0
$\begingroup$

The equation of motion of a damped forced oscillator is;

$$\ddot{x}(t)+\gamma\dot{x}(t)+\omega_0^2x(t)=F(t),$$

$$F(t) = F_0 \cos(\omega_dt);$$ also for the purpose of this problem we may set $\omega_d=\omega_0=\omega$, just not inside cos and sine arguments.

and $$x(t)=A(\omega_d)\cos(\omega_dt-\delta),$$ which is a steady-state solution. From my understanding, it is a particular solution only.

We need to find a full sol. ($x(t)=x_\text{characteristic} + x_\text{particular}$) to a damped forced oscillator with initial conditions, $x(0)=x'(0)=0$.

Here's my result;

$$𝑥(t)=\frac{F_0}{r\omega}𝑒^{−\gamma 𝑡/2}\sin(\omega 𝑡) + \frac{F_0}{r\omega}\cos\left(\omega _d 𝑡 + \frac{\pi}{2}\right)$$

Now, I want to calculate the total power on the system and integrate the result for further interpretation.

I know that power = force $\times$ velocity

Power due to external/driving force: $P = F(t) \cdot \frac{dx}{dt}$

Power due to damping force: $P = −\gamma \left(\frac{dx}{dt}\right)^2$

But which $x(t)$ sol. do I use for $\frac{dx}{dt}$?

$\endgroup$
  • $\begingroup$ What's the problem? You have $x(t)$, so $P(t)=x'(t)\times F(t)$. $\endgroup$ – Gert Oct 13 at 23:17
  • $\begingroup$ Yes, but do I use x(t) - the steady-state solution? Or should I use the solution I obtained through applying initial conditions? $\endgroup$ – Laura Oct 13 at 23:59
  • $\begingroup$ There is only one $x(t)$ and that's the one to use. Your 4th equation. $\endgroup$ – Gert Oct 14 at 2:29
  • $\begingroup$ What about the steady-state solution, i.e. the third equation? Would I use that in the absence of initial conditions to calculate the power? $\endgroup$ – Laura Oct 14 at 12:10
  • $\begingroup$ If you want to know $P(t)$ at steady state only, then yes. $\endgroup$ – Gert Oct 14 at 15:27

Your Answer

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

Browse other questions tagged or ask your own question.