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Oscillations, two masses on opposite ends of the same spring

by rubenhero
Tags: oscillation, spring
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rubenhero
#1
Nov24-11, 04:06 AM
P: 42
1. The problem statement, all variables and given/known data
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 ?
list of choices are attached

2. Relevant equations
ω2M=k, center of mass, xmax=A


3. The attempt at a solution
for mass 1 : ω2M=k, (2∏f)2M=k
center of mass 1: MA/M+4M = A/5
for mass 2: ω24M=k, (2∏f)24M=k, 4∏(f/2)24M=k
center of mass 2: 4MnA/M+4M = 4nA/5

my explanation:
-since both masses are on the same spring they have the same spring constant so for each of the masses it should be equal but for 4M the frequency had to be f/2 in order for it to be squared and cancel the 4 in 4M, however I'm not sure.
I used the set the equilibrium point as the reference point for center of mass, but got kind of confused.
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grzz
#2
Nov24-11, 04:18 AM
P: 950
There is no resultant force on the WHOLE system.
Hence
1. the centre of mass of the WHOLE system must remain at the same place.

2. At any time, the momentun of M to one side must be equal and opposite to the momentum of 4M to the opposite side.
rubenhero
#3
Nov24-11, 04:32 AM
P: 42
I understand that the the center of mass of the whole system remains in the same place. We weren't taught momentum for these type of questions. I don't understand how I can use center of mass to solve for the amplitude and frequency of mass 4M. I do think conceptually 4M's amplitude should be smaller than A. The frequency i got was wrong.

JHamm
#4
Nov24-11, 07:25 AM
P: 389
Oscillations, two masses on opposite ends of the same spring

If you know the center of mass doesn't move then you know that [itex]r_1m = -4r_2m [/itex] where [itex] r_1 [/itex] and [itex] r_2 [/itex] are the position vectors of the [itex] m [/itex] and [itex] 4m [/itex] particles respectively. Now since this relation has to be true, if the particle of mass [itex] m [/itex] suddenly came to a stop and started changing direction, would it be possible for the [itex] 4m [/itex] particle to keep going the way it was going?
More mathematically we see that [itex] \dot{r_1} = -4\dot{r_2} [/itex] so [itex] \dot{r_1} = 0 \Rightarrow \dot{r_2} = 0 [/itex]
rubenhero
#5
Nov24-11, 01:54 PM
P: 42
Since the forces on both masses must be equal and opposite in magnitude and direction respectively then that means mass 4M has to stop when mass M stops and if one of the masses move toward the equilibrium point again then the other has to do the same. The amplitude of mass 4M would have to be a fourth of the amplitude of mass M in order to balance out. The masses are oscillating on the same center of mass so the period would be the same, which means the frequency is the same. linear momentum (mvx) would be the same and velocity is the same because the frequency is the same. Is this conclusion right or close to being right?
JHamm
#6
Nov25-11, 04:21 AM
P: 389
The velocity of the two masses aren't the same, you can see from their equal momenta that their velocities differ by a multiple of 4.
rubenhero
#7
Nov25-11, 05:20 AM
P: 42
If the velocities have to be different would that mean: mvx = mvA = m(wA)A so in comparison of the two masses it would be m*(wA)*A = 4m*(wA/2)*A/2. I don't understand why you said the velocity would differ by a multiple of 4 when the amplitude has to be accounted for too.
JHamm
#8
Nov25-11, 08:21 AM
P: 389
Fair enough let me give it a crack :)
If the two momenta are the same at every point (in magnitude, the directions are opposite) in time we have that
[tex] p_1 = -p_2 [/tex]
[tex] m_1v_1 = -m_2v_2 [/tex]
Now since one of the masses is 4 times the other we can rewrite
[tex] mv_1 = -4mv_2 \Rightarrow v_1 = -4v_2 [/tex]
So since the momenta are equal in magnitude and the masses differ by a multiple of 4 the velocities must also differ by a multiple of 1/4 :)
Now for your question on amplitudes, this depends on the frame of reference you choose, if you take it to be the frame of the center of the spring then yes the amplitudes will be the same, but since the center of the spring moves we must change the velocities from our calculated ones (which would be a reference frame with the center of the spring motionless before any oscillations begin) so that they would in fact become equal.
In the frame we chose the distance each mass moves from the "original" center of the spring won't be the same.

If you have any more questions feel free to post away :)


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