Question

Loudspeakers can produce intense sounds with surprisingly small energy input in spite of their low efficiencies. Calculate the power input needed to produce a\(90.0\;{\rm{dB}}\)sound intensity level for a\({\rm{12}}{\rm{.0}}\;{\rm{cm}}\)diameter speaker that has an efficiency of\(1.00\% \). (This value is the sound intensity level right at the speaker.)

Short answer

The input power is\(1.767 \times {10^{ - 3}}\;{\rm{W}}\).

Step-by-step solution

Intensity of Sound

The intensity of sound depends on the amplitude, and the loudness is the effect of intensity on human ears.

Given Data

The sound intensity is\(90.0\;{\rm{dB}}\).

The diameter of the speaker is\(12.0\;{\rm{cm}}\).

The efficiency is\(1.00\% \).

Calculation of the power of the sound

The intensity of the unaided hearing is,

\({{\rm{d}}_{\rm{1}}}{\rm{ = 10log}}\frac{{{{\rm{I}}_{\rm{0}}}}}{{{\rm{1}}{{\rm{0}}^{{\rm{12}}}}}}\)and\({I_0} = {10^{12}}\;{\rm{W/}}{{\rm{m}}^{\rm{2}}}\).

Now,

\(\begin{array}{c}90 = 10\log \frac{{{I_0}}}{{{{10}^{ - 12}}}}\\9 = \log \frac{{{I_0}}}{{{{10}^{ - 12}}}}\\{I_0} = {10^9} \times {10^{ - 12}}\\{I_0} = {10^{ - 3}}\end{array}\)

The power output is,

\(\begin{array}{c}P = {I_0}A\\P = {10^{ - 3}} \times \pi \times {\left[ {\frac{{0.15}}{2}} \right]^2}\\P = 1.767 \times {10^{ - 5}}\;{\rm{W}}\end{array}\)

The input power is,

\(\begin{array}{c}P' = \frac{P}{{efficiency}}\\ = \frac{{1.767 \times {{10}^{ - 5}}}}{{0.01}}\\ = 1.767 \times {10^{ - 3}}\;{\rm{W}}\end{array}\)

The input power is\(1.767 \times {10^{ - 3}}\;{\rm{W}}\).