Driven Oscillator

$\ddot{x}+2\beta \dot{x}+{\omega }_0^{2}x=\frac{F_0}{m}\cos \omega t$

• Driven oscillation differential equation
• $\beta$ is the damping/resistivity constant
• ${\omega }_0$ is the angular frequency
• $F$ is the force supplied by the driven oscillator

$x=x_h+x_p$

$x_p=D\cos (\omega t-\delta ),D=\frac{F_0}{m\sqrt{({\omega }_0^2-{\omega }^2{)}^2+4{\beta }^2{\omega }^2}},\delta ={\tan }^{-1}\left(\frac{2\beta \omega }{{\omega }_0^2-{\omega }^2}\right)$

${\omega }_R=\sqrt{{\omega }_0^2-2{\beta }^2}$
The resonance frequency of a driven oscillation is given by the oscillator’s natural frequency and its damping coefficient
${\omega }_R$ is specifically for the amplitude resonance
- ${\omega }_E={\omega }_0$
${\omega }_E$ is the kinetic energy resonance frequency, the frequency of the oscillator that minimizes energy loss due to the
- $Q=\frac{{\omega }_R}{2\beta }\cong \frac{{\omega }_0}{\Delta \omega }$
$Q$ is the quality or Q factor of an oscillation
$\Delta \omega$ is the width of the frequency such that the square of the amplitude has decreased by half. The approximation is valid for $\beta \ll {\omega }_0$

Tells how underdamped an oscillator is
As $Q$ increases, the oscillator becomes more underdamped/less damped
Resonance: ${\omega }_0>\sqrt{2}\beta$
No resonance: ${\omega }_0<\sqrt{2}\beta$
Overdamping: ${\omega }_0<\beta$
$H(t_0)=\{\begin{array}{l}0,t<t_0\\ a,t>t_0\end{array}$
$I(t_0,t_1)=H(t_0)-H(t_1)=\{\begin{array}{l}0,t<t_0\\ a,t_0<t<t_1\\ 0,t>t_1\end{array}$
For $x(t^{\prime })=0,\dot{x}(t^{\prime })=0$
$G(t,t^{\prime })=\{\begin{array}{l}\frac{1}{m{\omega }_1}{e}^{-\beta (t-t^{\prime })}\sin ({\omega }_1(t-t^{\prime }))t\ge t^{\prime }\\ 0,t<t^{\prime }\end{array}$
$x_p(t)={\int }_{-\infty }^{t}F(t^{\prime })G(t,t^{\prime })d{t}^{\prime }$