Does This Lagrange's Equation for a Spring Pendulum Look Correct?

[superpig10000]

A pendulum consists of a mass m suspended by a massless spring with unextended length b and. spring constant k. Find Lagrange’s equations of motion

Here's how I set up my equation:

x = lsin(theta)
y = -lcos(theta)

(x=0 at equilibrium, y=0 at the point wehre the pendulum is hung from)

Kinetic energy = 1/2 * m * l^2 * (theta dot)^2
Potential energy = -mglcos(theta) + 1/2 kl^2
L = K - U

So after differentiating, I come up with:
theta double dot = w^2 sin (theta)

Does that look right to you?

I appreciate any help.

 

[superpig10000]

Can anyone help?

 

[Hargoth]

Since the mass is attached to a spring, the length l of the pendulum is not constant.

 

[Galileo]

With Lagrangian problems you should always consider the degrees of freedom of the system and decide on which coordinates to use to decribe the configuration of the system. In this case the angle theta is a good choice. For the other the distance the spring is stretched from equilibrium seems like another good one, call it u. So the length of your pendulum is the rest length of the spring plus u. Then express L in terms of these variables.

 

[Stephan Hoyer]

Your kinetic energy is also missing a term.

 

[ag91]

heres how i would solve the problem:
T = kinetic energy
U = potential energy

L is the displacement of the mass along the spring
O is the angular displacement of the pendulum

T = m*(L_dot^2)/2 + m*(L^2)(O_dot^2)/2
U = m*g*L(1-cos(O)) + k*(L^2)/2

plug into Lagrange's equation for two equations of motion (generalized coordinates are L and O)