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

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:)

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.