Intensity and Sound Level — Question 10

Question 10

A large standing-wave pressure does not imply net transmitted power

Equal opposite progressive waves in a medium with impedance $Z=400\,\mathrm{Pa\,s/m}$ form $p=2P\cos kx\cos\omega t$ and $u=(2P/Z)\sin kx\sin\omega t$. Let $P=0.20\,\mathrm{Pa}$ and $p_0=20\,\mu\mathrm{Pa}$.

1. Find the pressure level at a pressure antinode.

2. Compute cycle-averaged signed intensity $\langle pu\rangle$ at arbitrary $x$. Explain why substituting the antinode RMS pressure into $I=p_{\mathrm{rms}}^2/Z$ is incorrect.

Question 10 - Solution

A large standing-wave pressure does not imply net transmitted power

(a) Method and calculation

At an antinode, pressure amplitude is $2P=0.40\,\mathrm{Pa}$, so$$p_{\mathrm{rms}}=\frac{0.40}{\sqrt2}=0.28284\,\mathrm{Pa},\quad L_p=20\log_{10}\frac{0.28284}{20\times10^{-6}}=83.010\,\mathrm{dB}.$$

(b) Interpretation and further calculation

The local flux is$$pu=\frac{4P^2}{Z}\cos kx\sin kx\cos\omega t\sin\omega t.$$Its cycle average is zero because $\langle\cos\omega t\sin\omega t\rangle=0$. The progressive-wave formula assumes pressure and velocity are in phase; the standing field violates that assumption. Opposite traveling components carry equal and opposite mean powers.

Physical check

Pressure level characterizes a local pressure fluctuation. It does not by itself determine the direction or net magnitude of acoustic energy transport.