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I know I'm doing something stupid but I just can't find out where. Suppose that I have a sine wave whose angular frequency varies linearly with time. So:

$\omega = ct$

And the sine wave would be:

$S=sin(\omega t) =sin(ct^2)$

Now suppose I have $S$ and I would like to find its angular frequency. For that, I believe I should differentiate the angle of the sinusoid, which gives

$\omega=\frac{d}{dt}ct^2 = 2ct$

and this is twice the value I started with

$\omega = ct$

Where did the "2" come from?!

Ahmad KFUPM
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  • If you know "the angle of the sinusoid" $,f(t) = ct^2,$ then you know $,c = f(t) / t^2,$ and $,\omega=f(t)/t,$. If you don't know $,f(t),$ then what exactly is it that you know? – dxiv Aug 28 '21 at 05:28
  • I know the angle, but shouldn't $\omega$ be the time derivative of the angle? If you differentiate the angle, you will get $\omega=2ct$ – Ahmad KFUPM Aug 28 '21 at 05:31
  • "shouldn't ω be the time derivative of the angle" $;-;$ Only if $,\omega,$ is constant, which in your case it's not. Otherwise $\displaystyle \frac{d\left(\omega \cdot t\right)}{dt}= \frac{d \omega}{dt} \cdot t + \omega \cdot \frac{d t}{d t}$. – dxiv Aug 28 '21 at 05:32

1 Answers1

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You are using $\omega$ in two different ways. There is no relation between the $\omega$ in the first and second equations and the $\omega$ in the third. The $\omega$ in the third is the instantaneous rate of increase of the angle. It is, in fact, twice the $\omega$ of the first two equations, as you have shown. Give it a new name and the problem disappears.

Ross Millikan
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