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

$P_\mathrm{forward} = 0.75P$: the fraction of the power emitted to the forward direction

$P_\mathrm{backward} = 0.25P$: the fraction of the power emitted in towards the back

$\beta = 85~\mathrm{dB}$: sound intensity level observed by an observer in front of the observer by $x = 3.5~\mathrm{m}$ and below by $y = 2.5~\mathrm{m}$.
Useful Quantity
 $I_{0} = 1.0 \times 10^{12}~\mathrm{W/m^{2}}$: the reference intensity
Required Quantity
We calculate the power $P$ emitted by the loudspeaker.
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