Question

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?

Answer 1

Answer:

The decibel level of this second sound is 48.021 decibels.

Step-by-step explanation:

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

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

Where:

$I_{o}$ - Reference sound intensity, measured in watts per square meter.

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

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

$\frac{B_{dB}}{10} = \log_{10}\left(\frac{I}{I_{o}} \right)$

$\frac{I}{I_{o}} = 10^{\frac{B_{dB}}{10} }$

$I = I_{o}\cdot 10^{\frac{B_{dB}}{10} }$

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

$I = 1.585\times 10^{-8}\,\frac{W}{m^{2}}$

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

$I' = 6.340\times 10^{-8}\,\frac{W}{m^{2}}$

If we know that $I_{o} = 10^{-12}\,\frac{W}{m^{2}}$ and $I' = 6.340\times 10^{-8}\,\frac{W}{m^{2}}$, then the acoustic intensity of the second sound is:

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

$B_{dB} = 10\cdot \log_{10}\left(\frac{6.340\times 10^{-8}\,\frac{W}{m^{2}} }{10^{-12}\,\frac{W}{m^{2}} } \right)$

$B_{dB} = 48.021\,dB$

The decibel level of this second sound is 48.021 decibels.