Intensity and Sound Level — Question 5

Question 5

A radiation cone concentrates a fixed acoustic power

A source emits acoustic power $W=0.020\,\mathrm W$ uniformly within a cone of half-angle $60^\circ$, and zero outside it. The cone solid angle is $\Omega=2\pi(1-\cos\theta)$. Neglect absorption; use $I_0=10^{-12}\,\mathrm{W/m^2}$.

1. Find intensity and level inside the cone at radius $10.0\,\mathrm m$.

2. Compare with an isotropic source of the same total power. Verify conservation of power through the spherical cap.

Question 5 - Solution

A radiation cone concentrates a fixed acoustic power

(a) Method and calculation

The cone occupies $\Omega=2\pi(1-1/2)=\pi\,\mathrm{sr}$, so the cap area is $\Omega r^2$. Hence$$I=\frac W{\Omega r^2}=\frac{0.020}{100\pi}=6.3662\times10^{-5}\,\mathrm{W/m^2},$$$$\beta=78.039\,\mathrm{dB}.$$

(b) Interpretation and further calculation

The isotropic intensity is $W/(4\pi r^2)$, one quarter as large. The directed level is therefore higher by $10\log_{10}4=6.0206\,\mathrm{dB}$. Integrating over the cap confirms$$\int_{\mathrm{cap}}I\,dA=I\Omega r^2=W.$$Concentration changes angular distribution while conserving total power.

Physical check

As the solid angle decreases at fixed power, in-cone intensity increases. The ideal sharp cone edge is a supplied model, not a diffraction calculation.