A loudspeaker emits sound waves isotropically (uniformly in all directions). The average acoustic power delivered by the loudspeaker is P = 5.00 W . There are two 1. S_av listeners, one is at a distance R¸ =12.0 m , and the other is at a distance R, = 24.0 m. a. What is the intensity of the sound (in W / m² ) at the listener A's position, I ?

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A loudspeaker emits sound waves isotropically (uniformly in all directions). The
average acoustic power delivered by the loudspeaker is P = 5.00 W . There are two
1.
_av
listeners, one is at a distance R, =12.0 m, and the other is at a distance R, = 24.0 m.
a. What is the intensity of the sound (in W / m² ) at the listener A's position, I, ?
Transcribed Image Text:A loudspeaker emits sound waves isotropically (uniformly in all directions). The average acoustic power delivered by the loudspeaker is P = 5.00 W . There are two 1. _av listeners, one is at a distance R, =12.0 m, and the other is at a distance R, = 24.0 m. a. What is the intensity of the sound (in W / m² ) at the listener A's position, I, ?
Expert Solution
Step 1

Given that

The average acoustic power delivered by the loudspeaker is

Ps_av=5.00 W

and distance 

RA=12.0 mRB=24.0 m

Therefore cross sectional area for distance RA is given by

Cross sectional area=π×RA2Cross sectional area=3.14×12.0 m×12.0 mCross sectional area=452.16 m2

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