# Write Lagrangian of spring-mass system

## 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 $x$ as the stretch from equilibrium of the spring and $k$ its elastic constant. $M$ is the mass on the spring.

Take $\theta$ as the angular displacement from equilibrium of the pendulum, and $m$ as the mass on the end of the pendulum.

$$T = \frac{1}{2} M \dot{x}^2 + \frac{1}{2} m \dot{x}^2 + \frac{1}{2} m l^2 \dot{\theta}^2$$

$$V = \frac{1}{2} k x^2 + m g l (1-\cos \theta)$$

Now if I take $M=2m$ and $k=mg/l$ and $(1-\cos \theta) = \theta^2/2$ 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?

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diazona
Homework Helper

Yeah, I think you are. When you compute the kinetic energy for the small mass m, you can't just write separate terms for the spring motion $m\dot{x}^2/2$ and the pendulum motion $ml^2 \dot{\theta}^2/2$, because those two are not linearly independent. (By writing them as separate terms, you were kind of implicitly assuming that they are) Instead, I'd suggest finding expressions for the x and y coordinates of the small mass and differentiating them to get the velocity, which you can then plug into the kinetic energy formula:
$$T = \frac{1}{2}m v_x^2 + \frac{1}{2}m v_y^2$$

$$T = \frac{1}{2}(M+m)\dot{x}^2 + \frac{1}{2}m(l^2 \dot{\theta}^2 + 2 \dot{x} l \dot{\theta} \cos \theta)$$

$$V = \frac{1}{2} k x^2 - m g l \cos \theta$$

I want to find the normal mode frequencies. How do I handle the $\cos \theta$ factor in the kinetic energy matrix when constructing the eigenvalue problem?

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