Question

At a rock concert, a dB meter registered 135 dB when placed 2.2 m in front of a loudspeaker on the stage.

1) What was the power output of the speaker, assuming uniform spherical spreading of the sound and neglecting absorption in the air?

2) How far away would the sound level be a somewhat reasonable 95 dB?

Answer 1

Concepts and reason

The concept required to solve the given problem is power and intensity of sound.

Initially, use the expression of sound intensity level to find the intensity of sound in ${\rm{W/}}{{\rm{m}}^2}$ . Use the expression of power output i.e. $P = IA$ , find the area of uniform sphere, substitute the values in the equation and find output power.

In the next part, first find the intensity of sound. Use the same power and intensity to find the distance value r.

Fundamentals

The expression to measure sound intensity level is decibels is given as follows:

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

Here, I is the intensity of sound and ${I_0}$ is the reference intensity.

The power output of the speaker is calculated as follows:

$P = IA$

Here, I is the intensity and A is the area.

(1)

Substitute 135 dB for $\beta$ and ${10^{ - 12}}{\rm{ W/}}{{\rm{m}}^2}$ for ${I_0}$ in equation $\beta = 10{\log _{10}}\left( {\frac{{{I_1}}}{{{I_0}}}} \right)$ and find ${I_1}$ .

$\begin{array}{c}\\135{\rm{ dB}} = 10{\log _{10}}\left( {\frac{{{I_1}}}{{{{10}^{ - 12}}{\rm{ W/}}{{\rm{m}}^2}}}} \right)\\\\\frac{{{I_1}}}{{{{10}^{ - 12}}}} = {10^{13.5}}\\\\{I_1} = 31.62{\rm{ W/}}{{\rm{m}}^2}\\\end{array}$

The power output of speaker is,

$P = {I_1}A$

The area of the spherical waves is,

$A = 4\pi {r^2}$

Substitute equation $A = 4\pi {r^2}$ in $P = {I_1}A$ .

$P = {I_1}\left( {4\pi {r^2}} \right)$

Substitute $31.62{\rm{ W/}}{{\rm{m}}^2}$ for ${I_1}$ and 2.2 m for r in equation $P = {I_1}\left( {4\pi {r^2}} \right)$ .

$\begin{array}{c}\\P = \left( {31.62{\rm{W/}}{{\rm{m}}^2}} \right)\left( {4\pi {{\left( {2.2{\rm{ m}}} \right)}^2}} \right)\\\\ = 1922.2{\rm{ W}}\\\end{array}$

(2)

Substitute 95 dB for $\beta$ and ${10^{ - 12}}{\rm{ W/}}{{\rm{m}}^2}$ for ${I_0}$ in equation $\beta = 10{\log _{10}}\left( {\frac{{{I_2}}}{{{I_0}}}} \right)$ and find ${I_2}$ .

$\begin{array}{c}\\95{\rm{ dB}} = 10{\log _{10}}\left( {\frac{{{I_2}}}{{{{10}^{ - 12}}{\rm{ W/}}{{\rm{m}}^2}}}} \right)\\\\\frac{{{I_2}}}{{{{10}^{ - 12}}}} = {10^{9.5}}\\\\{I_2} = 3.162 \times {10^{ - 3}}{\rm{ W/}}{{\rm{m}}^2}\\\end{array}$

The power output of speaker is,

$P = {I_2}\left( {4\pi {r^2}} \right)$

Re-arrange equation $P = {I_2}\left( {4\pi {r^2}} \right)$ for r.

$r = \sqrt {\frac{P}{{{I_2}\left( {4\pi } \right)}}}$

Substitute 1922.2 W for P and $3.162 \times {10^{ - 3}}{\rm{ W/}}{{\rm{m}}^2}$ for ${I_2}$ in equation $r = \sqrt {\frac{P}{{{I_2}\left( {4\pi } \right)}}}$ .

$\begin{array}{c}\\r = \sqrt {\frac{{1922.2{\rm{ w}}}}{{\left( {3.162 \times {{10}^{ - 3}}{\rm{ W/}}{{\rm{m}}^2}} \right)\left( {4\pi } \right)}}} \\\\ = 220.0{\rm{ m}}\\\end{array}$

Ans: Part 1

The power output of speaker is 1922.2 W.

[Answer Choice]

Part B

The required distance is 220 m.

[Answer Choice]