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moonjob
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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 M. The length and mass of the pendula are l and m, respectively.
Homework Equations
It is given that one of the eigenfrequencies is \sqrt{\frac{g(M+2m)}{lM}}. 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\\
<br /> \begin{eqnarray*}<br /> x = s + l\sin\phi\\<br /> y = -l\cos\phi\\<br /> \end{eqnarray*}<br />
The derivatives are
<br /> \begin{eqnarray*}<br /> \dot x = \dot s + l\dot\phi\cos\phi\\<br /> \dot y = l\dot\phi\sin\phi<br /> \end{eqnarray*}<br />
The total kinetic energy is
<br /> \begin{eqnarray*}<br /> T = \frac{1}{2}M\dot s^2 + m(\dot x^2 + \dot y^2)\\<br /> = \frac{1}{2}M\dot s^2 + m((\dot s + l\dot\phi\cos\phi)^2 + (l\dot\phi\sin\phi)^2)\\<br /> = \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)\\<br /> = (\frac{1}{2}M+m)\dot s^2 + m( 2l\dot s\dot\phi\cos\phi + l^2\dot\phi^2)\\<br /> \end{eqnarray*}<br />
The potential energy from the height of the bobs
<br /> \begin{eqnarray*}<br /> U = 2mgy\\<br /> =-2mg\cos\phi<br /> \end{eqnarray*}<br />
The Largrangian is
<br /> \begin{eqnarray*}<br /> \mathcal{L} = T - U\\<br /> = (\frac{1}{2}M+m)\dot s^2 + m( 2l\dot s\dot\phi\cos\phi + l^2\dot\phi^2) + 2mg\cos\phi\\<br /> \end{eqnarray*}<br />
Find the Lagrangian equation relative to s
<br /> \begin{eqnarray*}<br /> \frac{\partial \mathcal{L}}{\partial \dot s} = (M+2m)\dot s + 2lm\dot\phi\cos\phi\\<br /> \frac{d}{dt}\left ( \frac{\partial \mathcal{L}}{\partial \dot s} \right )<br /> = (M+2m)\ddot s + 2lm(\ddot\phi\cos\phi - \dot\phi^2\sin\phi)\\<br /> \frac{\partial \mathcal{L}}{\partial s}=0\\<br /> \boxed{(M+2m)\ddot s + 2lm(\ddot\phi\cos\phi - \dot\phi^2\sin\phi) = 0}<br /> \end{eqnarray*}<br />
Find the Lagrangian equation relative to \phi
<br /> \begin{eqnarray*}<br /> \frac{\partial \mathcal{L}}{\partial \dot \phi} = m(2l\dot s\cos\phi + 2l^2\dot\phi)\\<br /> \frac{d}{dt}\left ( \frac{\partial \mathcal{L}}{\partial \dot \phi} \right )<br /> = 2lm(\ddot s\cos\phi - \dot s\dot\phi\sin\phi + l\ddot\phi)\\<br /> \frac{\partial \mathcal{L}}{\partial \phi} = -2m\sin\phi (l\dot s\dot\phi + g)\\<br /> \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}<br /> \end{eqnarray*}<br />
Now, recall the first equation
<br /> \begin{eqnarray*}<br /> (M+2m)\ddot s + 2lm(\ddot\phi\cos\phi - \dot\phi^2\sin\phi) = 0\\<br /> 2lm(\ddot s\cos\phi - \dot s\dot\phi\sin\phi + l\ddot\phi) + 2m\sin\phi (l\dot s\dot\phi + g) = 0<br /> \end{eqnarray*}<br />
Take the small angle approximation
<br /> \begin{eqnarray*}<br /> \frac{(M+2m)}{2lm}\ddot s + \ddot\phi - \dot\phi^2\phi = 0\\<br /> \ddot s + l\ddot\phi + \phi \frac{g}{l} = 0<br /> \end{eqnarray*}<br />
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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