Simple Harmonic Motion
Equation of motion
In solving the differential equation
- [math] m\ddot x =-kx [/math]energy is constant with time
and observe that the above differential equation is a special case of the more general differential equation
- [math] m\ddot x + c \dot x =-kx [/math]energy is constant with time
one could rewrite the above as
- [math] \left ( \frac{d}{dt}\frac{d}{dt} +\frac{d}{dt} + k \right ) x = 0 [/math]
One could cast the above differential equation into an analogous quadratic equation if you
let
- [math]O = \frac{d}{dt}[/math]
then the analogous equation becomes
- [math] \left ( mO^2 + aO + k \right ) x = 0 [/math]
where m, a, and k are constants
factoring this quadratic you would have
- [math] \left ( O - \gamma \right )\left ( O - \beta \right ) x = 0 [/math]
where a non-trivial solution would exist if one of the terms in the parentheses were zero
this basically reduces our 2nd order differential equation down to two first order differential equations
- [math] \left ( \frac{d}{dt} - \gamma \right ) \left ( \frac{d}{dt} + \beta \right ) x = 0 [/math]
one of the solutions would be
- [math] \frac{d}{dt} x - \gamma x= 0 [/math]
- [math] \frac{dx}{x} = \gamma dt [/math]
- [math] x = Ae^{\gamma t} [/math]
- [math] x = Ae^{\gamma t} + Be^{\beta t} [/math]
For the special case where there isn't a first derivative term (a=0)
You simply have
- [math] \left ( m\frac{d}{dt}\frac{d}{dt} + k \right ) x = 0 [/math]
or
- [math] \left ( mO^2 + k \right ) x = 0 [/math]
- [math] O= \pm \sqrt{-\frac{k}{m}} = \pm i \sqrt{-\frac{k}{m}}[/math]
- [math] \gamma \equiv +i\sqrt{\frac{k}{m}} = i \omega[/math]
- [math] \beta = -i\sqrt{\frac{k}{m}} = -i \omega[/math]
then you have
- [math] x = Ae^{\gamma t} + Be^{\beta t} [/math]
- [math] = Ae^{i \omega t} + Be^{-i\omega t} [/math]
Forest_UCM_Osc#Simple_Harmonic_Motion_.28SHM.29