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wglmb
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Homework Statement
A spring of negligible mass and spring constant k, hanging vertically with one end at a fixed point O, supports a mass m, and beneath it as second, identical spring carrying a second, identical mass.
Using a generalised coordinates the vertical displacements x and y of the masses from their positions with the springs unextended, write down the Lagrangian.
Find the position of equilibrium and the normal modes and frequencies of vertical oscillations.
Homework Equations
The Attempt at a Solution
KE: \frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2})
PE: mg(l+x)+mg(2l+y)+\frac{1}{2}kx^{2}+\frac{1}{2}k(y-x)^{2} =mg(3l+x+y)+kx^{2}+\frac{1}{2}ky^{2}-kxy
Lagrangian: L=\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2})-mg(3l+x+y)-kx^{2}-\frac{1}{2}ky^{2}+kxy
Equations of motion:
\frac{d}{dt}(\frac{\partial L}{\partial \dot{x}})-\frac{\partial L}{\partial x}=0 \Rightarrow m\ddot{x}+mg+2kx-ky=0
\frac{d}{dt}(\frac{\partial L}{\partial \dot{y}})-\frac{\partial L}{\partial y}=0 \Rightarrow m\ddot{y}+mg+ky-kx=0
Know that B_{ij}\ddot{q}_{j}+A_{ij}q_{j}=0
\Rightarrow B=\left( \begin{array}{cc}<br /> m & 0 \\<br /> 0 & m \end{array} \right), A=\left( \begin{array}{cc}<br /> 2k & -k \\<br /> -k & k \end{array} \right)
B^{-1}A=\left( \begin{array}{cc}<br /> \frac{1}{m} & 0 \\<br /> 0 & \frac{1}{m} \end{array} \right)\left( \begin{array}{cc}<br /> 2k & -k \\<br /> -k & k \end{array} \right)<br /> =\left( \begin{array}{cc}<br /> \frac{2k}{m} & \frac{-k}{m} \\<br /> \frac{-k}{m} & \frac{k}{m} \end{array} \right)
Eigenvalues turn out to be:
\lambda=\frac{k}{2m}(3+\sqrt{5}) and \lambda=\frac{k}{2m}(3-\sqrt{5})
This doesn't look very likely to be right.
Then when I work out the eigenvectors, I get (0,0) in both cases...
... where have I gone wrong?
