A sound measures 42 dB. The intensity of a second sound is four times as great. What is the decibel level of this second sound?

Question

asked Apr 13, 2022 99.6k views

1 Answer

Xiduzo · Answer 1 · 2022-04-19T06:02:36+0000

Answer:

The decibel level of this second sound is 48.021 decibels.

Explanation:

The acoustic intensity (
B_(dB) ), measured in decibels, is defined by the following formula:

$B_(dB) = 10\cdot \log_(10)\left((I)/(I_(o)) \right)$ (1)

Where:

I_(o) - Reference sound intensity, measured in watts per square meter.

- Real sound intensity, measured in watts per square meter.

If we know that
$B_(dB) = 42\,dB$ and
$I_(o) = 10^(-12)\,(W)/(m^(2))$ , then the real sound intensity of the first sound is:

$(B_(dB))/(10) = \log_(10)\left((I)/(I_(o)) \right)$

$(I)/(I_(o)) = 10^{(B_(dB))/(10) }$

$I = I_(o)\cdot 10^{(B_(dB))/(10) }$

$I = \left(10^(-12)\,(W)/(m^(2)) \right)\cdot 10^{(42\,dB)/(10) }$

$I = 1.585* 10^(-8)\,(W)/(m^(2))$

The real sound intensity of the second sound is four times greater, that is:

$I' = 6.340* 10^(-8)\,(W)/(m^(2))$

If we know that
$I_(o) = 10^(-12)\,(W)/(m^(2))$ and
$I' = 6.340* 10^(-8)\,(W)/(m^(2))$ , then the acoustic intensity of the second sound is:

$B_(dB) = 10\cdot \log_(10)\left((I')/(I_(o)) \right)$

$B_(dB) = 10\cdot \log_(10)\left((6.340* 10^(-8)\,(W)/(m^(2)) )/(10^(-12)\,(W)/(m^(2)) ) \right)$

$B_(dB) = 48.021\,dB$

The decibel level of this second sound is 48.021 decibels.