Sound Intensity - Online Tutor, Practice Problems & Exam Prep

Sound Intensity Level and the Decibel Scale

Hey, guys. So in a previous video, we talked about the intensity of a wave. The wave intensity equation was given by this equation over here. Remember that this equation actually works for all different kinds of waves, whether they are transverse waves or longitudinal. But we're going to take this equation, and we're going to talk about a related idea that has to do only specifically for sound, called the sound intensity level. So let's go ahead and check this out here. Basically, the idea is that humans can actually hear over a huge range of intensities. Now you don't need to memorize any of these numbers. I'm just using them to make this point here. But if you're walking through the park, the rustling of leaves has an intensity on average of about \(1 \times 10^{-11}\) watts per meter squared. If you're having a normal conversation with your friend, that's \(1 \times 10^{-6}\). And if you're standing close to a jet engine, that's actually \(1 \times 10^{1}\). Notice how the numbers are getting smaller as they go down, but they're actually negative exponents, so the number is actually getting bigger. They're more intense sounds. Now these numbers are pretty impractical to use in everyday language. So instead of using watts per meter squared, we're going to use a logarithmic scale when we talk about sound intensity. So this logarithmic scale is going to make our numbers a lot simpler. So here's the equation the textbooks are going to give you for sound intensity level. They're going to use the letter beta, the Greek letter beta, and it's going to be \(10 \times \log_{10} \left(\frac{I}{I_0}\right)\), where \(I\) is the intensity of the sound source, and \(I_0\) is a constant that has to do with the lower threshold of what humans can on average hear. It's just a number and it's always going to be \(1 \times 10^{-12}\). Alright. So it's just those two letters there. Really, there's only one variable because this is a constant and the units for this are called decibels. So this is where we get our decibel system from, which is a representation of how loud a sound is, and the units for this are going to be in dB. So basically, we can use this once we use this logarithmic scale, we can take these numbers that are pretty impractical and long, and we can turn them into much nicer numbers, like for instance, 10 decibels or 60 or 130. So that's all there is to it, guys. Let's go ahead and take a look at an example. So we have a siren that is producing a power of 9 milliwatts. So that's power \(P\), which equals 9 milliwatts, which equals 0.009. Now what happens is this siren is continuously producing a sound in 3 dimensions in all directions. We're going to calculate the sound level in decibels at a distance of 3 meters. So basically our \(r\) is equal to 3. Alright? So if we want to calculate the sound level, we're just going to use our new equation here, which is beta. So this is \(\beta = 10 \times \log_{10} \left(\frac{I}{I_0}\right)\). The great thing about this equation is that there's actually only just one unknown that you could possibly have, which is this intensity. So how do we actually find this out? Well, remember this equation for intensity is really just this, \(I = \frac{P}{4 \pi r^2}\). So this is going to be \(I = \frac{P}{4\pi r^2}\), and what you're going to get here is you're going to solve for \(0.009 \div 4\pi \times 3^2\). If you go ahead and work this out, what you're going to get is you'll get \(7.96 \times 10^{-5}\), which are watts per meters squared. So now what we do is we take this number here, and we're going to plug it inside for the intensity in our logarithm equation. Alright. So to finish things off, we're going to have that the beta is going to be \(10 \times \log_{10} \left(\frac{7.96 \times 10^{-5}}{I_0}\right)\), and we're going to just divide it by our \(I_0\). \(I_0\) again is just that lower threshold of human hearing and it's kind of like a reference point or a reference level, and basically this is going to be \(1 \times 10^{-12}\) like this. So if you go ahead and work this out, what you're going to get is exactly 79 decibels. And that's the answer. So that's it for this one, guys, and let me know if you have any questions.

A sound wave from a police siren has an intensity of 0.01 W/m² at a certain distance. A second sound wave from an ambulance has a sound intensity level 8 dB greater than the police siren, when measured at the same distance. What is the sound intensity level (in dB) of the sound wave due to the ambulance?

• A. 8.01 dB
• B. 108 dB
• C. 0.063 dB

Example 2

Hey, everyone. Welcome back. So let's take a look at this practice problem here. This problem tells us we have a sound source that emits equally in all directions. Just have a sound source like this, some P over here, and it's going to emit, so, like, a siren or something like that. Now what I'm told here is that I'm trying to figure out that if I move twice the distance away from the source, how many decibels does the sound intensity level drop by? So this is what's going on. Right? Imagine you go some distance away from this sound source. I'm going to call this r1. Basically, at this little circle over here, this distance, you're going to measure some intensity. I'm going to call this I1. And because of that intensity, I can plug this into the sound intensity level formula and get some beta. Right? So I'm going to measure some sound in decibels. Now let's say, basically, now you take that distance and you just double it. So if I walk further away and I end up at some distance r2, what I know is that the intensity is going to drop off because it gets weaker as you go farther from the source. So I'm going to get an I2. If I plug that in here, I'm going to get some other beta, some other decibel level that's going to be lower than the one that I started out with. So, basically, what I'm trying to find in this video is actually not beta 1 or beta 2. I'm actually trying to find out how many decibels it decreases by. So I'm not finding beta 1 or beta 2. I'm actually looking for Δbeta. It's the difference between these 2. And, really, what you could do is just take the 2 sound intensity levels and just subtract them. That's basically what we're going to do in this problem here. So I'm going to set up a little bit of an equation, but notice how we actually have no numbers that we can plug in. So this is going to be a little bit of a derivation. So let's check it out. Right? So this is going to be beta 1 minus beta 2. Right? Now I actually know what the beta formula is, so I can just replace those things with that formula. Alright? So let's just work this out. Let's expand this. This is just going to be 10 times logarithm base 10, and then I'm going to have I over I knots. Now because I have a beta 1 and a beta 2, because I have an intensity 1 and an intensity 2, then, basically, I'm just going to replace this not with just I, but I1 over I0. And then over here, this is just the first term, and the second term just becomes 10 times logarithm base 10, and this is going to be I2 divided by I0. Alright? Now what we can do here is we notice that this 10 multiplies both of them. So I can actually clean this up a little bit and sort of, like, sort of collapse them and and factor them or, you know, into the distributive property. So this is going to be 10, and then I'm going to have logarithm sorry. This is gonna be log base 10 of I1 over I0 minus log base 10 of I2 over I0. Delta beta = 10 * log 10 ( I 1 I 2 ) Alright, so notice how we really just went from, this is kind of like a and b, and we went from log a minus log b, and we just put them as a fraction. That's basically what we did here. Because of this, we can look at this. This is kind of like a weird complex fraction because there's, like, multiple stuff going on, but one of the things you'll notice is that we're actually multiplying or dividing by, sorry, we're dividing by the same factor of I0s. So remember, what happens is if you take a fraction over a fraction, you're going to have to flip the fraction, and you'll notice that these things actually will cancel out. Right? The I0s will cancel. So it's actually good that we don't really need it anymore. Alright? So what does this whole thing become? It really just becomes 10 times the logarithm base 10, and this is just going to be I1 divided by I2. Now we're almost done here. Basically, this is just going to be let's see. One of the things we can do here is we notice that we have a ratio of intensities. Right? Remember that the ratio of the intensity has to do with the inverse square law. One of the relationships that we saw here is that I1 over I2 is actually really just r2 squared over r1 squared. So here's what happens. We can basically just take this expression over here, which I'll highlight in blue, I'm actually just gonna replace it with this. So this whole thing actually just becomes, this is going to become 10 times logarithm base 10 of and this actually I'm just going to replace this with r2 squared over r1 squared. Now let's take a look look because we've actually just replaced intensity with radius. But if you look at this sort of scenario over here, you look at this diagram, I actually know what the ratio of the distances is. I don't know what either one of them are individually, but I do know that this distance r2 is just going to be twice the distance of r1. So in other words, as a relationship here, as a ratio, I know that r2 over r1 just equals 2. So I'm going to take this number here, just 2, and actually just pop it into this expression, except it's squared. So, really, what happens is this just becomes 10 times the logarithm base 10, and this just becomes 2 squared, or what you could do, this is going to be 10 times logarithm base 10 of 4. Now what you're going to get here, by the way, is you're actually going to get 6 decibels if you actually plug this in. So, in other words, what happens here is that the sound intensity level decreases by a factor of 6 dB if you were to move twice the distance away from the source. So this is actually kind of a pretty interesting quirk of sounds of sound levels and inverse square laws. But, basically, the whole point here is that whenever the distance from an inverse square law power source, I know it's kind of a mouthful, but, like, sort of like any sound that sort of drops off as as one over r squared, the sound intensity level actually decreases by a factor of 6 dB. So So it's actually not really linear. Each, sort of like, if you double the distance, you're decreasing by another 6 decibels. And then if you double that distance, it decreases by another 6 decibels. Hopefully, that made sense, folks. Let me know if you have any questions.