### Physics #25: Harmonic Motion

Hooke’s Law

An elastic object returns to its original shape after being stretched. To stretch a spring a distance x, apply a force proportional to x:

$F_a \equiv kx$

$\text{sgn}\{+\}\implies \text{stretching}$

$\text{sgn}\{-\}\implies \text{compression}$

The restoring force is the force the spring exerts on an object:

$F \equiv -kx.$

Simple Harmonic Motion

The stretch distance is given by

$x_i \equiv \frac{mg}{k}.$

Harmonic Displacement

$y \equiv A\sin\theta$

$y \equiv A \sin(\omega t)$

Given a period $T,$ the angular frequency is given as

$\omega \equiv \frac{2\pi}{T}$

in the units of radians per second.

This implies

$T \equiv \frac{2\pi}{omega}.$

The frequency, in cycles per second, is defined as:

$f \equiv \frac{1}{T} \implies Tf \equiv 1.$

From this, we see that

$\omega\equiv\left(\frac{2\pi}{T}\right)Tf \equiv 2\pi f \,\therefore\, y \equiv A\sin(2\pi ft).$

Harmonic Velocity

Harmonic velocity is given by the formula

$v \equiv A\omega \cos(\omega t)$

Note that

$\text{mag}(v) \equiv A\omega$

$\text{min}(v) \equiv -A\omega$

$\text{equil}(v) \equiv 0.$

Also note that

$\frac{\mathrm{d}(y\equiv A\sin(2\pi ft))}{\mathrm{d}t} \equiv v$

$\therefore \frac{\mathrm{d}}{\mathrm{d}t}\left(A\sin(\omega t)\right) \equiv A\omega\cos(\omega t).$

Harmonic Acceleration

Note that

$\frac{\mathrm{d}v}{\mathrm{d}t} \equiv a \equiv -A\omega^2\sin\theta$

Therefore,

$\iint -A\omega^2\sin\theta \sim A\sin(\omega t).$

Harmonic Dynamics

We find that, when mass is introduced into the equation,

$\omega \equiv \sqrt{\frac{k}{m}}$

$f \equiv \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

$T \equiv 2\pi\sqrt{\frac{m}{k}}$

Harmonic Energy

Elastic Kinetic Energy

$W \equiv \bar{F}s$

$W \equiv \left[\frac{1}{2}\left(F_f + F_i\right)\right](x_f - x_i)$

By Hooke’s Law:

$W \equiv \left[-\frac{1}{2}\left(kx_f + kx_i\right)\right](x_f - x_i)$

$W \equiv \frac{1}{2}kx_f^2 - \frac{1}{2}kx_i^2.$

Elastic Potential Energy

$E_p \equiv \frac{1}{2}kx^2.$

Pendulums

The torque of a pendulum is given as

$\tau \equiv -mgs$

If $L$ is the length of the string and $\theta$ is the angle of harmony, it follows that

$\tau\equiv -mgL\theta.$

If the quantity $mgL$ is constant, it is clear that the law is consistent with Hooke’s Law:

$F \equiv -[k]x$

$\tau \equiv -[mgL]\theta.$

Finally,

$\omega \equiv \sqrt{\frac{mgL}{I}} \equiv \sqrt{\frac{mgL}{mL^2}}$

$\therefore \omega \equiv \sqrt{\frac{g}{L}}$

$\therefore T \equiv 2\pi\sqrt{\frac{L}{g}}.$