Chapter 5, Problem 23CT

(a)

To determine

The sound, if two people shout at the same time, where shouts of a single person measure $80$ decibels. The decibel level, $D$ , of sound is given by the equation $D=10\log \left(\frac{I}{I_0}\right)$ , where $I$ is the intensity of the sound, and $I_0={10}^{-12}$ watt per square meter.

Answer to Problem 23CT

Solution:

The sound is approximately $83$ decibels if two people shout at the same time.

Explanation of Solution

Given information:

The decibel level, $D$ , of sound is given by the equation $D=10\log \left(\frac{I}{I_0}\right)$ , where $I$ is the intensity of the sound, and $I_0={10}^{-12}$ watt per square meter.

The equation for decibel level $D$ is given by the equation $D=10\log \left(\frac{I}{I_0}\right)$ .

Let $I_0={10}^{-12}$ watt per square meter.

Let intensity of single person be $I_1$ for sound $D_1$ decibel.

Implies that $D_1=80$ for $I=I_1$ .

Here, find $D_2$ when $I_2=2I_1$ .

  $D_1=10\log \left(\frac{I_1}{I_0}\right)$

  $D_2=10\log \left(\frac{I_2}{I_0}\right)$

Using $I_2=2I_1$ ,

  $\Rightarrow D_2=10\log \left(\frac{2I_1}{I_0}\right)$

By using $\log \left(A\cdot B\right)=\log A+\log B$ ,

  $\Rightarrow D_2=10\log 2+10\log \left(\frac{I_1}{I_0}\right)$

Since $D_1=10\log \left(\frac{I_1}{I_0}\right)=80$ ,

  $\Rightarrow D_2=10\log 2+80$

  $\Rightarrow D_2=3.010299957+80$

  $\Rightarrow D_2=83.010299957\approx 83$

Therefore, the sound is approximately $83$ decibels if two people shout at the same time.

(b)

To determine

The number of people in the crowd that need to shout at the same time for the resulting sound level to meet or exceed the pain threshold, where shouts of a single person measure $80$ decibels. The pain threshold for sound is $125$ decibels. The Athens Olympic Stadium $2004$ can seat $74,400$ people. The decibel level, $D$ , of sound is given by the equation $D=10\log \left(\frac{I}{I_0}\right)$ , where $I$ is the intensity of the sound, and $I_0={10}^{-12}$ watt per square meter.

Answer to Problem 23CT

Solution:

  $31623$ people need to shout to make pain threshold sound of $125$ decibels.

Explanation of Solution

Given information:

The shouts of a single person measure $80$ decibels. The pain threshold for sound is $125$ decibels.

The Athens Olympic Stadium $2004$ can seat $74,400$ people. The decibel level, $D$ , of sound is given by the equation $D=10\log \left(\frac{I}{I_0}\right)$ , where $I$ is the intensity of the sound, and $I_0={10}^{-12}$ watt per square meter.

The equation for decibel level $D$ is given by the equation $D=10\log \left(\frac{I}{I_0}\right)$ .

Let $I_0={10}^{-12}$ watt per square meter.

Let intensity of single person be $I_1$ for sound $D_1$ decibel.

Implies that $D_1=80$ for $I=I_1$ .

Here find $D_n=125$ when $I_n=n{I}_1$ .

  $D_1=10\log \left(\frac{I_1}{I_0}\right)$

  $D_n=10\log \left(\frac{I_n}{I_0}\right)$

Using $I_n=n{I}_1$ ,

  $\Rightarrow 125=10\log \left(\frac{n{I}_1}{I_0}\right)$

By using $\log \left(A\cdot B\right)=\log A+\log B$ ,

  $\Rightarrow 125=10\log n+10\log \left(\frac{I_1}{I_0}\right)$

Since $D_1=10\log \left(\frac{I_1}{I_0}\right)=80$ ,

  $\Rightarrow 125=10\log n+80$

  $\Rightarrow 125-80=10\log n$

  $\Rightarrow 45=10\log n$

  $\Rightarrow \frac{45}{10}=\log n$

  $\Rightarrow 4.5=\log n$

As $y={\log }_{a}x$ and $x=a^y,$ where $a>0$ and $a\ne 1,$ are equivalent,

  $\Rightarrow n={10}^{4.5}$

  $\Rightarrow n=31622.7766$

Therefore, $31623$ people need to shout to make pain threshold sound of $125$ decibels.