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I am looking at the problem of a coupled oscillator, whereby we have 3 springs connected between two walls in the following way: wall, then spring (k), then mass (m), then spring (2k), then mass (m), then spring (k), then wall.

I have calculated the characteristic frequencies (I think) by using the fact that we will have

$$m \dfrac{d^2x}{dt^2}=-kx+2k(y-x)=-3kx+2ky$$ and $$m\dfrac{d^2y}{dt^2}=-ky-2k(y-x)=-3ky+2kx.$$

Thus we have in a new coordinate system

$$m\pmatrix{\ddot{X}\\\ddot{Y}}=k\pmatrix{-3&-2\\2&-3}\pmatrix{x\\y}.$$

Evaluating the matrix in the previous expression I obtain Eigenvalues $\mu_1=-5$ and $\mu_2=-1$. The the unit Eigenvectors will be $$\hat{e}_1=\dfrac{1}{\sqrt{2}}\pmatrix{1\\1}$$ and $$\hat{e}_2=\dfrac{1}{\sqrt{2}}\pmatrix{1\\-1}.$$

So in the new coordinate system I have $$\pmatrix{\ddot{X}\\\ddot{Y}}=\dfrac{k}{m}\pmatrix{\mu_1&0\\0&\mu_2}\pmatrix{X\\Y}.$$ This implies that $$\ddot{X}=-\dfrac{5k}{m}X$$ and $$\ddot{Y}=-\dfrac{k}{m}Y.$$

Solving these equations we obtain $$X=Asin(\omega_1t+\alpha)$$ and $$Y=Bsin(\omega_2t+\beta),$$ where $\omega_1=\sqrt{\dfrac{5k}{m}}$ and $\omega_2=\sqrt{\dfrac{k}{m}}$.

Now I need to find the characteristic frequency and the mode of vibration. (Please do not give any answers as this is a homework assignment that I would like to solve myself.)

Q1. Am I correct in thinking that the characteristic frequency is simply $f=\omega/2\pi$, or am I misunderstanding what the characteristic frequency is?

Q2. What is the mode of vibration? I understand what this is for a longitudinal wave oscillating between two fixed points, but I think this system is a transverse one, and I am struggling to think what the modes of vibration could be.

Please can someone verify whether my understanding of what the characteristic frequency is is correct, and if not could you explain what it is. Also, can someone explain what the mode of vibration is in a transverse system like this one.

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  • $\begingroup$ fyi, In your Latex, you need to escape the "sin" using "\" so it looks right. $\endgroup$
    – Nasser
    Commented Jun 2, 2016 at 3:14
  • $\begingroup$ Thank you, however you are about half a year too late, realised this not too long ago ;) $\endgroup$
    – ODP
    Commented Jun 18, 2016 at 20:28

1 Answer 1

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The system of two coupled differential equations is characterized by its eigenvalues and its eigenvectors. You've found those. Now, let's get back to the physical part of the task.

The eigenvalues are frequencies of two normal modes of oscillations. Now, there may be different conventions, and one may define frequency as $\omega$ or $\frac{\omega}{2\pi}$, but in my opinion, $\omega$ will do.

The modes of oscillations have to do with the eigenvectors. The second question can be rephrased as "If the system is oscillating in one of its normal modes, how exactly the two masses move (in original coordinate system)?". Or, in other words, "What is the motion described by each of the eigenvectors?".

Hope that gives you enough hints.

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    $\begingroup$ OK thanks. So with my eigenvector (1,1), the two masses will be moving both right, and then both left, and so on (in SHM?). With my eigenvector (1,-1), the middle spring will be compressed, then stretched, then compressed, and so on. Does this interpretation make sense? $\endgroup$
    – ODP
    Commented Nov 4, 2015 at 18:10
  • $\begingroup$ @OllyPrice, yes, that's exactly what happens. $\endgroup$
    – svavil
    Commented Nov 4, 2015 at 21:33

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