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Q17.3-29PE

Expert-verified
Found in: Page 595

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# 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.)

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

See the 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$$1.767 \times {10^{ - 3}}\;{\rm{W}}$$.

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