Question

If you install special sound-reflecting windows that reduce the sound intensity level by 30.0 db , by what factor have you reduced the sound intensity?

Answer 1

Answer: The sound intensity was reduced by -14.77 dB.

Explanation:

The level of the intensity of the sound with respect to the reference value is called Sound Intensity level.

$L_I=10\log\frac{I}{I_o} dB$

I = Intensity of the sound

$I_o$ = Reference sound intensity

According to question, sound-reflecting windows that reduce the sound intensity level by 30.0 db .

Let the intensity of the sound be $I_o$ and reduced intensity be I.

$I=\frac{I_o}{30} dB$

$L_I=10\log\frac{I}{I_o} dB$

$L_I=10\log\frac{\frac{I_o}{30}}{I_o}=10\log\frac{1}{30} dB=-14.77 dB$

The sound intensity was reduced by -14.77 dB.