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Of all the pages I've read on this stack exchange, I've seen numerous proofs and comments on the complete solution to the period of an undamped pendulum, and they all involve complete elliptic integrals of the first kind. They all start with $\ddot{\theta}=-\frac{mgL}{I}sin(\theta)$, and end up at:

$$T=4\sqrt{\frac{l}{g}}\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{1-k^2sin^2(\theta)}}d\theta$$

But I can't seem to find anything on the entire internet for the solution to the differential equation for damped oscillators: $$\ddot{\theta}=-\frac{L}{I}(mgsin(\theta)+b\dot{\theta})$$ $$\ddot{\theta}+\frac{bL}{I}\dot\theta+\frac{mgL}{I}sin(\theta)=0$$

EDIT: I am looking for a solution that does not make the small angle approximation of $sin(\theta)\approx\theta$. Sorry for any inconvenience!

Does anyone have any solution? I haven't taken complex analysis and have a very surface-level understanding of elliptic integrals, but I'm very curious about the answer...

JBatswani
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2 Answers2

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We can write solution easily by considering small angle approximation, $\sin\theta =\theta$
So, the differential equation becomes,
$\ddot\theta + \frac{bL}{I}\dot\theta+\frac{mgL}{I}\theta = 0\tag{1}$
Now we can see that it is a differential equation with constant coefficients.
So, we can assume that solution might be $e^{\lambda t}$. Because then if we take derivatives of exponential function the coeffecient of $t$ in the exponent (which is a constant) comes in front of it.
So, plugging $\theta = e^{\lambda t}$ in $(1)$ we get
$(\lambda^2+\frac{bL}{I}\lambda+\frac{mgL}{I})e^{\lambda t} = 0$
As the above equation should hold $\forall t$. But $e^{\lambda t}\neq 0$ for any $t$.
So, $\lambda^2+\frac{bL}{I}\lambda+\frac{mgL}{I}=0$
We get a quadratic equation in $\lambda$.
Put $\gamma = \frac{bL}{I}$ (damping constant) and $\omega_o^2 = \frac{mgL}{I}$
$\lambda = \frac{-\gamma\pm\sqrt{\gamma^2-4\omega_o^2}}{2}$

Now we can get different solutions depending on the damping conditions.
i) Underdamped ($\gamma^2<4\omega_o^2$)
$\lambda = \frac{-\gamma\pm i\sqrt{\gamma^2-4\omega_o^2}}{2}$

Put $\frac{(\gamma^2-4\omega_o^2)}{4} = \omega^2$
We get $\lambda = -\frac{\gamma}{2}\pm i\omega$
So, $x(t)=e^{-\frac{\lambda}{2}}(c_1e^{i\omega t}+c_2e^{-i\omega t})$
We can also write $x(t)=e^{-\frac{\lambda}{2}}(c_1'\cos\omega t+c_2'\sin\omega t)$

Similarly we can write solution for critically damped and overdamped conditions

Iti
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The second order non-linear differential equations are generally not solvable. The reason why we can solve it without damping is that we know the first integral of the equation (the energy), which allows reducing it to a first order one (see, e.g., this answer for the general technique.)

Roger V.
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