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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 $\beta=\frac{b}{2m}$

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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 $$\sigma=\arctan \left(\frac{2\beta \omega_d}{\omega_0^2-\omega_d^2}\right)$$

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