# Two masses connected to a spring.

1. Feb 13, 2013

### Sefrez

1. The problem statement, all variables and given/known data
Two masses m1 and m2 are connected to a spring of elastic constant k and are free to move without friction along a linear rail. Find the frequency of oscillatory motion of this system.

2. Relevant equations
m1a1 = -m2a2

3. The attempt at a solution

There is equal and opposite force, so momentum is conserved. The force on each mass is proportional to k. I choose a coordinate system to aid in formulating the following differential equations:
m1x1''=-k(l0 - x2 + x1)
m2x2''= k(l0 - x2 + x1)

where x1,2 indicates the respective masses position and l0 is the spring equilibrium length.

Is this a correct approach so far? If so, I have never solved differential equations of this type. How would one find x1, x2? Perhaps I am going about this wrong all together.

2. Feb 13, 2013

### ehild

If you add them you get m1x1"+m2x2"=0 which means that the CM is stationary.

If you isolate x1" and x2" and subtract the first equation from the second, you get a differential equation for the variable u=x2-x1-lo ( the change of length of the spring). That equation for u is an SHM equation with "reduced" mass μ=1/(1/m1+1/m2).

ehild

3. Feb 14, 2013

### rude man

I am posting only to follow this thread. I will use Laplace xfrm to get at an answer.

4. Feb 14, 2013

### vela

Staff Emeritus
I'm nitpicking here, but this actually means the center of mass has a constant velocity, which makes sense because there's no net external force acting on the system.

5. Feb 14, 2013

### ehild

If the two masses were in rest when connected to the ends of the spring, the CM would be in rest, too, and it stayed in rest after releasing the masses. In general case, the CM moves with constant velocity.

Introducing the variable u=x2-x1-l0 u"=x2"-x1"

m1x1''=-k(l0 - x2 + x1) ---->x1"=(k/m1)u
m2x2''= k(l0 - x2 + x1)----->x2"=-(k/m2)u, and subtracting the equations above:

u"=-k(1/m1+1/m2)u

This is the equation of simple harmonic motion u=Acos(ωt+θ) of a particle with mass μ=1/(1/m1+1/m2): u"=-(k/μ)u. The two-body problem has been reduced for the equation of a single body, of mass equal to the reduced mass μ.

The motion of the individual particles can be obtained from the equations m1x1+m2x2=Vt, (or m1x1+m2x2=0 in the center of mass frame of reference ) and x2-x1-l0=u=Acos(ωt+θ).

ehild

6. Feb 14, 2013

### rude man

My answer is in accord with this observation.

7. Feb 14, 2013

### Sefrez

Thanks, people. I believe I have it now.