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?

Respuesta :

Answer:

"0.049 W" is the correct answer.

Explanation:

According to the given question,

[tex]r = \sqrt{(3.5)^2+(2.5)^2}[/tex]

  [tex]=\sqrt{8.5}[/tex]

[tex]SL=85[/tex]

As we know,

⇒  [tex]SL=10 \ log(\frac{I}{I_o} )[/tex]

     [tex]85=10 \ log(\frac{I}{10^{-12}} )[/tex]

      [tex]I=3.162\times 1^{-4} \ W/m^2[/tex]

Now,

⇒  [tex]P_{front} = I(2\pi r^2)[/tex]

               [tex]=(3.162\times 10^{-4})(2\pi\times 18.5)[/tex]

               [tex]=0.0368 \ W[/tex]

               [tex]=0.75 \ P[/tex]

or,

                [tex]=0.049 \ W[/tex]