Driven Oscillator

#Physics

$\ddot x + 2 \beta \dot x + \omega_0^2 x = \frac{F_0}{m}\cos \omega t$

• 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: $\omega_0 < \beta$
  $$H(t_0) = \begin{cases}
  0, \quad t < t_0\\a, \quad t > t_0
  \end{cases}$$
  $$I(t_0, t_1) = H(t_0) - H(t_1) = \begin{cases}
  0, \quad t < t_0\\a, \quad t_0 < t < t_1\\0, \quad t > t_1
  \end{cases}$$
  For $x(t') = 0, \dot x(t') = 0$
  $$G(t, t') = \begin{cases}
  \frac{1}{m\omega_1}e^{-\beta(t - t')}\sin (\omega_1(t - t')) \quad t \ge t'\\0, \quad t < t'
  \end{cases}$$
  $x_p(t) = \int_{-\infty}^t F(t')G(t, t') dt'$