Question

Write the equation of motion of a forced weakly damped harmonic oscillator and explain
the physical significance of each term. Obtain the steady-state solution of this equation.

Answer 1

Answer on Question 39277, Physics, Other

The equation of motion of a forced weakly damped oscillator is $\ddot{x} + 2\lambda \dot{x} + \omega_0^2 x = \frac{f}{m}\cos \gamma t$ , where $\ddot{x}$ is the acceleration, $2\lambda \dot{x}$ takes damping into account, and $\frac{f}{m}\cos \gamma t$ is the external periodic force.

Let us first obtain the solution of homogenous equation. $\ddot{x} + 2\lambda \dot{x} + \omega_0^2 x = 0$ . Let us look for a solution in form $x = e^{rt}$ . Plugging this into equation, obtain $r^2 + 2\lambda r + \omega_0^2 = 0$ . General solution is hence $x = a_1 e^{r_1 t} + a_2 e^{r_2 t}$ , where $r_{12} = -\lambda \pm \sqrt{\lambda^2 - \omega_0^2}$ . If $\lambda < \omega_0$ , then we have two complex solutions, which are complex conjugate one to each other. Thus, solution might be written as $x = ae^{-\lambda t} \cos (\omega t + \alpha)$ , where $\omega = \sqrt{(\omega_0^2 - \lambda^2)}$ .

Knowing the homogenous solution, one has to find the particular solution of $\ddot{x} + 2\lambda \dot{x} + \omega_0^2 x = \frac{f}{m}\cos \gamma t$ . Seek for it in form $x_p(t) = A\cos \gamma t + B\sin \gamma t$ . Plugging in into equation, obtain $A = \frac{-f}{m}\frac{(\gamma^2 - \omega_0^2)}{[(\omega_0^2 - \gamma^2) + 4\lambda^2\gamma^2]}$ and $B = \frac{f}{m}\frac{2\gamma\lambda}{[(\omega_0^2 - \gamma^2) + 4\lambda^2\gamma^2]}$ .

Particular solution $x_{p}(t) = A\cos \gamma t + B\sin \gamma t$ might be then rewritten as

$\sqrt{A^2 + B^2} (\cos \delta \cos \gamma t - \sin \delta \sin \gamma t)$ , where $\cos \delta = \frac{A}{\sqrt{A^2 + B^2}}$ , $\sin \delta = \frac{-B}{\sqrt{A^2 + B^2}}$ , thus $\tan \delta = \frac{-B}{A} = \frac{2\gamma\lambda}{\gamma^2 - \omega_0^2}$ .

Hence, $x_{p}(t) = C\cos (\gamma t + \delta)$ , where $C = \frac{f}{m\sqrt{(\omega_0^2 - \gamma^2)^2} + 4\lambda^2\gamma^2}$ .

Finally, solution of equation is $x = ae^{-\lambda t}\cos (\omega t + \alpha) + C\cos (\gamma t + \delta)$ .