0
$\begingroup$

In Classical Mechanics by Taylor, we find the solution to the differential equation of a damped oscillator with a sinusoidal driving force: $$\ddot{x} + 2\beta\dot{x} + \omega_0^2x = f_0\cos\left(\omega t\right).$$ My intuition tells me that there's a good reason for not considering the more general case, $$\ddot{x} + 2\beta\dot{x} + \omega_0^2x = f_0\cos\left(\omega t - \delta\right),$$ where the sinusoidal force has an arbitrary phase $\delta$ inserted. Something to do with redefining $t = 0$, but I don't know how to articulate it because I don't fully understand it. I'd appreciate it if someone could explain clearly why considering an arbitrary phase is superfluous or tell me that I'm wrong to think that the phase is superfluous.

$\endgroup$
  • $\begingroup$ I'm curious, what does the book say the solution is? $\endgroup$ – DanielSank Mar 27 at 21:45
  • $\begingroup$ @DanielSank The solution is $$x = A\cos\left(\omega t - \delta\right) + C_1 e^{r_1 t} + C_2 e^{r_2 t},$$ where $r_1 < 0 > r_2$, $$A = \frac{f_0}{\sqrt{\left(\omega_0^2 - \omega^2\right)^2 + 4\beta^2 \omega^2}},$$ and $$\delta = \arctan\left(\frac{2\beta\omega}{\omega_0^2 - \omega^2}\right).$$ $\endgroup$ – PiKindOfGuy Mar 28 at 12:33
2
$\begingroup$

If you make a change of variables, lets say $\omega t'=\omega t - \delta$ you get the same equation without the phase, and you don't lose generality.

$\endgroup$
  • $\begingroup$ Would you make this more explicit please. $\endgroup$ – PiKindOfGuy Mar 27 at 10:05
1
$\begingroup$

Below I have drawn two graphs.
One for $\cos(\omega \,t)$ (grey) and the other for $\cos(\omega \,t-\delta)$ (orange).

enter image description here

You will note that the graph of $\cos(\omega \,t-\delta)$ is the same as that of $\cos(\omega \,t)$ except that is lagging by a phase angle of $\delta$ or a time $\frac{\delta}{\omega}$.

So the peak of $\cos(\omega \,t)$ occurs at time $t=0$ and the peak of $\cos(\omega \,t-\delta)$ occurs later at a time $t = \frac {\delta}{\omega}$.

However I could have started time $T$ later than time $t$ where $\omega \, t - \delta = \omega T$ or $T = t+\frac{\delta}{\omega}$ and the orange graph with the orange axes is now a graph of $\cos (\omega\,T)$ against $T$.

So the clock which is measuring time $T$ is delayed relative to the clock which is measuring time $t$ by a time $\frac{\delta}{\omega}$.

$\endgroup$
0
$\begingroup$

The steady state solution to the equation $$\ddot x + 2 \beta \dot x + \omega_0^2 x = A \cos(\Omega t + \theta)$$ is $$x(t) = \text{Re} \left[ - e^{i( \Omega t + \theta)}\frac{A}{\Omega^2 - 2 i \beta \Omega - \omega_0^2} \right] \, .$$ So if we define a new time coordinate by the equation $$\Omega t + \phi = \Omega t' \Longrightarrow t' = t + \phi/\Omega$$ Then the phase would disappear from the differential equation and its solution.

The physical reason that this is possible is that there's time translation symmetry in the original differential equation. In other words, shifting the drive by any amount in time shifts the solution by the same amount.

$\endgroup$
  • $\begingroup$ If you want a complete proof of the solution I wrote there, I can send you a writeup that derives it with no hand-waving, as long as you don't mind doing a pretty simple contour integral. $\endgroup$ – DanielSank Mar 27 at 21:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.