Question

A loudspeaker, mounted on a tall pole, is engineered to emit 75%75 \% of its sound energy into the forward hemisphere, 25%25 \% toward the back. You measure an 85 dB85 \mathrm{~dB} sound intensity level when standing 3.5 m3.5 \mathrm{~m} in front of and 2.5 m2.5 \mathrm{~m} below the speaker. What is the speaker's power output?

Solution

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Given Quantities

  • Pforward=0.75PP_\mathrm{forward} = 0.75P: the fraction of the power emitted to the forward direction

  • Pbackward=0.25PP_\mathrm{backward} = 0.25P: the fraction of the power emitted in towards the back

  • β=85 dB\beta = 85~\mathrm{dB}: sound intensity level observed by an observer in front of the observer by x=3.5 mx = 3.5~\mathrm{m} and below by y=2.5 my = 2.5~\mathrm{m}.

Useful Quantity

  • I0=1.0×1012 W/m2I_{0} = 1.0 \times 10^{-12}~\mathrm{W/m^{2}}: the reference intensity

Required Quantity

We calculate the power PP emitted by the loudspeaker.

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