Question

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 output same sound energy per second, what will be the noise level if the number of students is increased to 100?

Answer 1

The noise level if the number of students is increased to 100 is $\tau^{\prime}=53 d B$

Explanation:

The cause of noise is class members. This is an unreasonable Joke.

But, yes students in the class yell a lot, which is very hard to quiet them and inspire them toward studies there.

now, Intensity is directly proportional with the sound source.

Thus,

$\frac{I}{I^{\prime}}=\frac{50}{100}$

$\frac{I}{I^{\prime}}=\frac{1}{2}$

Now, Using the formula,

Sound Level = $10 \log \left[\frac{l}{l_{0}}\right]$

Thus,

$\tau=10 \log \left[\frac{I}{I_{0}}\right]$

$\tau^{\prime}=10 \log \left[\frac{I^{\prime}}{I_{0}}\right]$

$\tau^{\prime}-\tau=10 \log \left[\frac{I^{\prime}}{I}\right]$

$\tau^{\prime}-50=10 \log [2]$

$\tau^{\prime}=50+3$

$\tau^{\prime}=53 d B$

Answer 2

The noise level if the number of students is increased to 100 is 53 dB

Given:

Noise level = 50 dB

Students = 50

To find:

Noise level if students is increased to 100 = ?

Solution:

Let 'I' be intensity of each student.

Noise level when number of students is 50:

$\beta_{A}=10 \log _{10} \frac{50 I}{I_0}$

Noise level when number of students is 100:

$\beta_{B}=10 \log _{10}\left(\frac{100I}{I_0}\right)$

Now, on subtracting noise level of 50 from noise level of 100, we get,

$\beta_{B}-\beta_{A}=10 \log _{10}\left(\frac{100I}{I_0}\right)-10 \log _{10} \left(\frac{50I}{I_0}\right)$

$\Rightarrow \beta_{B}-\beta_{A}=10 \log \left( \frac{100I}{I_0} - \frac{50I}{I_0}\right)$

Since, the above equation is similar to equation given below, we get,

$\log A-\log B =\log \frac{A}{B}$

Now, the equation becomes,

$\Rightarrow \beta_{B}-\beta_{A}=10 \log\left( \frac{100I}{I_0} \times \frac{I_0}{50I}\right)$

$\Rightarrow \beta_{B}-\beta_{A}=10 \log _{10} 2$

$\therefore \beta_{B}-\beta_{A}=3$

Now, the noise level of 100 students will be,

$\beta_B=\beta_A+3=50+3=53 \ dB$