Two masses connected to a spring.

[Sefrez]

Homework Statement

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.

Homework Equations

m₁a₁ = -m₂a₂

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 x_1,2 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.

 

[ehild]

Your equations are correct.
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

 

[rude man]

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

 

[vela]

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

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.

 

[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

 

[rude man]

Your equations are correct.
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

My answer is in accord with this observation.

 

[Sefrez]

Thanks, people. I believe I have it now.