Chapter 3.5, Problem 48E

a .

To determine

To find: the number of decibels $\beta$ of the sound of intensity $I={10}^{-11}{\text{ watt per m}}^2$ .

Answer to Problem 48E

  $10dB$

Explanation of Solution

Given:

The number of decibels $\beta$ of a sound with an intensity of I watts per square meter is given by:

  $\beta =10\log \left(\frac{I}{I_0}\right)$.........(1)

Where $I_0={10}^{-12}{\text{ watt per m}}^2$

Formula used:

1. $\frac{a^m}{a^n}=a^{m-n}$

2. ${\log }_a{a}^x=x$

Substitute $I={10}^{-11}{\text{ watt per m}}^2$ and $I_0={10}^{-12}{\text{ watt per m}}^2$ in the equation (1), to find the number of decibels $\beta$ ,

  $\begin{array}{l}\beta =10\log \frac{I}{I_0}\\ \beta =10\log \left(\frac{{10}^{-11}}{{10}^{-12}}\right)\\ \beta =10\log \left({10}^{-11-\left(-12\right)}\right)\end{array}$

  $\begin{array}{l}\beta =10\log \left({10}^1\right)\\ \beta =10\times 1\\ \beta =10\end{array}$

Conclusion:

The number of decibels $\beta$ is $10dB$

b.

To determine

To find: the number of decibels $\beta$ of the sound of intensity $I={10}^2{\text{ watt per m}}^2$ .

Answer to Problem 48E

  $140dB$

Explanation of Solution

Given:

The number of decibels $\beta$ of a sound with an intensity of I watts per square meter is given by:

  $\beta =10\log \left(\frac{I}{I_0}\right)$.........(1)

Where $I_0={10}^{-12}{\text{ watt per m}}^2$

Formula used:

1. $\frac{a^m}{a^n}=a^{m-n}$

2. ${\log }_a{a}^x=x$

Substitute $I={10}^2{\text{ watt per m}}^2$ and $I_0={10}^{-12}{\text{ watt per m}}^2$ in the equation (1), to find the number of decibels $\beta$ ,

  $\begin{array}{l}\beta =10\log \frac{I}{I_0}\\ \beta =10\log \left(\frac{{10}^2}{{10}^{-12}}\right)\\ \beta =10\log \left({10}^{2-\left(-12\right)}\right)\end{array}$

  $\begin{array}{l}\beta =10\log \left({10}^{14}\right)\\ \beta =10\times 14\\ \beta =140\end{array}$

Conclusion:

The number of decibels $\beta$ is $140dB$

c .

To determine

To find: the number of decibels $\beta$ of the sound of intensity $I={10}^{-4}{\text{ watt per m}}^2$ .

Answer to Problem 48E

  $80dB$

Explanation of Solution

Given:

The number of decibels $\beta$ of a sound with an intensity of I watts per square meter is given by:

  $\beta =10\log \left(\frac{I}{I_0}\right)$.........(1)

Where $I_0={10}^{-12}{\text{ watt per m}}^2$

Formula used:

1. $\frac{a^m}{a^n}=a^{m-n}$

2. ${\log }_a{a}^x=x$

Substitute $I={10}^{-4}{\text{ watt per m}}^2$ and $I_0={10}^{-12}{\text{ watt per m}}^2$ in the equation (1), to find the number of decibels $\beta$ ,

  $\begin{array}{l}\beta =10\log \frac{I}{I_0}\\ \beta =10\log \left(\frac{{10}^{-4}}{{10}^{-12}}\right)\\ \beta =10\log \left({10}^{-4-\left(-12\right)}\right)\end{array}$

  $\begin{array}{l}\beta =10\log \left({10}^8\right)\\ \beta =10\times 8\\ \beta =80\end{array}$

Conclusion:

The number of decibels $\beta$ is $80dB$

d.

To determine

To find: the number of decibels $\beta$ of the sound of intensity $I={10}^{-6}{\text{ watt per m}}^2$ .

Answer to Problem 48E

  $60dB$

Explanation of Solution

Given:

The number of decibels $\beta$ of a sound with an intensity of I watts per square meter is given by:

  $\beta =10\log \left(\frac{I}{I_0}\right)$.........(1)

Where $I_0={10}^{-12}{\text{ watt per m}}^2$

Formula used:

1. $\frac{a^m}{a^n}=a^{m-n}$

2. ${\log }_a{a}^x=x$

Substitute $I={10}^{-6}{\text{ watt per m}}^2$ and $I_0={10}^{-12}{\text{ watt per m}}^2$ in the equation (1), to find the number of decibels $\beta$ ,

  $\begin{array}{l}\beta =10\log \frac{I}{I_0}\\ \beta =10\log \left(\frac{{10}^{-6}}{{10}^{-12}}\right)\\ \beta =10\log \left({10}^{-6-\left(-12\right)}\right)\end{array}$

  $\begin{array}{l}\beta =10\log \left({10}^6\right)\\ \beta =10\times 6\\ \beta =60\end{array}$

Conclusion:

The number of decibels $\beta$ is $60dB$