# Long term solution for a driven harmonic oscillator

Let $$F(t)= \cos (\omega_d t)$$ be the driving force of a harmonic oscillator of mass $$m$$ which is damped with a damping constant $$b$$ such that $$F= -bv$$ is the damping force and the spring exerts a force $$F=-kx$$

A 2nd D.E. is obtained of the form: $$\ddot{x}+2\beta \dot{x}+\omega_0^2 x = \frac{F}{m}\cos(\omega_dt)$$ where $$\beta=\frac{b}{2m}, \omega_0=\sqrt{\frac{k}{m}}$$ the natural frequency of the oscillator.

My professor gave the long term solution to this as: $$x(t)=\frac{F_0 \cos (\omega_d t + \sigma)}{\sqrt{(\omega_0^2-\omega_d^2)^+(\beta \omega_d)^2}}$$ where $$\sigma=\arctan \left(\frac{\beta \omega_d}{\omega_0^2-\omega_d^2}\right)$$ I wonder whether he made a mistake in the expression for $$\beta$$ and it should be $$\beta=\frac{b}{m}$$ instead of $$\beta=\frac{b}{2m}$$

Your professor started with $$\beta=\frac{b}{2m}$$ but finished with $$\beta=\frac{b}{m}$$. Note that the equation without parameters is $$\ddot{x}+\frac{b}{m} \dot{x}+\frac{k}{m}x = \frac{F}{m}cos(\omega_d t)$$ and now with $$\beta=\frac{b}{2m}, \omega_0=\sqrt{\frac{k}{m}}$$: $$\ddot{x}+2\beta \dot{x}+\omega_0^2x = \frac{F}{m}cos(\omega_d t)$$ And the solution is: $$x(t)=\frac{F_0 \cos (\omega_d t + \sigma)}{\sqrt{(\omega_0^2-\omega_d^2)^+(2\beta \omega_d)^2}}$$ where $$\sigma=\arctan \left(\frac{2\beta \omega_d}{\omega_0^2-\omega_d^2}\right)$$