oscillation as if it were independent of the larger mass'. The spring energy is 1/2 kx^2 where x is the difference in length of the spring from its equilibrium length, not the difference in equilibrium position of the individual masses.

Au contraire, au contraire. Sorry, doing 1/2 kx^2 for each piece and adding it up won't give you the correct total energy. Why? Because x1^2 + x2^2 does not equal (x1+x2)^2.

Remo's solution, otoh, will work. What's the difference??? Note that a fraction of a spring is stiffer than the entire spring (ie a half a spring is twice as stiff and a third of a spring is three times as stiff). If you take that into account, then yes, you can add up the energies.

But you're still just working in circles and making a mountain out of a molehill of a trivial problem.

The reasoning is good in parts of the solution. For example, since the same spring force acts on both masses at any point in time (in opposite directions), it follows that

m1 x1" + m2 x2" = 0

m1 x1' + m2 x2' = constant = 0 initially

m1 Δx1 + m2 Δx2 = 0

Therefore the heavier mass has half the amplitude as the lighter one.

The given solution though has some dubious terminology which eventually results in an error. For example "The maximum spring energy of the lighter mass is ...". There is only one spring energy. Neither mass has any claim to it. In the end that solution made a similar error to

(a+b)^2 = a^2 + b^2

If he had put the kinetic energies as being 2:1 he would have been correct. Instead he put the contributions to spring energy as being 2:1 and ended up in a messy situation.

I hate reviewing answers and looking for errors, but I would solve all energy, momentum, etc., as follows:

Cut the spring at the center of mass and pretend you then tied each side to a fixed point. That means 1/3rd is on the heavier side, 2/3rds is on the lighter side. Make a new spring constant for both. You will note that both objects now have the same period. All energy, amplitude, etc., calculations for each mass can be derived from there.

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