Acceleration amplitude of a damped harmonic oscillator

In summary, the acceleration amplitude of a damped harmonic oscillator is given by a formula involving the parameters Q, F_o, and m. As the angular frequency, omega, tends to infinity, the acceleration amplitude approaches the ratio of F_o and m. This can be seen by examining the denominator and noticing that it becomes negligible as omega increases.
  • #1
TheBigDig
65
2

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.
 
Physics news on Phys.org
  • #2
Look at the denominator. What does it become when ##\omega## becomes very large? In other words, what is its ##\omega## dependence?
 
Back
Top