Intensity and Sound Level — Question 8

Question 8

Pressure level and intensity level depend on compatible references

Define pressure level $L_p=20\log_{10}(p_{\mathrm{rms}}/p_0)$ with $p_0=20\,\mu\mathrm{Pa}$, and intensity level $L_I=10\log_{10}(I/I_0)$ with $I_0=10^{-12}\,\mathrm{W/m^2}$. A progressive wave has $p_{\mathrm{rms}}=0.020\,\mathrm{Pa}$ in a medium of impedance $Z=800\,\mathrm{Pa\,s/m}$.

1. Calculate both levels using $I=p_{\mathrm{rms}}^2/Z$.

2. Derive $L_I-L_p$ and find the impedance for which the two numerical levels agree.

Question 8 - Solution

Pressure level and intensity level depend on compatible references

(a) Method and calculation

The pressure ratio is $0.020/(20\times10^{-6})=1000$, so $L_p=60.0\,\mathrm{dB}$. But$$I=\frac{(0.020)^2}{800}=5.0\times10^{-7}\,\mathrm{W/m^2},\qquad L_I=56.990\,\mathrm{dB}.$$

(b) Interpretation and further calculation

Subtract the logarithms explicitly:$$L_I-L_p=10\log_{10}\left(\frac{p_0^2}{ZI_0}\right).$$Equality requires $Z=p_0^2/I_0=400\,\mathrm{Pa\,s/m}$. At $Z=800$, the difference is $10\log_{10}(1/2)=-3.0103\,\mathrm{dB}$. The pressure reference is fixed here for comparison; references must be stated when comparing different media.

Physical check

The difference is independent of measured pressure amplitude for fixed impedance and references.