MHB What are Lagrange's equations of motion for a pendulum suspended by a spring?

[Ackbach]

Here is this week's POTW:

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

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[Ackbach]

Congratulations to logan3 for correctly solving this week's POTW! The solution follows:

Spoiler

SOLUTION
m = mass
k = spring constant
θ = angular displacement of spring pendulum from rest position
b = unextended length of spring pendulum
x = extended length of spring pendulum

$F_{mass} = mg\cos(\theta) - k(x - b) = ma_{mass} = m(\ddot x - x \dot \theta^2) \Rightarrow m \ddot x - mx \dot \theta^2 - mg\cos(\theta) + k(x - b) = 0$

$F_{\theta} = -mg\sin(\theta) = ma_{\theta} = m(x \ddot \theta + 2 \dot x \dot \theta) \Rightarrow x \ddot \theta + 2 \dot x \dot \theta + g\sin(\theta) = 0$

or

$T = KE = \frac {1} {2} m (\ddot x^2 + x^2 \dot \theta^2)$
$U = PE = -mgx\cos(\theta) + \frac {1} {2} k(x - b)^2$

Lagrangian $(L) = T - U = \frac {1}{2} m (\ddot x^2 + x^2 \dot \theta^2) + mgx\cos(\theta) - \frac {1}{2} k(x - b)^2$

Euler-Lagrangian for x:
$\frac {\partial L}{\partial x} = mx \dot \theta^2 + mg\cos(\theta) - k(x - b)$
$\frac {d}{dt} \frac {\partial L}{\partial \dot x} = m \ddot x$

$\frac {d}{dt} \frac {\partial L}{\partial \dot x} - \frac {\partial L}{\partial x} = m \ddot x - mx \dot \theta^2 - mg\cos(\theta) + k(x - b) = 0$

Euler-Lagrangian for θ:
$\frac {\partial L}{\partial \theta} = -mgx\sin(\theta)$
$\frac {d}{dt} \frac {\partial L}{\partial \dot \theta} = m x^2 \ddot \theta + 2mx \dot x \dot \theta$

$\frac {d}{dt} \frac {\partial L}{\partial \dot \theta} - \frac {\partial L}{\partial \theta} = x \ddot \theta + 2 \dot x \dot \theta + g\sin(\theta) = 0$