# Simple Harmonic Motion Derivation of frequency

• Jun 9th 2011, 08:27 AM
imagemania
Simple Harmonic Motion Derivation of frequency
Ok i know how to deduce the angular frequency for the basic problem of:
$\displaystyle \frac{d^{2}x}{dt^{2}} = -\frac{kx}{m}$
You simple do:
$\displaystyle x = -\alpha cos(\omega t) + \beta sin(\omega t)$
x ' =...
x'' = ...

Hence$\displaystyle x'' = -\omega^{2}x$
So $\displaystyle \omega = \sqrt{\frac{k}{m}}$

But how do i do the same for the equation:
$\displaystyle \frac{d^{2}x}{dt^{2}} = -\frac{kx}{m} - \frac{b}{m}\frac{dx}{dt} + \frac{F_{0}}{m}Cos(\alpha t)$
I know it should be:

$\displaystyle \omega = \sqrt{\frac{k}{m}-b^{2}/4m^{2}}$

However, with the damping expression i don't think i'll be able to use the same approach

any help is much appreciated :)
• Jun 10th 2011, 09:38 PM
CaptainBlack
Quote:

Originally Posted by imagemania
Ok i know how to deduce the angular frequency for the basic problem of:
$\displaystyle \frac{d^{2}x}{dt^{2}} = -\frac{kx}{m}$
You simple do:
$\displaystyle x = -\alpha cos(\omega t) + \beta sin(\omega t)$
x ' =...
x'' = ...

Hence$\displaystyle x'' = -\omega^{2}x$
So $\displaystyle \omega = \sqrt{\frac{k}{m}}$

But how do i do the same for the equation:
$\displaystyle \frac{d^{2}x}{dt^{2}} = -\frac{kx}{m} - \frac{b}{m}\frac{dx}{dt} + \frac{F_{0}}{m}Cos(\alpha t)$
I know it should be:

$\displaystyle \omega = \sqrt{\frac{k}{m}-b^{2}/4m^{2}}$

However, with the damping expression i don't think i'll be able to use the same approach

any help is much appreciated :)

The solution at large times is a sinusoid at the driving frequency.

CB