Physics of Sound and Music HELPP!!?

Question

If a loudspeaker can produce a sound pressure level of 83 dB for 0.5 W of input power at a distance of 0.75 m, estimate its efficiency in converting electrical power to acoustic power. Assume the sound is uniformly radiated into a hemisphere.

Answer 1

Acoustic power is proportional to sound pressure^2, and from ref. 1, one watt is equivalent to 112 dB SPL at one meter (when radiated into a hemisphere). This SPL value is a ratio (pressure/ref. pressure) of 10^(112/20) = 398107.
83 dB SPL equals a pressure ratio (pressure/ref. pressure) of 10^(83/20) = 14125. Sound pressure is inversely proportional to distance, so this pressure ratio at .75 m is equivalent to .75/1 * 14125 = 10594 at one meter. Thus the acoustic power is 1*(10594/398107)^2 = .000708 w. Efficiency is .000708/.5 = .001416.
This can be checked by the ref. 2 calculator. Enter 83 in the first field, click calculate and you see that the power intensity = 0.000195 w/m^2. Multiply by 2*pi*.75^2 to get the total (hemispheric) power and you have .000689 w, agreeing within 3%. The small disagreement is because 120 dB SPL is defined as exactly 1 w/m^2, so the 112 dB equivalence to 1/(2*pi) w/m^2 is not exact. Thus the calculator solution is accurate, and the efficiency of the test speaker is .000689*2 = .001378.

Answer 2

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