I've been doing this problem for a good while now and just coming to grips with it. I'd like to know if my solution is acceptable. Thanks in advance.
A forced and damped harmonic oscillator satisfies $$\ddot{x}+2k\dot{x}+\omega_0^2 x = F\cos\omega t$$ where $\omega_0^2 > 2k^2.$ Determine the amplitude of the long term solution as a function of $\omega$.
The associated homogeneous equation is
$$\ddot{x}+2k\dot{x}+\omega_0^2 x =0$$
The characteristic polynomial of our homogenous equation is
$$P(\lambda)= \lambda^2 + 2k\lambda +\omega_0^2$$
and has roots
$$\lambda_{\pm}= -k\pm \sqrt{k^2-\omega_0^2} =-k\pm i\sqrt{\omega_0^2-k^2}$$
The question tells us that $\omega_0^2 > 2k^2.$ This is the under-damped case ($\omega_0^2 > k^2$). The solution to the homogeneous equation is then
$$x_h(t)=e^{-kt}(A\cos\omega' t+ B\sin\omega' t)$$
where $\omega'=\sqrt{\omega_0^2-k^2}$
Now, to find the particular solution we use the complex method.
We need to solve
$$\ddot{z}+2k\dot{z}+\omega_0^2 z = Ae^{i\omega t}$$
By using the trial function $z(t)= ae^{i\omega t}$ we get
$$\ddot{z}+2k\dot{z}+\omega_0^2 z=[(i\omega)^2+2k(i\omega)+\omega_0^2]ae^{i\omega t}=Ae^{i\omega t}$$
Let's define $P(i\omega) = [(i\omega)^2+2k(i\omega)+\omega_0^2] $ since it's our characteristic equation.
The complex solution is
$$z(t) = \dfrac{1}{P(i\omega)}Ae^{i\omega t}$$.
The question want us to find the long-term solution. This is also called the steady-state solution. This is the real solution of $z(t)$. Letting $x_p(t)$ be the real particular solution, then what we are looking for is simply
$$x_p(t) = Re(z(t))$$.
What is the real part of $z(t)$? We can find this using complex polar coordinates.
$$\begin{align} P(i\omega)& = (i\omega)^2 + 2k(i\omega)+\omega_0^2\\ & = (\omega_0^2-\omega^2) +2k\omega i \end{align}$$
Recall that $R = \sqrt{x^2 +y^2} = \sqrt{(\omega_0^2-\omega^2)^2 +4k^2\omega^2}$
Now we can rewrite $P(i\omega)$ as
$$P(i\omega) = Re^{i\theta}=R(\cos \theta + i\sin \theta)$$
Now we only want the real part, we don't want $i\sin \theta$
$$\begin{align}x_p(t) & = \dfrac{1}{Re^{i\theta}}Ae^{i\omega t} \\ & = \dfrac{1}{R}Ae^{i(\omega t - \theta)} \end{align}$$
The real solution to $z(t)$ is
$$x_p(t) = \dfrac{1}{R}A\cos(\omega t-\theta)$$
letting $G(\omega) = \dfrac{1}{R}$ we have
$$x_p(t) = G(\omega)A\cos(\omega t - \theta)$$.
the amplitude of the long-term solution as a function of $\omega$ can be written as
$$H(\omega) = G(\omega)A = \dfrac{A}{\sqrt{(\omega_0^2-\omega^2)^2 +4k^2\omega^2}}$$
where $G(\omega)A$ is the amplitude of $x_p$
Is this too long an answer? I might get something like this in the exam and I think this could take a long time to write.
What is a resonance curve and if I was expected to sketch one how would I go about it. I think I saw that it's the amplitude against $\omega$. Thanks very much.
