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Homework Statement
Spring-mass system on a frictionless surface. A pendulum hangs from the mass of the spring-mass system. Write the Lagrangian.
The Attempt at a Solution
Take [itex]x[/itex] as the stretch from equilibrium of the spring and [itex]k[/itex] its elastic constant. [itex]M[/itex] is the mass on the spring.
Take [itex]\theta[/itex] as the angular displacement from equilibrium of the pendulum, and [itex]m[/itex] as the mass on the end of the pendulum.
[tex]T = \frac{1}{2} M \dot{x}^2 + \frac{1}{2} m \dot{x}^2 + \frac{1}{2} m l^2 \dot{\theta}^2[/tex]
[tex]V = \frac{1}{2} k x^2 + m g l (1-\cos \theta)[/tex]
Now if I take [itex]M=2m[/itex] and [itex]k=mg/l[/itex] and [itex](1-\cos \theta) = \theta^2/2[/itex] I get two normal modes but in one only the spring moves, and in the other only the pendulum moves. Am I missing an interaction term between the spring and pendulum?