3

This is an example problem from my textbook. I read through its solution, but was confused by the following step.

$$x=\frac{\omega^{2}A\cos(\omega t-\delta)}{\sqrt{(\omega_{0}^{2}-\omega^{2})^{2}+\omega^{2}\gamma^{2}}}+A\cos\omega t$$

The textbook proceeds to say that, quote

Since $x$ is a superposition of two cosine terms in $\omega t$, we can write it as $$x=C(\omega)\cos(\omega t-\alpha)$$, where $$[C(\omega)]^2=\frac{A^{2}(\omega_{0}^{4}+\omega^{2}\gamma^{2})}{(\omega_{0}^{2}-\omega^{2})^{2}+\omega^{2}\gamma^{2}}$$

I am confused how it gets there

Sirou Ewei
  • 199

1 Answers1

4

First, recall that $\cos(\omega t - \delta) = \cos \omega t \cos \delta + \sin \omega t \sin \delta$. This lets you write the expression as $M \cos \omega t + N \sin \omega t$, where $M$ and $N$ may depend on $\omega$ and $\delta$ but are independent of $t$.

Then, set $M \cos \omega t + N \sin \omega t = C \cos(\omega t - \alpha) = C \cos \omega t \sin \alpha + C \sin \omega t \cos \alpha$, so you can then solve $C \sin \alpha = M$ and $C \cos \alpha = N$ for $C$ and $\alpha$.

ConMan
  • 24,300