Question

A noisy grinding machine in a factory produces a soundintensity of 1.0 * 10-5 W/m2 .

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asked 2021-01-13

A noisy grinding machine in a factory produces a soundintensity of \(1.0 \cdot 10-5 W/m2\) . What is thesound level of this machine? A second identical machine is brought into the factory. Whatis the new sound intensity? What is the new sound level?

Answers (1)

2021-01-14

The sound level in decibles (dB) is given by the equation:
beta (in dB) =\(\displaystyle{10}{\log{{\left(\frac{{I}}{{I}_{{o}}}\right)}}}\)
where...
beta = sound level in dB
I = intensity of the sound in \(\displaystyle\frac{{W}}{{m}^{{{2}}}}\)
I_o = intensity of a reference value* = 1.0 x \(\displaystyle{10}^{{-{12}}}\frac{{W}}{{m}^{{{2}}}}\)
Plugging in these values we get:
\(\displaystyle\beta{\left(\in{d}{B}\right)}={10}{\log{{\left({1.0}{x}{10}^{{-{5}}}\frac{{W}}{{m}^{{{2}}}}/{1.0}{x}{10}^{{-{12}}}\frac{{W}}{{m}^{{{2}}}}\right)}}}={70}{d}{B}\)
The intensity of a sound is additive. This means that ifa second machine were brought into the factory the combinedintensity of the new sound would be the sum of the individualintensities.
So, "I" now equals:
\(\displaystyle{1.0}{x}{10}^{{-{5}}}+{1.0}{x}{10}^{{-{5}}}={2.0}{x}{10}^{{-{5}}}\)
Plugging this value back into our formula for soundlevel:
beta (in dB) =\(\displaystyle{10}{\log{{\left({2.0}{x}{10}^{{-{5}}}\frac{{W}}{{m}^{{{2}}}}/{1.0}{x}{10}^{{-{12}}}\frac{{W}}{{m}^{{{2}}}}\right)}}}={73}{d}{B}\)
* The \(I_o\) value is a reference value and can be any value youchoose. Generally it is considered the minimum intensity ofsound that a human ear can hear, or \(\displaystyle{1.0}{x}{10}^{{-{12}}}\frac{{W}}{{m}^{{{2}}}}\).
Hope this helps!

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