# I Driven forced oscillation wmax for Amax stuck on maths ><

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1. Aug 16, 2016

### NMS2

Hello

Need some help simplifying this. In relation to a driven oscillator I am looking for wmax for which the amplitude is maximum. I understand the theory and I am just missing something with the maths and I know its probably something so stooooopid Im missing >< I'm taking the derivative but I end up with something pretty convoluted and I know this is meant to be relatively simple but I just cant get it... Im staring at it too long, need someone else to look at it.

I have worked out the Amplitude as:

A= fo/m/sqrt((wo2-w2)22w2)

Then I attempted dA/dw to get wmax for which A is max and set to zero but it all got a little out of hand

-((fo/m)(2)(wo2-w2)(-2w)+2γw2))/((2)(wo2-w2)22w2)3/2)=0

I want to work with this expression because I understand where it came from and it makes sense to me. I just can not remember how to simplify something like this and it is going to keep me awake at night

I know that its supposed to boil down to this:(but I can not get my dA/dw to look like this. )

wmax= (wo2-(γ2/2))1/2

my goal is to figure out how to make my expression look like this

Thank you

2. Aug 16, 2016

### Twigg

To make life easier, instead of trying to maximize $\frac{1}{\sqrt{(\omega_{0}^{2} - \omega^{2})^{2} + \gamma^{2} \omega^{2}}}$, instead try minimizing $(\omega_{0}^{2} - \omega^{2})^{2} + \gamma^{2} \omega^{2}$. I guarantee you'll sleep easier.

If you're dead-set on getting your expression to work, then try substituting $f(\omega) = (\omega_{0}^{2} - \omega^{2})^{2} + \gamma^{2} \omega^{2}$ and using the chain rule to show that they share a stationary point.