Two Masses on a spring

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Homework Statement


Two masses, M and 4M, are on opposite ends of a massless spring, sitting on a frictionless, horizontal table. The spring is compressed, and the spring-and-masses system is released from rest. If mass M oscillates with amplitude A and frequency f, then mass 4M will oscillate with (HINT: Consider the center of mass!)


Homework Equations



F=-kx
xcom = 1/MƩximi

ω2=k/m

f=ω/2∏

The Attempt at a Solution



There is no net external force and i'm assuming the system is compressed to its center of mass which means there is a displacement for M of 4L/5 (L being the spring length) and a displacement of L/5 for 4M. So the amplitude of 4M is A/4 because it was displaced 1/4 the distance of M.

Using ω2=k/m
and f=ω/2∏ I found the frequency of 4M to be f/2. This is wrong acc. to webassign tho so i'm confused if my whole approach is wrong or just my calculations? Thanks
 

Answers and Replies

  • #2
Doc Al
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Using ω2=k/m
and f=ω/2∏ I found the frequency of 4M to be f/2. This is wrong acc. to webassign tho so i'm confused if my whole approach is wrong or just my calculations? Thanks
That formula gives the frequency for a mass at the end of spring where the other end is fixed. In any case, you are given the frequency of one mass. How must the frequency of the other mass relate to it? Again, consider the center of mass.
 
  • #3
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So my amplitude is correct though?


As for the frequency, I am guessing at this point I have to find a relationship whereby the frequency of the COM will be zero?

Thanks
 
  • #4
Doc Al
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So my amplitude is correct though?
Yep.
As for the frequency, I am guessing at this point I have to find a relationship whereby the frequency of the COM will be zero?
Well, the COM doesn't move, if that's what you mean. Hint: This should require no calculation at all.
 
  • #5
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Ok thanks. Yes, that is what I meant. I am not fully understanding though why their frequencies should be equal. If the center of mass isn't moving, then don't their frequencies have to be weighted by their masses, and so the frequency of the larger one should be smaller?

Or is the weighing only with velocities, not frequencies? And so since the amplitude of the larger mass is smaller, with its frequency the same its velocity is smaller, keeping the center of mass at zero momentum?

Thanks
 
  • #6
Doc Al
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I am not fully understanding though why their frequencies should be equal. If the center of mass isn't moving, then don't their frequencies have to be weighted by their masses, and so the frequency of the larger one should be smaller?
Think about it. In order for the COM not to move, the two masses better move in phase. When M moves in, so must 4M. (Different distances, of course.)

If they had different frequencies they would get out of phase, which is inconsistent with a fixed COM.

Another way to look at this: Imagine the COM is fixed and two smaller springs are connected to it. These smaller springs have spring constants > k. (Figure out those spring constants.) Then you can treat each mass as being on the end of a spring with the other end fixed.
 
  • #7
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Ok I get it, thanks. Is that in phase strictly speaking though, or is that 180 degrees out of phase? Because they are going in different directions.

Also for the smaller springs what do you mean by 'figure out those spring constants?' Assign arbitrary values? Or solve for them? If the latter, I thought constants have to be determined via experiment or given. Or is there some formula for: given an original k, chopping it in two along its length at a certain distance will give you a fixed k1 and k2 for each new spring? Never heard of that, but maybe that's what you mean?

Thanks
 
  • #8
Doc Al
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Ok I get it, thanks. Is that in phase strictly speaking though, or is that 180 degrees out of phase? Because they are going in different directions.
Yes, they move in opposite directions.
Also for the smaller springs what do you mean by 'figure out those spring constants?' Assign arbitrary values? Or solve for them? If the latter, I thought constants have to be determined via experiment or given. Or is there some formula for: given an original k, chopping it in two along its length at a certain distance will give you a fixed k1 and k2 for each new spring? Never heard of that, but maybe that's what you mean?
Yes, that's what I mean. If you have a spring of length L and spring constant k, what would be the spring constant for half a spring (length L/2)?
 
  • #9
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We never learned this in my class, but after doing some research it seems that there are N coils in the denominator of the k equation. So the small mass would have a spring constant of 1.25k and the large mass would have a spring constant of 5k, right?
 
  • #10
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Yes, they move in opposite directions.

So they do move in opposite directions, but is that in phase or antiphase? After thinking about it more I think its actually in phase, not antiphase, because if you rotate one around and align it with the other they are in phase. Just because they are moving in opposite directions doesn't seem to determine in phase vs. antiphase. That's just determined by where there are in the oscillation, regardless of how they are oriented in space, right?
 
  • #11
Doc Al
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We never learned this in my class, but after doing some research it seems that there are N coils in the denominator of the k equation. So the small mass would have a spring constant of 1.25k and the large mass would have a spring constant of 5k, right?
Yes. An easy way to see that is to consider that if a given tension stretches the entire spring by an amount X, the two pieces would stretch by amounts (4/5)x and x/5. Then you can solve for the effective spring constants; for example: F = kx = k'(4/5)x, thus k' = (5/4)k.
 

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