The Speed of Sound — Question 8

Question 8

A temperature gradient produces a different travel-time law

In a vertical column, $T(z)=T_0(1+\beta z/H)$ and $c(z)=c_0\sqrt{T(z)/T_0}$. Let $H=100\,\mathrm m$, $c_0=340\,\mathrm{m/s}$ and $\beta=0.21$. A pulse travels vertically upward; neglect wind and bending.

1. Find the one-way travel time by integration.

2. Find the constant speed giving the same travel time, and compare it with the speed at the mean temperature.

Question 8 - Solution

A temperature gradient produces a different travel-time law

(a) Method and calculation

Integrating reciprocal speed gives$$t=\frac H{c_0}\int_0^1\frac{du}{\sqrt{1+\beta u}}=\frac{2H}{c_0\beta}(\sqrt{1+\beta}-1).$$Rationalizing yields $t=2H/[c_0(1+\sqrt{1+\beta})]=200/714=0.28011\,\mathrm s$.

(b) Interpretation and further calculation

The equivalent speed is$$c_{\mathrm{eff}}=\frac Ht=\frac{c_0+c(H)}2=\frac{340+374}2=357\,\mathrm{m/s}.$$At mean temperature, $c(\overline T)=340\sqrt{1.105}=357.405\,\mathrm{m/s}$. Equality with the endpoint-speed mean is special to this square-root profile; substituting an average temperature is not the exact travel-time calculation.

Physical check

The limit $\beta\to0$ is finite and gives $H/c_0$; rationalization avoids an apparent zero divided by zero.