Physics #25: Harmonic Motion

by Qubit Factory

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

Additionally,

\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}}.

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