Acceleration amplitude of a damped harmonic oscillator

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SUMMARY

The acceleration amplitude of a damped harmonic oscillator is defined by the equation $$A_{acc}(\omega) = \frac{QF_o}{m} \frac{\omega}{\omega _o} \sqrt{\it{R}(\omega)}$$. As the frequency approaches infinity, the limit of the acceleration amplitude simplifies to $$\lim_{\omega\to\infty} A_{acc}(\omega) = \frac{F_o}{m}$$. The function $$R(\omega)$$ is given by $$\it{R}(\omega) = \frac{(\gamma \omega)^2}{(\omega _o ^2 - \omega^2)^2 +(\gamma \omega)^2}$$, where $$Q = \frac{\omega _o}{\gamma}$$. The discussion highlights the importance of analyzing the behavior of the denominator as $$\omega$$ increases to understand the limit.

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Homework Statement


The acceleration amplitude of a damped harmonic oscillator is given by
$$A_{acc}(\omega) = \frac{QF_o}{m} \frac{\omega}{\omega _o} \sqrt{\it{R}(\omega)}$$
Show that as ##\lim_{\omega\to\infty}, A_{acc}(\omega) = \frac{F_o}{m}##

Homework Equations


$$\it{R}(\omega) = \frac{(\gamma \omega)^2}{(\omega _o ^2 - \omega^2)^2 +(\gamma \omega)^2} $$
$$ Q = \frac{\omega _o}{\gamma}$$

The Attempt at a Solution


From substituting in the above equations into the formula and cancelling off, I've gotten this far
$$\frac{F_o}{m} \frac{\omega ^2}{\sqrt{(\omega _o ^2 - \omega^2)^2 +(\gamma \omega)^2}}$$
I'm fairly certain I haven't made any mistakes in my cancelling off. I don't see how the equation will tend to ##\frac{F_o}{m}## as surely ##\frac{\omega ^2}{\sqrt{(\omega _o ^2 - \omega^2)^2 +(\gamma \omega)^2}}## will go to zero.
 
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Look at the denominator. What does it become when ##\omega## becomes very large? In other words, what is its ##\omega## dependence?
 

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