Normal modes of four coupled oscillating masses (Kleppner and Kolenkow)

[dford]

Homework Statement

Four identical masses $m$ are joined by three identical springs, of spring constant $k$, and are constrained to move on a line, as shown.

There is a high degree of symmetry in this problem, so that one can guess the normal mode motions by inspection, without a lengthy calculation. Once the relative amplitudes of the normal mode motions are known, the normal vibrational frequencies follow directly.

Because of the symmetry, the normal mode amplitudes must obey $x_1 = \pm x_4$ and $x_2 = \pm x_3$. Another condition is that the center of mass must remain at rest. The possibilities are:

$(x_4 = x_1)$ and $(x_3 = x_2)$

$(x_4 = -x_1)$ and $(x_3 = -x_2)$

The normal mode equations lead to three possible non-trivial vibrational frequencies and three corresponding normal modes. Find the normal mode frequencies. It is convenient to use the dimensionless parameter $\beta = \omega^2/\omega_0^2$, where $\omega$ is the frequency to be found and $\omega_0 \equiv \sqrt{k/m}$.

Relevant Equations

Hooke's Law and harmonic oscillation: $m\ddot x = -k(x-x_0)$, where $x_0$ is the equilibrium position for a spring. The general solution to this ODE is $x(t) = A\sin(\omega t + \phi)$, where $\omega = \sqrt{k/m}$.

This exercise comes from Kleppner and Kolenkow, 2nd ed., problem 6-3. I'm using a solution key as a study reference, but the solution key is coming to a pretty different conclusion. Mostly the issue is in the equations of motion for this system. I'm not sure if there's something I'm fundamentally misunderstanding, or if the solution key is wrong.

Each $x_i$ is the displacement of the $i$th mass from its equilibrium. So for example if $x_1 = x_2$, then the first spring is unstretched even though the first two masses are displaced from the system's equilibrium.

I beieve that the equations of motion are:
$$
\begin{cases}
m\ddot x_1 = k(x_2 - x_1), \\
m\ddot x_2 = -k(x_2 - x_1) + k(x_3 - x_2) = k(x_1 - 2x_2 + x_3), \\
m\ddot x_3 = -k(x_3 - x_2) + k(x_4 - x_3) = k(x_2 - 2x_ + x_4), \\
m\ddot x_4 = -k(x_4 - x_3).
\end{cases}
$$
Each mass oscillates according to the equation $\ddot x_i = -\omega^2 x_i$ in a normal mode, where $\omega$ is the frequency shared by all masses. Letting $\omega_0 = \sqrt{k/m}$ and $\beta = \omega^2/\omega_0^2$, and making the substitution $\ddot x_i = -\omega^2 x_i$ and dividing by $\omega_0^2 = -k/m$, the above equations become:
$$
\begin{cases}
\beta x_1 = x_1 - x_2, \\
\beta x_2 = 2x_2 - x_1 - x_3, \\
\beta x_3 = 2x_3 - x_2 - x_4, \\
\beta x_4 = x_4 - x_3
\end{cases}
$$
And the idea is now to use the symmetry in the cases $x_4 = x_1, x_2 = x_3$ and $x_4 = -x_1, x_2 = -x_3$ to find $\beta$.

However: The solution key I'm using as a reference has different equations of motion and different conclusion and I'm not following it. I quote it below (they're missing some negative signs but these cancel out):

For the harmonic motion of each mass, $\ddot x_i = \omega^2 x_i$. The equation of motion for mass 1 is, for example,
$$
m\ddot x_1 = k(x_1 - x_2) \Longrightarrow \omega^2 x_1 = \frac k m (x_1 - x_2) \omega_0^2(x_1 - x_2)
$$
$$
\beta x_1 = (x_1 - x_2)
$$
where $\omega_0 = \sqrt{k/m}$ and $\beta = \omega^2/\omega_0^2$. Hence
$$
\beta x_1 = (x_1 - x_2) \quad \beta x_2 = (x_2 - x_1 - x_3)
$$
$$
\beta x_3 = (x_3 - x_2 - x_4) \quad \beta x_4 = (x_4 - x_3)
$$​

But this to me suggests that their equation of motion for the second mass, for example, would be $m\ddot x_2 = k(x_2 - x_1 - x_3)$. Shouldn't the second mass feel the effect of both springs on either side, as I observed in my equations of motion? Or did I miss a coefficient somewhere?

UPDATE:

As someone pointed out in the link below, yes, the solution key is wrong. Thanks all

 

[BvU]

Homework Statement: Four identical masses $m$ are joined by three identical springs, of spring constant $k$, and are constrained to move on a line, as shown.

There is a high degree of symmetry in this problem, so that one can guess the normal mode motions by inspection, without a lengthy calculation. Once the relative amplitudes of the normal mode motions are known, the normal vibrational frequencies follow directly.

Because of the symmetry, the normal mode amplitudes must obey $x_1 = \pm x_4$ and $x_2 = \pm x_3$. Another condition is that the center of mass must remain at rest. The possibilities are:

$(x_4 = x_1)$ and $(x_3 = x_2)$

$(x_4 = -x_1)$ and $(x_3 = -x_2)$

The normal mode equations lead to three possible non-trivial vibrational frequencies and three corresponding normal modes. Find the normal mode frequencies. It is convenient to use the dimensionless parameter $\beta = \omega^2/\omega_0^2$, where $\omega$ is the frequency to be found and $\omega_0 \equiv \sqrt{k/m}$.

Relevant Equations: Hooke's Law and harmonic oscillation: $m\ddot x = -k(x-x_0)$, where $x_0$ is the equilibrium position for a spring. The general solution to this ODE is $x(t) = A\sin(\omega t + \phi)$, where $\omega = \sqrt{k/m}$.

-------------

Note: I edited the above to get the MathJax input sorted out. Good thing you know $\TeX$ ! In PF MatJax the delimiter for inline math is ## and for displayed math you have $$ as delimiter. Apparently you know already, because you did well in

I'm just seeing if this will work: x¨=−kx, or x¨=−kx.

which is a start. I don't see $x$ in the exercise, only $x_1$ etc.

Check out the treatment of a slightly simpler configuration here

( as you can see physicists are lazy buggers -- if possible they let google do their work for them)

$\$

 

[dford]

Note: I edited the above to get the MathJax input sorted out. Good thing you know $\TeX$ ! In PF MatJax the delimiter for inline math is $and for displayed math you have as delimiter. Apparently you know already, because you did well in which is a start. I don't see$x$in the exercise, only$x_1## etc.

Check out the treatment of a slightly simpler configuration here

Yes indeed; even math.stackexchange uses slightly different mathjax syntax. I think I figured it out. Thanks :)

 

[dford]

Check out the treatment of a slightly simpler configuration here

Aha, and in fact your link answered my confusion. Good to know I understood that much at least

 

[BvU]

The thread of this thread is now a bit hard to follow -- not unusual within PF, but what can one do ?

Well:

• mark/indicate major edits
• don't add big chunks in a post if subsequent posts then become unhinged
• In a 'reply', remove irrelevant parts to keep the volume of repetitious stuff limited
• use common sense

$\$