# Sound Intensity level (Decibels) Problems and Solutions

In this article, we are going to solve some problems on sound intensity level and decibels aimed at college prep courses. First, we will introduce the necessary formulas in short and then practice some questions.

The sound level, intensity level, or decibel level is a measure of the relative intensity of a sound and formulated as below $\beta=10\log\left(\frac{I}{I_0}\right)$ where $I_0=1\times 10^{-12}\,\rm W/m^2$ is the sound intensity at the hearing threshold, called reference intensity.

$I$ is the intensity of a sound (in $\rm W/m^2$), and $\beta$ is the corresponding sound level measured in decibels (dB).

## Intensity level and decibels: Problems

Problem (1): The intensity of a sound is $1.5\times 10^{-6}\,\rm W/m^2$.
(a) What is the intensity level of that sound?
(b) If the sound intensity is doubled, what is the new sound intensity level?

Solution:
(a) Usually, in all intensity level problems, an intensity is given and asks us to find its intensity level. Keep in mind that the intensity level is nothing than a logarithmic comparison of a given sound intensity $I$ with the reference sound intensity $I_0$ and then multiply the result by ten as follows \begin{align*} \beta &=10\log\left(\frac{I}{I_0}\right) \\\\ &=10\log\left(\frac{1.5\times 10^{-7}}{10^{-12}}\right) \\\\ &= 10\log(1.5\times 10^5) \\\\ &= 51.7\,\rm dB \end{align*} According to the table of intensity levels for common sounds, this sound is categorized as quiet.

(b) The intensity is doubled $I'=2I$, so the sound level of the new intensity becomes \begin{align*} \beta &=10\log\left(\frac{I'}{I_0}\right) \\\\ &=10\log\left(\frac{2\times 1.5\times 10^{-7}}{10^{-12}}\right) \\\\ &= 10\log(3\times 10^5) \\\\ &= 54.7\,\rm dB \end{align*} As you can see, the intensity of the sound is doubled but the corresponding sound level slightly changes. Consequently, a doubling of intensity corresponds to an increase of $3\,\rm dB$ in sound intensity level.

Problem (2): The pain level for humans has a intensity level of $120\,\rm dB$.
(a) What is the intensity of such sound in $\rm W/m^2$?
(b) Compare that with a sound level of a whisper at $20\,\rm dB$.

Solution: another typical question on decibels is when the sound level is given and asks us to find the intensity corresponding to that level.

(a) In this case, substitute the intensity level $120\,\rm dB$ into the decibel formula and divide both sides by $10$. \begin{gather*} \beta=10\log\left(\frac{I}{I_0}\right) \\\\ 120 =10\log\left(\frac{I}{10^{-12}}\right) \\\\ 12=\log \left(\frac{I}{10^{-12}}\right) \\\\ 10^{12}=\frac{I}{10^{-12}} \\\\ \rightarrow \quad \boxed{I=1\,\rm W/m^2} \end{gather*} In above, we used the definition of logarithms as $a=\log b \longleftrightarrow b=10^a$ Note that $\rm 1\, W/m^2$ is the loudest sound that human ear can tolerate without being hurt and is called the threshold of pain. This corresponds to a sound that you might perceive in a rock concert.

(b) This part is also very common in all problems. Here, the level of two sounds is given and wants the ratio of their corresponding intensities.

In such cases, the best thing to do is to use the following formula in which the reference level has been removed. $\beta_2-\beta_1=\log\frac{I_2}{I_1}$ Here, $\beta_1=120\,\rm dB$ and $\beta_1=20\,\rm dB$. Substitute these known values into the above formula and solve for the ratio of intensities \begin{gather*} 120-20=10\log{\frac{I_{120}}{I_{20}}} \\\\ 100=10\log{\frac{I_{120}}{I_{20}}} \\\\ 10^{10}=\frac{I_{120}}{I_{20}} \\\\ \rightarrow \quad \boxed{I_{120}=10^{10} I_{20}} \end{gather*} This result is roughly logical since a $20\,\rm dB$ sound corresponds to the sound of rustling leaves which is barely audible. On the other hand, a $120\,\rm dB$ sound is perceived when you are at a distance of $60\,\rm m$ from a jet engine during takeoff. Now, compare these two situations!

Problem (3): The intensity of a sound is tripled. By how many decibels does it increase?

Solution: In this question, there is nothing about the magnitude of the sound intensity.

Assume the original intensity to be $I$ and the new intensity $I_{new}=kI$, where $k$ is a multiple that indicates how much the sound intensity changed. Substitute this into the decibel formula and apply the logarithms relation $\log(ab)=\log a+\log b$ to reach the following result \begin{align*} \beta_{new}&=10 \log\left(\frac{I_{new}}{I_0}\right) \\\\ &=10\log\left(\frac{k\times I}{I_0}\right) \\\\ &=10\log k+ 10\log\left(\frac{I}{I_0}\right) \\\\ &=10\log k+\beta \end{align*} Consequently, the change in the sound level is found to be $\Delta \beta =10\log k$ This is a general result and you can memorize it for faster solving in MCAT test.

Here, $k=3$ so the change in the sound intensity becomes $\Delta \beta=10\log 3=4.77\rm dB$ This shows that a tripling in the sound intensity corresponds to nearly $4.77\,\rm dB$ in the intensity level.

Problem (4): By what factor will the intensity change for a $\pm 6\,\rm dB$ change in the out level of a loudspeaker?

Solution: The change in sound level is given as $\beta_2-\beta_1=6\,\rm dB$, and the change in the intensity is asked us.

Recall that the difference in intensity (sound) level is related to the change in the intensity as $\beta_2-\beta_1=10\log\frac{I_2}{I_1}$ Substituting the numerical values into that and solved for the fraction $\frac{I_2}{I_1}$. \begin{gather*} \Delta \beta=10\log\frac{I_2}{I_1} \\\\ 6=10\log\frac{I_2}{I_1} \\\\ \Rightarrow \frac{I_2}{I_1}=10^{0.6}=4 \end{gather*} Where in the last line we used the fact that $\log b=a \longleftrightarrow b=10^a$ Therefore, a $\pm 6\,\rm dB$ change corresponds to four times the intensity of the sound.

Problem (5): We are standing a certain distance from four noisy juicers and hear a sound level of $77\,\rm dB$ in a juice shop. What sound level would this person experience if three of them are turned off?

Solution: Assume each juicer produces a sound intensity $I=?$ and a sound level $\beta=75\,\rm dB$.

First of all, find the sound intensity that four equally noisy juicers produce using the decibel formula below \begin{gather*} \beta=10\log\left(\frac{I}{I_0}\right) \\\\ 77 =10\log\left(\frac{I}{10^{-12}}\right) \\\\ 7.5=\log \left(\frac{I}{10^{-12}}\right) \\\\ 10^{7.7}=\frac{I}{10^{-12}} \\\\ \rightarrow \quad \boxed{I=5.01\times 10^{-5}\,\rm W/m^2} \end{gather*} This is the intensity that all four juicers produce simultaneously. Dividing that by the number of juicers gives the sound intensity of one juicer as $I_{one}=\frac{I}{4}=1.25\times 10^{-5}\,\rm W/m^2$ Now that the intensity of a single juicer has been determined, again use the sound level formula and find the decibel level for a single machine \begin{align*} \beta&=10\log\left(\frac{I}{I_0}\right) \\\\ &=10\log\left(\frac{1.25\times 10^{-5}}{10^{-12}}\right) \\\\ &=10\log \left(1.25\times 10^7 \right) \\\\ &=10(\log 1.25+\log 10^7) \\\\ &=0.9+70 \\\\ \rightarrow \quad \beta_1&=\boxed{70.9\,\rm dB} \end{align*} We used the following useful rule of logarithms for the last line. $\log(ab)=\log a+\log b$ This explicitly shows the difference between sound intensity and intensity level. The sound intensity of four machines running at the same time is four times the intensity of a single machine, but their sound level is nearly $6$ decibels less than a single machine.

Problem (6): The sound level near an operating air conditioner is $80\,\rm dB$. What would be the sound level of two of them operating side by side at the same distance?

Solution: The decibel level of a machine is given as $I_1=80\,\rm dB$ and wants the sound level of two such machines operating simultaneously $I$. We can't simply add them and say the total sound level would have been $160\,\rm dB$. This corresponds to the decibel level of a jet engine at a near distance. Thus, that conclusion is not logical.

Keep in mind that sound intensity is directly related to the energy that a typical sound can have. Thus, adding two similar machines lead to a doubling of energy spread in the environment.

Thus, substitute $I=2I_1$ into the decibel formula \begin{align*} \beta&=10\log\left(\frac{I}{I_0}\right) \\\\ &=10\log\left(\frac{2I}{10^{-12}}\right) \\\\ &=10 \log{\left(2\times \frac{I}{10^{-12}}\right)} \\\\ &=10(\log{2})+10\left(\log \frac{I}{10^{-12}}\right) \\\\ &=10(0.3)+\beta_1 \\\\ &=10(0.3)+80=\boxed{83\,\rm dB}\end{align*} In the fourth line, we used one of the fundamental relations for logarithms $\log(ab)=\log a+ \log b$ and set $\log 2=0.3$. Recall that $\beta_1$ is the sound level for a single operating air conditioner.

Problem (7): A person standing $1\,\rm m$ from a loudspeaker hears its sound at an intensity of $9\times 10^{-3}\,\rm W/m^2$.
(a) Find its corresponding sound level for this intensity.
(b) Assuming sound propagates from the source as a spherical wave, calculate the sound intensity at a distance of $45\,\rm cm$ from the person and its corresponding decibel level.

Solution: (a) As before, we have \begin{align*} \beta&=10\log\left(\frac{9\times 10^{-3}}{10^{-12}}\right) \\\\ &=10\log(9\times 10^9) \\\\ &=10\log 9+10\log 10^9 \\\\ &=9.5+90=\boxed{99.5\,\rm dB} \end{align*} where we set $\log 9=0.95$ and used the logarithm relation $\log 10^a=a$
(b) The intensity of a spherical sound wave produced by a point source at a distance of $r$ is calculated by the following formula $I=\frac{P_{av}}{4\pi r^2}$ where $P_{av}$ is the average power emitted by the source.

Usually, in all sound level problems, the average power emitted by a source is constant. The only thing that changes as inversely is the intensity at different locations, $I\propto \frac{1}{r^2}$.

We are given the intensity at distance $r_1=1\,\rm m$ as $I_1$ and asked us to find it at distance $r_2=45\,\rm cm$. Constructing the ratio of intensities at those locations gives us \begin{gather*} \frac{I_2}{I_1}=\left(\frac{r_1}{r_2}\right)^2 \\\\ \frac{I_2}{9\times 10^{-3}}=\left(\frac{1}{0.45}\right)^2 \\\\ \frac{I}{9\times 10^{-3}}=4.9 \\\\ \rightarrow \quad \boxed{I=44.1\times 10^{-3}\,\rm W/m^2} \end{gather*} As you can see, as we get closer to the loudspeaker, the intensity increases which is logical.

The sound level for this intensity is found to be $\beta=106.4\,\rm dB$. When we are moving closer to the source, the energy we will receive increases five times, but the sound level only changes by $6.9\,\rm dB$.

As a side note, we are going to derive the difference in decibel levels $\beta_1$ and $\beta_2$ of a sound source at two different distances from it.

Assuming a point source and the wave emitted propagates spherically, at distance $r$ the intensity and its corresponding sound level are given by the following formula $I=\frac{P_{av}}{4\pi r^2} \ , \ \beta=10\log\left(\frac{I}{I_0}\right)$ The difference in decibel levels are computed as \begin{align*} \beta_2-\beta_1&=10\log\left(\frac{I_2}{I_0}\right)-10\log\left(\frac{I_1}{I_0}\right) \\\\ &=10 \log\left(\frac{I_2}{I_1}\right) \\\\ &=10\log\left(\frac{r_1}{r_2}\right)^2 \end{align*} In above, the following famous logarithms relation has been used $\log a-\log b=\log\frac{a}{b}$ For the last line recall that $I\propto \frac{1}{r^2}$. Now, use the rule $\log 10^n=n$ to find the final result as below $\beta_2-\beta_1=20\log\left(\frac{r_1}{r_2}\right)$ Now we are in a position to find the change in sound levels for the previous problem. \begin{align*} \beta_2-\beta_1&=20\log \left(\frac{r_1}{r_2}\right) \\\\ &=20\left(\frac{1}{0.45}\right) \\\\ &=6.9 \end{align*} Exactly matches the result obtained directly by calculating the intensities.

Problem (8): If we made the amplitude of a sound wave $4.2$ times greater then,
(a) By what factor will the intensity change?
(b) By how many decibels will the sound level change?

Solution: the only given data is the ratio of amplitudes, $A_2=4.2 A_1$. The intensity $I$ of a wave is directly proportional to the square of the wave amplitude, $I\propto A^2$.

(a) Forming the ratio of intensities gives us the desired increase in intensity \begin{align*} \frac{I_2}{I_1}&=\left(\frac{A_2}{A_1}\right) \\\\ &=(4.2)^2 \\\\ \rightarrow I_2&=17.64 I_1 \end{align*} Thus, the sound intensity increases by a factor of $17.64$.
(b) Given the ratio of intensities, we can find the change in the decibel level as below $\beta_2-\beta_1=10\log\left(\frac{I_2}{I_1}\right)$ Substituting the numerical values into that, we will have \begin{align*} \beta_2-\beta_1&=10\log 17.64 \\ &=12.46 \end{align*} Consequently, the sound level increases by $12.46\,\rm dB$.

## Summary:

In this article, you gained some insight into how to solve sound or intensity level problems. All questions are answered by the following formulas \begin{gather*} \beta=10\log\left(\frac{I}{I_0}\right) \\\\ I=\frac{P_{av}}{4\pi r^2} \\\\ \frac{I_2}{I_1}=\left(\frac{r_1}{r_2}\right)^2 \\\\ \beta_2-\beta_1=20\log \left(\frac{r_1}{r_2}\right) \end{gather*}

Author: Dr. Ali Nemati
Published: Nov. 8, 2022