Two masses connected by spring, find period of oscillation

[Aziza]

Homework Statement

Two masses are connected by spring and slide freely without friction along horizontal track. What is period of oscillation?

Homework Equations

The Attempt at a Solution

My solution:
let x1 be position of mass 1 (m1) and x2 be position of mass 2 (m2) and L be length of spring in equilibrium.
Then, the total stretch of the spring is x2-x1-L. Also, F1 = -F2. Thus:

m1(x1)'' = -k(x1 - x2 + L)
m2(x2)'' = -k(x2 - x1 - L)

Solving for x2 from first eqn and substituting back into second eqn yield:

$\frac{d^2}{dt^2}[\frac{m1 m2}{k}(x1)''+(m1+m2)x1]$ = 0

I am unsure how to proceed from here, any hints? I would like to just multiply both sides by (dt^2)/d^2 but I am unsure if this is mathematically correct? It does simplify the problem though and gives me right answer...

 

[vela]

You should be well aware that $\frac{d^2}{dt^2}$ isn't a fraction.

 

[bigerst]

I don't think you really need differential equations for this one, since there is no external force there is a point on the spring that neither stretches nor compresses. on either side of this point is 2 springs with different spring constants, so effectively you have two different oscillators, with the same period.

 

[vela]

Do you know about reduced mass and how to convert a two-body problem into a one-body problem?

 

[paranormal]

To find the period of oscillation, you can use the formula T = 2π√(m/k), where T is the period, m is the total mass (m1 + m2), and k is the spring constant. In this case, since the masses are connected by the spring, the total mass is m1 + m2.

Another approach is to use the conservation of energy principle. At the equilibrium point, the total energy is all potential energy stored in the spring. As the masses oscillate back and forth, this potential energy is converted into kinetic energy and back to potential energy. The period of oscillation is the time it takes for the system to complete one full cycle of converting all potential energy into kinetic energy and back to potential energy.

To use this approach, you can set up the following equations:

At the equilibrium point:
Potential energy = 1/2k(x2-x1-L)^2

At the maximum displacement point:
Kinetic energy = 1/2m1(x1')^2 + 1/2m2(x2')^2 = 1/2m1(x1')^2 + 1/2m2(-x1')^2 = 1/2(x1')^2(m1+m2)

Equating these two energies and solving for x1' gives:
x1' = ±√(k/m1)(x2-x1-L)

Substituting this back into the equation for kinetic energy at the maximum displacement point gives:
Kinetic energy = 1/2(x1')^2(m1+m2) = 1/2(k/m1)(x2-x1-L)^2(m1+m2)

Setting this equal to the potential energy at the equilibrium point and solving for the period T gives:
T = 2π√(m1+m2)/k

Both methods should give you the same result for the period of oscillation.