What Are the Normal Frequencies in a Three-Spring System with Masses m and αm?

[Daron]

Homework Statement

Two walls are positioned a distance of 3L apart. Three identical springs of rest length L and spring constant k are connected in series between the walls. Two particles of mass m and αm are positioned at the junctures of the springs, respectively.

Write down the Largrangian for the system.
Use the Lagrangian to derive the equations of motion for the system.
Find the normal frequencies ω_i
Assume α is very large. Expand ω_i up to order O(¹/_α) and explain the result obtained.
Assume α is very small. Expand ω_i up to order O(1) and explain the result obtained.

The Attempt at a Solution

I have already found correct answers for the first three questions. The normal frequencies are √[(k ± 1/σ)/m]. The first corresponds to the rate of change of the distance between the two particles. The second to them moving back and forth while their distance remains the same.

So if we start with start with basis
(x = position of particle of mass m,0) and (0, y = position of particle of mass αm)
And change from (x,0); (y,0) to (x,y);(x,-y) the Lagrangian simplifies and the frequencies above drop out.

This gives the correct result for α = 1.

When we assume α is large expand the frequencies up to order 1, nothing changes, but the quotient of α makes very little contribution, so the frequencies are almost identical.

Likewise for when we assume α is small and expand to order zero, the α drop off and the frequencies are exactly zero. (I do not know why you would eliminate the higher order terms for small α)

In either case the frequencies are almost equal.

This is what I think will happen in those cases: the frequencies of the together//apart and back/forth together motions are the same. So the particles move together while moving to the right, and then apart while moving to the left, or vice versa. This corresponds to a "punching motion", where the particles take turns moving away from the centre while the other stays still.

The motion is the same for either case, because in either case one of the masses becomes negligible.

Am I right?

 

[anathema]

Thank you for your post and for sharing your thoughts on the problem. I would like to provide my own response to your questions.

First, I agree with your solution for the first three parts of the problem. The Lagrangian for this system can be written as L = 1/2 m (ẋ1)^2 + 1/2 αm (ẏ1)^2 - 1/2 kx1^2 - 1/2 k(y1-y2)^2 - 1/2 k(y2-y3)^2 - 1/2 kx3^2. From this, we can derive the equations of motion using the Euler-Lagrange equation, and the normal frequencies can be found by solving the eigenvalue problem for the Hessian matrix of the Lagrangian.

For the next part, where we assume α is very large, we can see that the term 1/2 αm (ẏ1)^2 dominates the Lagrangian. This means that the motion of the system will be primarily determined by the motion of the particle with mass αm. The frequencies will still be √[(k ± 1/σ)/m], but the contribution from the first term will be negligible compared to the second term, so the frequencies will be almost equal.

Similarly, when we assume α is very small, the term 1/2 m (ẋ1)^2 will dominate the Lagrangian, and the motion of the system will be primarily determined by the motion of the particle with mass m. This explains why the frequencies become exactly zero when we expand to order O(1) in this case.

In summary, your intuition is correct. The motion of the system will be the same in both cases, with one particle essentially becoming negligible compared to the other. This is because the system is symmetric, so the particles will move together and apart in a synchronized manner, resulting in the same frequencies for both particles.

I hope this helps clarify your understanding of the problem. Keep up the good work in your studies!