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Write Lagrangian of spring-mass system

  • Thread starter CNX
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  • #1
CNX
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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?
 

Answers and Replies

  • #2
diazona
Homework Helper
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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 [itex]m\dot{x}^2/2[/itex] and the pendulum motion [itex]ml^2 \dot{\theta}^2/2[/itex], 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:
[tex]T = \frac{1}{2}m v_x^2 + \frac{1}{2}m v_y^2[/tex]
 
  • #3
CNX
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[tex]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)[/tex]

[tex]V = \frac{1}{2} k x^2 - m g l \cos \theta[/tex]

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