Wave far from a speaker that produces a wave whose frequency changes with time

Question

Suppose there is a speaker at $x=0$ that produces sine waves with a frequency of $\omega(t)$. The sound waves right at the speaker can be described by $f(t,0)=\sin(\omega(t)t)$, and suppose that the speed of sound is 1 m/s. Now what do I hear at f.e. $x=5$m. Well I should hear what someone heard right at the spreaker 5 seconds ago. So the „wave function“ at x=5m should be $f(t,5)=f(t-5,0)=\sin(\omega(t-5)(t-5))$. Or more general $f(x,t)=\sin(\omega(t-x)(t-x))$

But now if we consider the phase of the wave for every time t at x: $\phi(t,x)$ such that $f(t,x)=\sin(\phi(t,x))$, then $\omega(t)$ tells us by how much this phase grows per unit time so $d\phi=\omega(t)dt \Rightarrow \phi(t)=\int_0^t\omega(t‘)dt‘$, and since $\phi(t,x)=\phi(t-x,0)$ (the speed of sound is one) this leads me to believe:

$f(t,x)=\sin(\int_0^{t-x}\omega(t‘)dt‘)$

But these two formulas are only the same if $\omega$ is constant. I know I‘m probably ignoring something very trivial but I can‘t see it right now, so I‘d be very happy if someone could help me:)

Answer 1

You are getting somewhat tripped up by looking at $x=5$. Stay at $x=0$ for a little longer.

You say that $\omega$ is the rate of change of the phase $$\phi(t)=t~\omega(t)$$ with respect to time. But it just obviously is not. The actual rate of change is $$ \frac{\mathrm d\phi}{\mathrm dt}=\omega(t)+t~\frac{\mathrm d\omega}{\mathrm dt}=\omega+t\dot\omega,$$by the product rule. Your integral should therefore be$$ \phi(t,x)=\int_0^{t-x} \mathrm dt'~\big( \omega(t')+t'~\dot\omega(t')\big). $$ Integrating the second term by parts, raising $\dot\omega(t')~\mathrm dt'$ and lowering $t'$, gives$$ \phi(t,x)=t'~\omega(t')\Big|_0^{t-x}+ \int_0^{t-x} \mathrm dt'~\big( \omega(t')-\omega(t')\big) $$as desired.