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

Answer:

"0.049 W" is the correct answer.

Explanation:

According to the given question,

$r = \sqrt{(3.5)^2+(2.5)^2}$

  $=\sqrt{8.5}$

$SL=85$

As we know,

⇒  $SL=10 \ log(\frac{I}{I_o} )$

     $85=10 \ log(\frac{I}{10^{-12}} )$

      $I=3.162\times 1^{-4} \ W/m^2$

Now,

⇒  $P_{front} = I(2\pi r^2)$

               $=(3.162\times 10^{-4})(2\pi\times 18.5)$

               $=0.0368 \ W$

               $=0.75 \ P$

or,

                $=0.049 \ W$