Periodic motion

Simple harmonic motion (SHM)

\begin{align*} \omega &= 2\pi f = \frac{2\pi}{T} \\ f &= \frac{\omega}{2\pi} \\ T &= \frac{1}{f} = \frac{2\pi}{\omega} \end{align*}

\begin{align*} x &= A\cos(\omega t + \phi) \\ v_x &= -\omega A\sin(\omega t + \phi) \\ v_{max} &= \omega A \\ a_x &= -\omega^2 A\cos(\omega t + \phi) \\ a_{max} &= \omega^2 A \\ A &= \sqrt{x_0^2 + \frac{v_0^2}{\omega^2}} \end{align*}

Phase angle

\phi

If $x_0>0$

\phi = \tan^{-1}\left(-\frac{v_0}{wx_0}\right)

If $x_0<0$

\phi = \tan^{-1}\left(-\frac{v_0}{wx_0}\right)+\pi

It tells the displacement of the particle $x_0$ at $t = 0$

Block-spring	Angular oscillator	Simple pendulum	Physical pendulum
$\omega = \sqrt{\dfrac{k}{m}}$	$\omega = \sqrt{\dfrac{\kappa}{I}}$	$\omega = \sqrt{\dfrac{g}{L}}$	$\omega = \sqrt{\dfrac{mgd}{I}}$
$k$ : spring constant	$\kappa$ : torsion constant	$L$ : length of the pendulum	$d$ : distance from the pivot to its CG

\begin{align*} E &= K + U = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 \\ E &= U_{max} = \frac{1}{2}kA^2 \end{align*}

\begin{align*} x &= Ae^{-\left(\frac{b}{2m}\right)t}\cos(\omega't+\phi) \\ \omega' &= \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}} \end{align*}

info

$b < 2\sqrt{km}$ : under damping (oscillation dies off)
$b = 2\sqrt{km}$ : critical damping (no oscillation, return to equilibrium in shortest time)
$b > 2\sqrt{km}$ : over damping (no oscillation)

\begin{align*} A = \dfrac{F_{max}}{\sqrt{(k-m\omega_d^2)^2 + b^2\omega_d^2}} \end{align*}

Resonance

When $\omega_d \rightarrow \omega$ (natural angular frequency of an undamped oscillator), $A$ reaches maximum (resonance peak)
Smaller $b$ (lighter damping) gives a sharper peak
If $b \ge 2\sqrt{km}$ , the peak disappears completely