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The noise level in a classroom in absence of the teacher is \[50\,dB\] when \[50\] students are present. Assuming that on the average each student outputs the same sound energy per second, what will be the noise level if the number of students is increased by \[100\] ?

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Last updated date: 24th Jul 2024
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Answer
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Hint: Study the definition of decibel. Use the formula to find the intensity of the sound. From there find the noise level due to the said number of students.Noise is defined as the unpleasant sound that causes disturbance.

Formula used:
The formula to find the sound energy in units of decibel is given by,
\[\beta = 10{\log _{10}}\dfrac{{{S_1}}}{{{S_2}}}\]
where, is the sound energy in units of \[dB\] \[{S_1}\] and \[{S_2}\] are the intensity of sound from two sources.

Complete step by step answer:
We have given here that a class full of \[50\] students are making noise in absence of the teacher is \[50\,dB\].The output of each student is the same sound energy per second. So, let the intensity of sound by each student is \[I\] with respect to some source \[{I_0}\]. Hence, the sound energy produced by \[50\] students is \[50I\]. Hence the sound energy produced by \[100\] students will be, \[100I\].

The mathematical expression for the sound energy in units of decibel is given by,
\[\beta = 10{\log _{10}}\dfrac{{{S_1}}}{{{S_2}}}\]
where, $\beta$ is the sound energy in units of \[dB\] \[{S_1}\] and \[{S_2}\]are the intensity of sound from two sources.
Now, in units of sound energy decibel the noise by \[50\] students is,
\[50 = 10{\log _{10}}\dfrac{{50I}}{{{I_0}}}\]..…(i)
Now, let the sound produced by \[100\] students in units of decibel is \[\beta \]. So, the noise produced by \[100\] students will be,
\[\beta = 10{\log _{10}}\dfrac{{100I}}{{{I_0}}}\]……(ii)

So, Subtracting equation(ii) from equation(i) we get,
\[\beta - 50 = 10{\log _{10}}\dfrac{{100I}}{{{I_0}}} - 10{\log _{10}}\dfrac{{50I}}{{{I_0}}}\]
Upon simplifying we will have,
\[\beta - 50 = 10 \times {\log _{10}}\dfrac{{100I}}{{50I}}\]
\[\Rightarrow \beta = 50 + 10{\log _{10}}2\]
\[\Rightarrow \beta = 50 + 3.01\]
\[\therefore \beta = 53.01\]

Hence, the noise output by \[100\] students in units of decibel in absence of teacher is \[53.01\,dB\].

Note: The noise output in units of bel is not proportional to the energy output by the students. It is so since the logarithm curve follows the curve of exponent here the curve is of the form \[y = {10^x}\]. So, it is not linear so the increase in noise is less than the increase in intensity of sound.