Sound Intensity - Video Tutorials & Practice Problems

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concept

Sound Intensity Level and the Decibel Scale

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Hey guys. So in a previous video, we talked about the intensity of a wave. So the wave intensity equation was given by this equation over here. Now remember that this equation actually works for all different kinds of waves, whether they are transverse waves or longitudinal. We're going to take this equation. We're gonna 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 one times 10 of the minus 11 watts per meter squared. If you're having a normal conversation with your friend, that's one times 10 to the minus six. And if you're standing close to a jet engine, it's actually one times 10 to the one notice how the numbers are getting smaller as they go down, but they're actually negative exponents. So the number is actually getting bigger, there's a more intense sounds. Now, these numbers are pretty impractical to use in everyday language. So instead of using watts per meter squared, we're gonna use a logarithmic scale when we use or when we talk about sound intensity. So this logarithmic scale is gonna make our numbers a lot simpler. So here's the equation. The textbooks are gonna give you for sound intensity level. They're gonna use the letter beta, the Greek letter beta and it's gonna be 10 times the logarithm of base 10 times uh the intensity divided by this I knot here I over I zero. So this intensity is really just the intensity of the sound source. And that's just this equation over here that we used in that previous video, right? There's no difference is I equals P over four pi squared R A pi R squared. And this letter over here, this I zero is really just a constant that has to do with the lower threshold of what humans can. On average here. It's just a number and it's always gonna be one times 10 to the minus 12. All right. So it's just those two letters there really, there's only one variable because this is a constant and the units for this is called the decibels. So this is where we get our decibel system for or from which is kind of a representation of how loud a sound is. And the units for this are gonna be in DB. So basically we can use, 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 100 and 30. 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 nine milliwatts. So that's power P which equals nine milliwatts, which equals 0.009. Now, what happens is this siren is continuously producing a sound in three dimensions, in all directions. And we're gonna calculate the sound level in decibels at a distance of 3 m. So basically, RR is equal to three. All right. So if we want to calculate the sound level, we're just gonna use our new equation here, which is beta. So this is beta equals, we're gonna use 10 times the logarithm of base 10. And we just need our intensity divided by this I zero here. The great thing about this equation, this is that there's actually only just one unknown that you could possibly have, which is dizzy intensity. All right. So now all we have to do is if you'll notice we actually don't have the, the intensity is all we were given was the power and the R. So how do we actually find this out? Well, remember this equation for intensity is really just this P over four pi R squared. So this is gonna be I equals P over four pi R squared. And what you're gonna get here is you're gonna solve for 0.0 09, divided by four pi times the radius, which is three squared. If you go ahead and work this out, what you're gonna get is you'll get 7.96 times 10 to the minus five. This is gonna be watts per meter squared. So now what we do is we take this number here and we're gonna plug it inside for the intensity in our logarithm equation. All right. So to finish things off, we're gonna have that the beta is gonna be 10 times the log base 10 of 7.96 times 10 to the minus five. And then we're gonna just divide it by our I zero I zero again, is just that lower threshold of human here. And it's kind of like a reference point or a reference level. Um And basically, this is gonna be one times 10 to the minus 12 like this. So if you go ahead and work this out, what you're gonna 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.

2

Problem

Problem

A sound wave from a police siren has an intensity of 0.01 W/m^{2} 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

3

example

Example 2

Video duration:

6m

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Hey, everyone, welcome back. So let's take a look at this practice problem here. So this problem tells us we have a sound source that emits in equally in all directions. So if 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, what I'm told here is that um I'm gonna try to figure out that if I move twice the distance away from the source, how many decibels this does 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 gonna call this R one basically at this little circle over here, this distance, you're gonna measure some intensity. I'm gonna call this I one. And because of that intensity, uh I can plug this into the sound intensity level formula and get some beta, right? So I'm gonna 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 R two, what I know is that the, the intensity is gonna drop off because it gets weaker as you go farther from the source. So I'm gonna get an I two. I plug that in here. I'm gonna get some other beta and some others decibel level that's gonna 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 one or beta two. I'm actually trying to find out how many decibels it decreases by. So I'm actually trying to find the change in the beta. So I'm not finding beta beta one or beta two. I'm actually looking for delta beta. It's the difference between these two. And really what you can do is just take the two sound intensity levels and just subtract them. That's basically what we're gonna do in this problem here. So I'm gonna set up a little bit of another equation but notice how we actually have no numbers that we can plug in. So this is gonna be a little bit of a derivation. So let's check it out, right? So this is gonna be beta one minus beta two right now. I actually know what the beta formula is. So I can just replace those things with uh that formula. All right. So let's just work this out. Let's expand this, this is just gonna be 10 times logarithm base 10. And then I'm gonna have I over I knots now because I have a beta one. And a beta two because I have an intensity one and an intensity two. Then basically, I'm just gonna replace this, not with just I, but I one over I knots. And then over here, this is just the, this is the first term and the second term just becomes 10 times logarithm base 10. And this is gonna be I two divided by I knots. All right. Now, what we can do here is we noticed 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 distributive property. So this is gonna be 10. Uh And then I'm gonna have lori this is gonna be log base 10 of I one over I knots minus log base 10 of I two over I knots. All right. So that's what this whole thing becomes because the 10 sort of multiplies both of them. So you can pull it out as a great common factor. All right. Now, if you take a look at this expression over here, we have a logarithm subtracting another logarithm. And one of the things we could actually here is we can use a property of logarithms that you may remember from math. Basically, if you ever have a log that's minus another log and they're both the same base like we have 10 and 10 here, then you can basically just put them as a division instead. So we can actually do that here. So what I'm gonna do is I'm gonna take this 10 and I'm gonna take this whole expression over here. And what this says here, this rule allows me to do is I can say, well, this expression is actually just if I take log 10 and actually just put the two arguments in the parentheses as division. So basically, this just becomes a really, really big division and it becomes I one over I knots divided by I two over I knots. All right. So notice how we really just went from, uh 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. All right now because of this, we actually 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 I knots. So remember what happens is if you take a fraction over a fraction, you're gonna have to flip the fraction and you'll notice that these things actually will cancel out, right? The I knots will cancel. So it's actually good that we don't really need it anymore. All right. So what is this whole thing become? It really just becomes 10 times the logarithm base 10. And this is just gonna be I one divided by I two. All right. Now, we're almost done here. Uh Basically, this is just gonna be, let's see. Um 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 I one over I two is actually really just R two squared over R one squared. All right. So here's what happens. We could basically just take this expression over here, which I'll highlight in blue and I'm actually just gonna place it with this. All right. So this whole thing actually just becomes uh this is gonna become 10 times logarithm base 10 of. And this actually, I'm just going to replace this with R two squared over R one squared. All right. 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 R two is just gonna be twice the distance of R one. So in other words, as a relationship here, as a ratio, I know that R two R over R one just equals two. All right, because I'm told that I'm gonna move twice the distance away. So I'm actually what this means, what I'm gonna, I'm gonna do is I wanna put this up here because that's really what this means. That means that R two over R one is equal to two. So I'm gonna take this number here, just two 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 two squared uh or what you could do, this is gonna be 10 times logarithm base 10 of four. Now, what you're gonna get here, by the way is you're actually gonna get uh six 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 six decibels if you were to move twice the distance away from the source. All right. So this is actually kind of a pretty interesting quirk of sounds of sounds levels and uh and inverse square laws. Um But basically, the whole point here is that whenever the distance from an inverse square law, how source I know it's kind of a mouthful but like sort of like uh any sound that sort of drops off as, as one of R squared, the sound intensity level actually decreases by a factor of six decibels. So it's actually not really linear. Um Each sort of like if you, if you double the distance, you're decreasing by another six decibels and then if you double that distance, it decreases by another six decibels. All right. So hopefully, that makes sense. Folks. Let me know if you have any questions.