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Why do two masses, connected to each other by a spring, and each connected to a wall by a spring, have the same frequencies of oscillation when perturbed? In solving for the motion of the masses, under and assumption of SHM for each, the displacement of each can be guessed as $Ae^{i(\omega t + \theta)}$, where each mass has a different amplitude, and they have the same frequency. That the frequencies are the same seems important, since the next step is to lay out the equations of motion so as to get a matrix multiplication like this,

$m\ddot x_1 + \kappa x_1 + \beta x_1 -\beta x_2 = 0$

$m\ddot x_2 + \kappa x_2 + \beta x_2 -\beta x_1 = 0$

$$ \begin{pmatrix} \omega_1^2 - \frac \kappa m-\frac \beta m & \frac \beta m\\ \frac \beta m & \omega_2^2 - \frac \kappa m-\frac \beta m\\ \end{pmatrix} $$

where the top left and bottom right terms are identical. Then, finding the determinant to find a non-trivial solution, you can solve for $\omega$. This works because $\omega_1$ = $\omega_2$, or the equation would be difficult, and maybe impossible, to solve. You might get a solution for the product of the two but not for either. In the end, you get two $\omega$ values, one for a fast oscillation and one for a slow one; in one mode the masses move together, and in one they are half a phase apart. In both they have equal amplitude, and the motion of the masses, whatever is observed, is a combination of these two motions (I think this result is saying that they only motions the masses can undergo are those which are linear combinations of these two motions).

Visually, I am trying to picture the masses moving with different frequencies. I can picture it until I get to the region of spring which connects them, which could not move at both frequencies simultaneously. Is this disconnect in reality and picture why the frequencies are different? Or is it possible for the frequency of oscillation to vary along a spring?

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  • $\begingroup$ " I can picture it until I get to the region of spring which connects them, which could not move at both frequencies simultaneously." Why not : the one end could oscillate with $\omega_1$ and the other end with $\omega_2 \ne \omega_1$. The spring is not a rigid rod moving along its direction. $\endgroup$
    – Frobenius
    Dec 4 '18 at 10:08
  • $\begingroup$ I guess it could. We seem to be just choosing to look for solutions where $w_1$ = $w_2$, with the certainty that other motions can be constructed from these. $\endgroup$ Dec 4 '18 at 17:21
  • $\begingroup$ "For the special case of a common spring constant $\:k_{\rho}=k \: (\rho=1,2,\cdots, n+1) \:$ and common particle mass $\:m_{\rho}=m \: (\rho=1,2,\cdots, n) \:$, equation (08) gives \begin{equation} \ddot{\mathbf{x}}+\omega_{o}^{2}\,\mathrm{\Xi}\,\mathbf{x}=0 \tag{17} \end{equation}" - this is because $M^{-1}K$ is then a constant. How does this help when it is not because $\kappa$ is different for each spring, and I am still asking how the $\omega$ cme to be equal in the equation as written? I think it happens to be because we're assuming we can build every solution out of $\endgroup$ Dec 5 '18 at 7:09
  • $\begingroup$ normal modes, and just finding those from the beginning. And normal modes are those where $\omega$ is the same for all the masses. Is this true? $\endgroup$ Dec 5 '18 at 7:11
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In the end, you get two $\omega$ values, one for a fast oscillation and one for a slow one; in one mode the masses move together, and in one they are half a phase apart. In both they have equal amplitude, and the motion of the masses, whatever is observed, is a combination of these two motions.

The oscillations are called the modes of the system and to simplify things assume that all three springs are the same.

The slower oscillation $\omega_{\rm slow}$ looks like this with the two masses moving in phase with one another.

enter image description here

The faster oscillation $\omega_{\rm fast}$ looks like this with the two masses moving exactly $\pi$ out of phase with one another.

enter image description here

Any oscillation of the system can be described by a linear combination of these two modes.

For example suppose that you started off having just one of the masses displaced by a certain amount and at rest and the other mass at rest at static equilibrium position.
In effect you are starting with the two masses in the mode with frequency $\omega_{\rm slow}$ both displaced by $-a$ from the equilibrium position and the two masses in the mode with frequency $\omega_{\rm fast}$ with one mass displaced by $-a$ from the equilibrium position and the other displaced by $+a$ from the equilibrium position.

On releasing the masses the motion of those two masses would look like this.

enter image description here

Here are a few frames from the gif file to show the complexity of the motion.

enter image description here

An analysis of the motion with an example which results is the following graph to show the motion of the two particles is done here.

enter image description here

A VPython simulation is shown in this video.

The animations come from this website.

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  • $\begingroup$ Thanks for the animations, which are detaikef, but I don't understand still about the frequency. This, though, is an explanation of normal mores. Thanks for that. $\endgroup$ Dec 6 '18 at 2:48
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    $\begingroup$ @RukiyaMeria In general the masses move in a complex way being the sum of two different frequency shms and the springs move appropriately. $\endgroup$
    – Farcher
    Dec 6 '18 at 11:04
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The equations of motion are two coupled linear second order differential equations. Since the system is linear, any linear combination of solutions is still a solution, hence the set of solutions is a linear space.

It can be proved that the solutions of a system of $n$-th order linear differential equations is a linear space of dimension $n$. This means that if we can find $n$ specific, linearly independent solutions, then we can find any solution as a linear combination of those $n$ solutions.

Our system is linear second order, then we have to find two linearly independent solutions. We do so by trying simple guess solutions. Our guess is that the problem is solved by some kind of normal mode. We try and we do find two normal modes that solve the problem and are linearly independent. So we know that we have solved the problem.

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