Driven Harmonic Oscillator: Proving that the max power is given by ω_r = ω_0

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SUMMARY

The discussion centers on proving that the maximum power, represented by the equation \(\bar{P} = \frac{1}{2} \gamma m \omega_r^2 A_{(\omega)}^2\), occurs at the resonant frequency \(\omega_r = \omega_0\). The amplitude \(A_{(\omega)}\) is defined as \(A_{(\omega)} = \frac{\frac{F_0}{m}}{\sqrt{{(\omega_0^2-\omega_r^2)}^2+{(\gamma \omega_r)}^2}}\). Participants suggest substituting this amplitude into the power equation and simplifying it by dividing both the numerator and denominator by \(\omega_r^2\) to facilitate the proof without extensive calculus. The consensus is that this approach leads to a clearer understanding of the relationship between the frequencies.

PREREQUISITES
  • Understanding of harmonic oscillators and resonant frequency
  • Familiarity with power equations in physics
  • Basic knowledge of calculus for differentiation
  • Concept of amplitude in oscillatory systems
NEXT STEPS
  • Study the derivation of power equations in driven harmonic oscillators
  • Learn about the implications of resonance in physical systems
  • Explore advanced calculus techniques for simplifying complex equations
  • Investigate the role of damping (\(\gamma\)) in oscillatory motion
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Students studying physics, particularly those focusing on oscillatory motion, as well as educators seeking to clarify concepts related to driven harmonic oscillators and resonance.

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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.
 
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Try dividing numerator and denominator of ##\bar{P}## by ##\omega_{r}^2## and simplifying. I don't think you'll need calculus to see the result.
 

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