Intensity and Sound Level — Question 9

Question 9

An inclined receiving surface intercepts projected acoustic area

A uniform plane wave has intensity $I=2.0\times10^{-3}\,\mathrm{W/m^2}$. It strikes a flat plate of actual area $A=0.50\,\mathrm{m^2}$ at $60^\circ$ to its normal. The plate absorbs $30\%$ of incident energy; the rest is reflected or transmitted.

1. Find incident and absorbed powers.

2. Find absorbed energy in $40\,\mathrm s$ and compare with normal incidence. Why must the angle be measured from the normal?

Question 9 - Solution

An inclined receiving surface intercepts projected acoustic area

(a) Method and calculation

Intensity is flux through a surface perpendicular to propagation. The projected receiving area is $A\cos\theta$, so$$W_{\mathrm{in}}=IA\cos60^\circ=5.0\times10^{-4}\,\mathrm W,$$$$W_{\mathrm{abs}}=0.30W_{\mathrm{in}}=1.5\times10^{-4}\,\mathrm W.$$

(b) Interpretation and further calculation

The absorbed energy is$$E=W_{\mathrm{abs}}\Delta t=(1.5\times10^{-4})(40)=0.0060\,\mathrm J.$$Normal incidence doubles this to $0.0120\,\mathrm J$. The surface flux is $\mathbf I\cdot\hat{\mathbf n}$, which explains the cosine of the normal angle; using the surface angle requires a sine instead.

Physical check

At grazing incidence the projected area tends to zero in this geometric model. Finite-edge diffraction is neglected.