- #1
SandyMan32
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I'm trying to figure out how to derive the equations for Energy from the differential equation corresponding to the (simple and damped) harmonic oscillator. Please note that I don't want to start with the expressions for kinetic and potential energy, I want to derive them. The references that I have referred to seem to have left out a step critical for my understanding.
For the simple oscillator my source starts with the differential equation
[tex] \dfrac{d^2}{dt^2}x(t) +\omega_0^2 x(t) =0 [/tex]
multiplying by [tex] \dot{x} = \dfrac{d}{dt}x(t) [/tex] gives
[tex] \dfrac{d}{dt} \left[ \dfrac{1}{2}( \dot{x}^2+ \omega_0^2 x^2) \right]=0 [/tex]
Integrating, gives the constant of motion
[tex] \dfrac{E}{m}=\dfrac{1}{2}(\dot{x}^2+ \omega_0^2 x^2) [/tex]
Similarly, for the damped oscillator my source starts with the differential equation
[tex] \ddot{x} + \gamma \dot{x} +\omega_0^2 x=0 [/tex]
multiplies by [tex]\dot{x}[/tex] giving
[tex] \dot{x}\ddot{x} + \gamma \dot{x} ^2 +\omega_0^2 \dot{x}x=0 [/tex]
then arrives at the answer
[tex] \dfrac{d}{dt}E = -m\gamma\dot{x}^2[/tex]
Unfortunately, in both of the previous examples I do not understand how to arrive at the last line given. Any help would be much appreciated.
For the simple oscillator my source starts with the differential equation
[tex] \dfrac{d^2}{dt^2}x(t) +\omega_0^2 x(t) =0 [/tex]
multiplying by [tex] \dot{x} = \dfrac{d}{dt}x(t) [/tex] gives
[tex] \dfrac{d}{dt} \left[ \dfrac{1}{2}( \dot{x}^2+ \omega_0^2 x^2) \right]=0 [/tex]
Integrating, gives the constant of motion
[tex] \dfrac{E}{m}=\dfrac{1}{2}(\dot{x}^2+ \omega_0^2 x^2) [/tex]
Similarly, for the damped oscillator my source starts with the differential equation
[tex] \ddot{x} + \gamma \dot{x} +\omega_0^2 x=0 [/tex]
multiplies by [tex]\dot{x}[/tex] giving
[tex] \dot{x}\ddot{x} + \gamma \dot{x} ^2 +\omega_0^2 \dot{x}x=0 [/tex]
then arrives at the answer
[tex] \dfrac{d}{dt}E = -m\gamma\dot{x}^2[/tex]
Unfortunately, in both of the previous examples I do not understand how to arrive at the last line given. Any help would be much appreciated.