- #1
Showaddywaddy
- 2
- 0
Homework Statement
Prove that that the power given by [itex] \bar{P} = \frac{1}{2} \gamma m \omega_r^2 A_{(\omega)}^2 [/itex] is at a maximum for [itex] \omega_r = \omega_0 [/itex]
Only variable is [itex] \omega_r [/itex]
[itex] \omega_r [/itex] is the resonant frequency of the external force while [itex] \omega_0 [/itex] is the eigen frequency of the system.
Homework Equations
[tex] A_{(\omega)} = \frac{\frac{F_0}{m}}{\sqrt{{(\omega_0^2-\omega_r^2)}^2+{(\gamma \omega_r)}^2}} [/tex]
The Attempt at a Solution
I have subbed in the equation for the amplitude into the equation for the power and then differentiated it and let it equal to zero to find the max power. The differentiated gets pretty messy but I still can't seem to get to the right answer that [itex] \omega_r = \omega_0 [/itex] no matter how many times I differentiate it. My notes from lectures has the proof but my lecturer has differentiated the equation in a really unusual way so I can't follow it.