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

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?

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bonexptip 4 years 2021-08-15T11:28:44+00:00 1 Answers 274 views 0

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  1. Orla
    0
    2021-08-15T11:29:47+00:00
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    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.

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