Forest UCM Osc SHM: Difference between revisions
Jump to navigation
Jump to search
Created page with " Forest_UCM_Osc#Simple_Harmonic_Motion_.28SHM.29" |
No edit summary |
||
| Line 1: | Line 1: | ||
==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]] | [[Forest_UCM_Osc#Simple_Harmonic_Motion_.28SHM.29]] | ||
Revision as of 14:27, 4 October 2014
Equation of motion
In solving the differential equation
- [math]\displaystyle{ 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]\displaystyle{ m\ddot x + c \dot x =-kx }[/math]energy is constant with time
one could rewrite the above as
- [math]\displaystyle{ \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]\displaystyle{ O = \frac{d}{dt} }[/math]
then the analogous equation becomes
- [math]\displaystyle{ \left ( mO^2 + aO + k \right ) x = 0 }[/math]
where m, a, and k are constants
factoring this quadratic you would have
- [math]\displaystyle{ \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]\displaystyle{ \left ( \frac{d}{dt} - \gamma \right ) \left ( \frac{d}{dt} + \beta \right ) x = 0 }[/math]
one of the solutions would be
- [math]\displaystyle{ \frac{d}{dt} x - \gamma x= 0 }[/math]
- [math]\displaystyle{ \frac{dx}{x} = \gamma dt }[/math]
- [math]\displaystyle{ x = Ae^{\gamma t} }[/math]
- [math]\displaystyle{ 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]\displaystyle{ \left ( m\frac{d}{dt}\frac{d}{dt} + k \right ) x = 0 }[/math]
or
- [math]\displaystyle{ \left ( mO^2 + k \right ) x = 0 }[/math]
- [math]\displaystyle{ O= \pm \sqrt{-\frac{k}{m}} = \pm i \sqrt{-\frac{k}{m}} }[/math]
- [math]\displaystyle{ \gamma \equiv +i\sqrt{\frac{k}{m}} = i \omega }[/math]
- [math]\displaystyle{ \beta = -i\sqrt{\frac{k}{m}} = -i \omega }[/math]
then you have
- [math]\displaystyle{ x = Ae^{\gamma t} + Be^{\beta t} }[/math]
- [math]\displaystyle{ = Ae^{i \omega t} + Be^{-i\omega t} }[/math]