A NOISY GRINDING MACHINE GOAL Working with watts and decibels. PROBLEM A noisy grinding machine in a factory produces a sound intensity of 1.00×10^−5 W/m² at a certain loca tion. Calculate (a) the decibel level of this machine at that point and (b) the new intensity level when a second, identical

Question

A NOISY GRINDING MACHINE

GOAL Working with watts and decibels.

PROBLEM A noisy grinding machine in a factory produces a sound intensity of [latex]1.00 imes 10^{-5} \mathrm{~W} / \mathrm{m}^{2}[/latex] at a certain loca tion. Calculate (a) the decibel level of this machine at that point and (b) the new intensity level when a second, identical machine is added to the factory. (c) A certain number of addi tional such machines are put into operation alongside these two machines. When all the machines are running at the same time the decibel level is [latex]77.0 \mathrm{~dB}[/latex]. Find the sound intensity. (Assume, in each part, that the sound intensity is measured at the same point, equidistant from all the machines.)

STRATEGY Parts (a) and (b) require substituting into the decibel formula, Equation 14.7,

[latex]\beta ≡ 10 \log \left(\frac{I}{I_0} ight)[/latex] [14.7]

with the intensity in part (b) twice the intensity in part (a). In part (c), the intensity level in decibels is given, and it's necessary to work backwards, using the inverse of the logarithm function, to get the intensity in watts per meter squared.

Accepted Answer

(a) Calculate the intensity level of the single grinder.

Substitute the intensity into the decibel formula:

[latex]\begin{aligned}\beta &=10 \log \left(\frac{1.00 imes 10^{-5} \mathrm{~W} / \mathrm{m}^{2}}{1.00 imes 10^{-12} \mathrm{~W} / \mathrm{m}^{2}} ight)=10 \log \left(10^{7} ight) \&=70.0 \mathrm{~dB}\end{aligned}[/latex]

(b) Calculate the new intensity level when an additional machine is added.

Substitute twice the intensity of part (a) into the decibel formula:

[latex]\beta=10 \log \left(\frac{2.00 imes 10^{-5} \mathrm{~W} / \mathrm{m}^{2}}{1.00 imes 10^{-12} \mathrm{~W} / \mathrm{m}^{2}} ight)=73.0 \mathrm{~dB}[/latex]

(c) Find the intensity corresponding to an intensity level of [latex]77.0 \mathrm{~dB}[/latex].

Substitute [latex]77.0 \mathrm{~dB}[/latex] into the decibel formula and divide both sides by 10 :

[latex]\begin{aligned}\beta &=77.0 \mathrm{~dB}=10 \log \left(\frac{I}{I_{0}} ight) \7.70 &=\log \left(\frac{I}{10^{-12} \mathrm{~W} / \mathrm{m}^{2}} ight) \end{aligned}[/latex]

Make each side the exponent of 10. On the right-hand side, [latex]10^{\log u}=u[/latex], by definition of base 10 logarithms.

[latex]\begin{aligned}10^{7.70}&=5.01 imes 10^{7}=\frac{I}{1.00 imes 10^{-12} \mathrm{~W} / \mathrm{m}^{2}} \I &=5.01 imes 10^{-5} \mathrm{~W} / \mathrm{m}^{2}\end{aligned}[/latex]

REMARKS The answer is five times the intensity of the single grinder, so in part (c) there are five such machines operating simultaneously. Because of the logarithmic definition of intensity level, large changes in intensity correspond to small changes in intensity level.

Question 14.2: A NOISY GRINDING MACHINE GOAL Working with watts and decibel...

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