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## Homework Statement

This problem is from the 1992 GRE. A tube is free to slide on a frictionless wire. On each end of the tube is attached a pendulum. The mass of the tube is [itex]M[/itex]. The length and mass of the pendula are [itex]l[/itex] and [itex]m[/itex], respectively.

## Homework Equations

It is given that one of the eigenfrequencies is [itex]\sqrt{\frac{g(M+2m)}{lM}}[/itex]. I want to verify this.

## The Attempt at a Solution

I found the Lagrangians and made the suitable small angle approximation. Here is my work:

The position of either pendulum bob, using the initial position of its pivot as an origin is\\

[itex]

\begin{eqnarray*}

x = s + l\sin\phi\\

y = -l\cos\phi\\

\end{eqnarray*}

[/itex]

The derivatives are

[itex]

\begin{eqnarray*}

\dot x = \dot s + l\dot\phi\cos\phi\\

\dot y = l\dot\phi\sin\phi

\end{eqnarray*}

[/itex]

The total kinetic energy is

[itex]

\begin{eqnarray*}

T = \frac{1}{2}M\dot s^2 + m(\dot x^2 + \dot y^2)\\

= \frac{1}{2}M\dot s^2 + m((\dot s + l\dot\phi\cos\phi)^2 + (l\dot\phi\sin\phi)^2)\\

= \frac{1}{2}M\dot s^2 + m(\dot s^2 + 2l\dot s\dot\phi\cos\phi + l^2\dot\phi^2\cos^2\phi + l^2\dot\phi^2\sin^2\phi)\\

= (\frac{1}{2}M+m)\dot s^2 + m( 2l\dot s\dot\phi\cos\phi + l^2\dot\phi^2)\\

\end{eqnarray*}

[/itex]

The potential energy from the height of the bobs

[itex]

\begin{eqnarray*}

U = 2mgy\\

=-2mg\cos\phi

\end{eqnarray*}

[/itex]

The Largrangian is

[itex]

\begin{eqnarray*}

\mathcal{L} = T - U\\

= (\frac{1}{2}M+m)\dot s^2 + m( 2l\dot s\dot\phi\cos\phi + l^2\dot\phi^2) + 2mg\cos\phi\\

\end{eqnarray*}

[/itex]

Find the Lagrangian equation relative to [itex]s[/itex]

[itex]

\begin{eqnarray*}

\frac{\partial \mathcal{L}}{\partial \dot s} = (M+2m)\dot s + 2lm\dot\phi\cos\phi\\

\frac{d}{dt}\left ( \frac{\partial \mathcal{L}}{\partial \dot s} \right )

= (M+2m)\ddot s + 2lm(\ddot\phi\cos\phi - \dot\phi^2\sin\phi)\\

\frac{\partial \mathcal{L}}{\partial s}=0\\

\boxed{(M+2m)\ddot s + 2lm(\ddot\phi\cos\phi - \dot\phi^2\sin\phi) = 0}

\end{eqnarray*}

[/itex]

Find the Lagrangian equation relative to [itex]\phi[/itex]

[itex]

\begin{eqnarray*}

\frac{\partial \mathcal{L}}{\partial \dot \phi} = m(2l\dot s\cos\phi + 2l^2\dot\phi)\\

\frac{d}{dt}\left ( \frac{\partial \mathcal{L}}{\partial \dot \phi} \right )

= 2lm(\ddot s\cos\phi - \dot s\dot\phi\sin\phi + l\ddot\phi)\\

\frac{\partial \mathcal{L}}{\partial \phi} = -2m\sin\phi (l\dot s\dot\phi + g)\\

\boxed{2lm(\ddot s\cos\phi - \dot s\dot\phi\sin\phi + l\ddot\phi) + 2m\sin\phi (l\dot s\dot\phi + g) = 0}

\end{eqnarray*}

[/itex]

Now, recall the first equation

[itex]

\begin{eqnarray*}

(M+2m)\ddot s + 2lm(\ddot\phi\cos\phi - \dot\phi^2\sin\phi) = 0\\

2lm(\ddot s\cos\phi - \dot s\dot\phi\sin\phi + l\ddot\phi) + 2m\sin\phi (l\dot s\dot\phi + g) = 0

\end{eqnarray*}

[/itex]

Take the small angle approximation

[itex]

\begin{eqnarray*}

\frac{(M+2m)}{2lm}\ddot s + \ddot\phi - \dot\phi^2\phi = 0\\

\ddot s + l\ddot\phi + \phi \frac{g}{l} = 0

\end{eqnarray*}

[/itex]

It is clear that the goal is in sight, but my diffeqs skills are somewhat rusty. I can't see exactly how to show that the system has the stated eigenfrequency. Any help would be appreciated :)

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