Forest UCM Osc SHM

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