Question

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 dB sound intensity level when standing 3.5 m in front of and 2.5 m below the speaker. What is the speaker’s power output?

Answer 1

"0.049 W" is the correct answer.

Step-by-step explanation:

According to the given question,

`r = √((3.5)^2+(2.5)^2)`

`=√(8.5)`

`SL=85`

As we know,

⇒
`SL=10 \ log((I)/(I_o) )`

`85=10 \ log((I)/(10^(-12)) )`

`I=3.162* 1^(-4) \ W/m^2`

Now,

⇒
`P_(front) = I(2\pi r^2)`

`=(3.162* 10^(-4))(2\pi* 18.5)`

`=0.0368 \ W`

`=0.75 \ P`

or,

`=0.049 \ W`