Traveling Sound Waves — Question 5

Question 5

Equal pressure amplitudes can hide unequal displacement amplitudes

Two separate progressive plane waves in the same fluid have equal pressure amplitude $P=1.0\,\mathrm{Pa}$ but frequencies $f_1=250\,\mathrm{Hz}$ and $f_2=1000\,\mathrm{Hz}$. The fluid has $Z=400\,\mathrm{Pa\,s/m}$ and $c=320\,\mathrm{m/s}$.

1. Compare particle-velocity and displacement amplitudes, including numerical values.

2. Compare their strain amplitudes $ka$. Explain why the higher frequency does not make the compression amplitude larger here.

Question 5 - Solution

Equal pressure amplitudes can hide unequal displacement amplitudes

(a) Method and calculation

For each wave, $u_m=P/Z$ and $a=u_m/(2\pi f)$. Therefore$$u_{m1}=u_{m2}=0.00250\,\mathrm{m/s},$$$$a_1=1.592\,\mu\mathrm m,\qquad a_2=0.3979\,\mu\mathrm m,\qquad a_2/a_1=1/4.$$

(b) Interpretation and further calculation

Since $k=\omega/c$,$$ka=\frac{\omega}{c}\frac{P}{Z\omega}=\frac{P}{Zc}=7.8125\times10^{-6}.$$Both waves have the same strain and density-perturbation amplitudes. Higher $k$ is exactly offset by smaller $a$ when pressure amplitude and medium are fixed.

Physical check

$B=Zc=128000\,\mathrm{Pa}$, so $ka=P/B$ is independently consistent with the compression law.