Consider the underdamped, forced pendulum governed by the ODE
$$\frac{d^2\theta}{dt^2}+\eta\frac{d\theta}{dt}+\omega_0^2\theta=F_0\cos(\omega_Ft)$$
where $2\omega_0>\eta,\omega_F>0$ and $F_0>0$.
Find the general solution assuming $\omega_0\neq\omega_F$.
I derived the complementary equation as follows:
$$\theta(t)=Ae^{-\frac{\eta}{2}t}\cos(\omega t)+Be^{-\frac{\eta}{2}t}\sin(\omega t)\quad\text{ where }\omega=\frac{\sqrt{4\omega_0^2-\eta^2}}{2}$$
Using the method of undetermined coefficients, we guess that $\theta_p$ is of the form
$$\theta_p(t)=A\cos(\omega_F t)+B\sin(\omega_F t)$$
The first and second order derivatives are
\begin{align*}
\frac{d\theta_p}{dt}&=-A\omega_F\sin(\omega_F t)+B\omega_F\cos(\omega_F t)\\
\frac{d^2\theta_p}{dt^2}&=-A\omega_F^2\cos(\omega_F t)-B\omega_F^2\sin(\omega_F t)
\end{align*}
Plugging these into the ODE, we can find $A$ and $B$.
\begin{align*}
F_0\cos(\omega_F t)&=-A\omega_F^2\cos(\omega_F t)-B\omega_F^2\sin(\omega_F t)\\
&\quad\quad\quad\quad\quad+\eta\left(-A\omega_F\sin(\omega_F t)+B\omega_F\cos(\omega_F t)\right)+\omega_0^2\left(A\cos(\omega_F t)+B\sin(\omega_F t)\right)\\
\end{align*}
I then found that the solution is
$$\theta(t)=Ae^{-\frac{\eta}{2}t}\cos(\omega t)+Be^{-\frac{\eta}{2}t}\sin(\omega t)+\frac{F_0\eta\omega_F\sin(\omega_Ft)}{\omega_F^4+\omega_F(\eta^2-2\omega_0)+\omega_0^4}-\frac{(F_0\omega_F^2-F_0\omega_0^2)\cos(\omega_F t)}{\omega_F^4+\omega_F(\eta^2-2\omega_0)+\omega_0^4}$$
where $\omega=\frac{\sqrt{4\omega_0^2-\eta^2}}{2}$.
The part that I am having trouble with is finding the particular solution given $\theta(0)=\theta'(0)=0$. This requires finding the derivative of $\theta(t)$ which is quite messy. I was wondering if I had computed the correct solution to this problem. Thanks
Best Answer
You general and particular solution look correct. Though the $\omega_F$ term should be $\omega_F^2$ and the $\omega_0$ term should be $\omega_0^2$ in the denominator. Otherwise the forms all look correct. As for initial conditions,
$$\theta(0)= A -\frac{F_0(\omega_F^2 - \omega_0^2)}{\omega_F^4 + \omega_F^2(\eta^2 - 2\omega_0^2) +\omega_0^4} =0 $$ so $$A= \frac{F_0(\omega_F^2 - \omega_0^2)}{\omega_F^4 + \omega_F^2(\eta^2 - 2\omega_0^2) +\omega_0^4}$$
And $$\theta'(0) = -\frac{\eta}{2}A + \omega B + \frac{F_0 \eta \omega_F^2}{\omega_F^4 + \omega_F^2(\eta^2 - 2\omega_0^2) +\omega_0^4} = 0$$ so
$$\omega B = \frac{\eta}{2}A -\frac{F_0 \eta \omega_F^2}{\omega_F^4 + \omega_F^2(\eta^2 - 2\omega_0^2) +\omega_0^4} $$ Simply plug in $A$ from above and divide by $\omega$ and you should have everything.