Question

Here @John_ES

Answer 1

Can you give some explanations too if it won't take too long to write?

Answer 2

I would propose a system of reference solidary with the pendulum, then
\[ml\ddot{\theta}+mg\sin\theta+k l \dot{\theta}=0\]Now you must suppose small oscillations,
\[\sin\theta\approx\theta\]So,
\[\ddot{\theta}+k\dot{\theta}+\frac{g}{l}\theta=0\]That is the differential equation we must solve.

Answer 3

Well, I miss the mass,
\[\ddot{\theta}+\frac{k}{m}\dot{\theta}+\frac{g}{l}\theta=0\]Or,
\[\ddot{\theta}+\gamma\dot{\theta}+\omega^2\theta=0\]

Answer 4

Differential equation!!!!!! I'm still taking that class and I didn't do good in my test yesterday :/

Answer 5

Oh, well, in that case, the way to reach to the solution is the following. First, we will redefine our gamma constant, that way,
\[\ddot{\theta}+2\gamma\theta+\omega^2=0\]
You can do that without problem. Now gamma is,
\[\gamma=\frac{k}{2m}\]
You must propose a solution of the form,
\[\theta(t)=Ae^{rt}\]And substitute, then you obtain,
\[r^2+\gamma r+\omega^2 =0\]And solve for r,
\[r=\frac{-2\gamma\pm\sqrt{4\gamma^2-4\omega^2}}{2}=-\gamma\pm\sqrt{\gamma^2-\omega^2}\]

Answer 6

These solutions can be named that way,
\[\omega_1=-\gamma +\sqrt{\gamma^2-\omega^2}\]\[\omega_2=-\gamma -\sqrt{\gamma^2-\omega^2}\] Now you would have three cases,

- Underdamping,
\[γ^2<ω^2\]The solutions are complex, and then,
\[θ(t)=e^{−γt}[A_1e^{iω_1t}+A_2e^{−iω_1t}]=Ae^{−γt}\cos(ω_1t−δ)\]
so it is a decreasing oscillation.

Answer 7

- Critical damping,
\[\beta^2=\omega^2\]
And then, there is only one solution,
\[\theta(t)=(A+Bt)e^{-\gamma t}\]

Answer 8

Where I put beta I mean gamma. Sorry.

-Overdamped motion,
\[\gamma^2>\omega^2 \]Then, there are two real solutions,
\[\theta(t)=e^{-\gamma t}[A_1e^{\omega_2t}+A_2e^{\omega_2 t}]\]with
\[\omega_2=\sqrt{\gamma^2-\omega^2}\]

Answer 9

I put wrong the first frequencies in the underdamping motion, at first. It must be,
\[\omega_1=\sqrt{\omega^2-\gamma^2}\]

Answer 10

If you need more details in the derivation, you can see the book of the Schaum series in differential equations.

Answer 11

Ah ok I'll take some time ti understand this. Can you see the last question again? I am confused about something.

Answer 12

Ah, yes, I miss a minus sign, in the second exponencial shoud be,
\[A_2e^{-\omega_2t}\]