I have this differential equation :
$\ddot{x} + \omega_0^2x = F_0\cos(\omega t + \Delta)$ with $\omega_0^2 = k/m \enspace .$
I take the complex equation
$ \ddot{z} + \omega_0^2 z = F_0e^{i(\omega t + \Delta)} \enspace ,$
assuming an exponential solution : $\alpha e^{ut + v}$
I find :
$z_p = \alpha e^{ut + v} \Leftrightarrow \alpha u^2 e^{ut + v} + \omega_0^2 \alpha e^{ut + v} = F_0 e^{i(\omega t + \Delta)} $
$\Leftrightarrow \alpha e^{ut + v} = \frac{F_0}{u^2 + \omega_0^2} e^{i (\omega t + \Delta)}$
So $ z_p = \frac{F_0}{\omega_0^2 – \omega^2}[\cos(\omega t + \Delta) + i \sin(\omega t + \Delta)]$
and I end up with :
$x_p = Re[z_p] = \frac{F_0}{\omega_0^2 – \omega^2}\cos(\omega t + \Delta) $
But the book states that a particular solution is :
$ x_p(t) = \frac{F_0/m}{\omega_0^2 – \omega^2}\cos(\omega t + \Delta) $
So where did the "m" come from ? Where did I go wrong ?
Best Answer
The general one degree-of-freedom motion equation for the mass-spring system has the form
$m\ddot{x}+kx=F(t)$
Dividing this equation by $m$, we get
$\ddot{x}+\omega_n^2x=F(t)/m=f(t)$
where $\omega_n^2=\frac{k}{m}$
For harmonic excitation, $F(t)=F_0\cos(\omega t+\Delta)$. In standard form, then
$\ddot{x}+\omega_n^2x=f_0\cos(\omega t+\Delta)$
The particular answer to this differential equation is
$x_p(t) = \frac{f_0}{\omega_n^2-\omega^2}\cos(\omega t+\Delta)$
So it all depends on how you write your force function. Either your equation should be
$\ddot{x}+\omega_n^2x=\frac{F_0}{m}\cos(\omega t+\Delta)$
to get the same answer you are looking for, or, for the exact equation you typed, then your answer is absolutely correct.
This "divide or not by m" thing always confuses the students. You should pay attention to how you derive the equation of motion and its response in standard form (with the $\omega_n$ parameter) or general form (with the $m$ and $k$ parameters).