idk if that kind of question is allowed here, if not, my bad. i alway seen the massless pendulum equation, where at end of all deduction you have something like
$$\frac{d^2\theta}{dt^2}+\frac{g}{L}\sin\theta=0$$
where $g$ is the local gravity value and $L$ is it length. My question is, what would the equation of pendulum be if we have a scenario of a wire that can extend by a certain amount and have a certain amout of mass looks like?
I am offering two ways of solving such question.
The first method is the Lagrangian equation:
$\frac{d}{dt}(\frac{\partial L}{\partial \dot q})=\frac{\partial L}{\partial q} \ and \ L= T-V$ where q and $\dot q$are general coordinates.
In this case, if we set the length of the pendulum string as $l$, and the angle between the pendulum string and the verticle line as $\theta$, the we can get that $L= \frac{1}{2}m((\dot l)^2+(l\dot \theta)^2)-mgl(1-\cos(\theta))-\frac{1}{2}k(l-l_0)^2$ Then you can plug $L$ into the lagrangian equation and get the final differential equation.
The second method is rather more Newtonian:
You can differentiate the total energy to get the relationship between $\theta \ and\ l$. Nevertheless, it would be really hard to solve the differential equation after differentiating $d E$ where $E= \frac{1}{2}m((\dot l)^2+(l\dot \theta)^2)+mgl(1-\cos(\theta))+\frac{1}{2}k(l-l_0)^2$.So, I think the first method would be better.
