# Sound Intensity level (Decibels) Problems and Solutions

In this article, we will delve into solving problems related to sound intensity levels and decibels, specifically tailored for college prep courses. First, we’ll introduce the essential formulas concisely, followed by practice questions.

## Understanding Sound Levels and Decibels

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

$I$ is the intensity of a sound (measured 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) In most intensity-level problems, we are given intensity and asked to find its corresponding intensity level. It’s essential to understand that the intensity level involves a logarithmic comparison between a given sound intensity ($I$) and the reference sound intensity ($I_0$). We then multiply the result by ten. Here’s how it works: \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 falls into the “quiet” category.

(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.

Isn’t this content as valuable as a $10 private class? Please support me here. I worked hard to prepare this article. Problem (2): The pain level for humans has an 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. It’s known as the threshold of pain and is similar to what you might experience at 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, we are not given specific information about the magnitude of the sound intensity. However, we can assume an original intensity ($I$) and a new intensity ($I_{new} = kI$), where ($k$) represents the multiple indicating how much the sound intensity changed. Substitute this into the decibel formula and apply the logarithms relation$\log(ab)=\log a+\log bto 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 given by: $\Delta \beta =10\log k$ This result is general and can be memorized for faster solving, especially in MCAT tests. Here,k=3$so the change in the sound intensity becomes: $\Delta \beta=10\log 3=4.77\rm dB$ This demonstrates that tripling the sound intensity corresponds to an increase of 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 nearly6$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 leads to a doubling of energy spread in the environment. Thus, substitute$I=2I_1into 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$45\,\rm cmfrom the person and its corresponding decibel level. Solution: (a) As before, we have \begin{align*} \beta&=10\log\left(\frac{9\times 10^{-3}}{10^{-12}}\right) \\\\ &=10\log(9\times 10^9) \\\\ &=10\log 9+10\log 10^9 \\\\ &=9.5+90=\boxed{99.5\,\rm dB} \end{align*} Here, we set\log 9=0.95$and used the logarithm relation $\log 10^a=a$ (b) The intensity of a spherical sound wave produced by a point source at distance$r$is calculated by the following formula: $I=\frac{P_{av}}{4\pi r^2}$ where$P_{av}$represents the average power emitted by the source. Usually, in all sound-level problems, the average power emitted by a source remains constant. The only thing that changes inversely is the intensity at different locations,$I\propto \frac{1}{r^2}$. Given the intensity at a distance$r_1=1\,\rm m$as$I_1$, we need to find it at a distance$r_2=45\,\rm cm$. Constructing the ratio of intensities at these locations, we have: \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 move closer to the source, the energy we receive increases fivefold, but the sound level only changes by$6.9\,\rm dB$. As a side note, let’s 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 emitted wave propagating 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 thatI\propto \frac{1}{r^2}$. Now, use the rule$\log 10^n=nto 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 wave4.2$times greater then, (a) By what factor will the intensity change? (b) By how many decibels will the sound level change? Solution: the only given data is the ratio of amplitudes,$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 of17.64. (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 by12.46\,\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