Wave Intensity - Video Tutorials & Practice Problems

On a tight schedule?

Get a 10 bullets summary of the topic

1

concept

Wave Intensity

Video duration:

2m

Play a video:

Hey guys. So for this video, I want to introduce a new concept and a new variable called the wave intensity, which has to do with the amount of energy the wave produces over a certain distance. Let's go ahead and check this out. So the idea here is that we've seen one dimensional waves that move to the rights and they also carry some amount of energy, right. So one dimensional waves just move along to the, they carry energy from point A to point B. But you can also have two and three dimensional waves as well. If you've ever dropped a rock in a pond, the ripples travel outwards not just in one direction, that's two dimensional. And the example that we're going to work out down here loudspeakers, which produce sound waves actually radiate in all directions. That's a three dimensional wave. So what happens here is that one dimensional waves carry energy just along the straight line, but two and three dimensional waves radiate energy in all directions and they spread it out over a surface area which we use the letter A for. So the idea here is we actually define the wave intensity the intensity of that wave is just the energy per time, which remember is equal to power, it's going to be the power divided by the surface area. So the idea here is that if you have a three dimensional wave that travels outwards like the sound source here at the center of our diagram, it's gonna radiate energy outwards in all dimensions like this. And so at a certain distance at any distance of R away from that source, you can sort of imagine the sphere, the surface area is four pi R squared and the intensity of the wave is going to be P divided by that surface area. So the equation for this is P divided by four pi R squared and the units for this are going to be in watts per meter squared. All right. So let's go and take a look at our example here, we have a source that produces 500 watts of power. So our loudspeaker here produces 500 watts of power. We want to calculate what the wave intensity is at a distance of 10 m. What this means here is that R equals 10. So basically at some distance away of 10, we can sort of draw this surface area which is our sphere and the intensity at this point is going to be I equals P over A. So it's going to be P divided by. Um so sorry, I'm actually use, I'm gonna use four pi R squared here. So four pi R squared. So we're gonna use 500 divided by four pi times 10 m squared. If you go and work this out, what you're gonna get is 0.4 and that's watts per meter squared. So watts per meter squared. All right. So that's it for this one, guys. Very straightforward example. And uh let's go ahead and take a look at some practice pros.

2

Problem

Problem

A sound source radiates sound waves in all directions. At a distance of 4 m from the source, you measure the wave intensity to be 0.06 W/m^{2}. How much sound energy does the source emit in 1 hour if the power output is constant?

A

724 J

B

10,860 J

C

1.07 J

D

43,400 J

3

example

Example 1

Video duration:

4m

Play a video:

Welcome back everyone. So this is kind of a cool problem because it actually has to do with real, real life information, real data here about the wave intensity of sunlight. So we're told that wave intensity also applies to sunlight, which is an electromagnetic wave. Uh We're told that it takes approximately 500 seconds for light from the sun to reach the earth. And the intensity when it reaches earth is 1360 watts per square meters. And we're asked to find what's the total power output or the average power output of our sun? All right. So basically what happens is we're looking for P average, right? So we're looking for P average, we're gonna use our wave intensity equation uh because we're also told some information about that, the wave intensity. All right. So let's go ahead and figure this out here. So I know that my wave intensity, I'm just gonna start off with my equation, I equals P average divided by uh A in which really what happens here is if you sort of model the solar system, what happens is that the sun is going to be in the center like this? So this is gonna be like the sun and it radiates light in all directions equally. And earth as it travels along in its orbit, basically travels around in a circle. So in other words, what happens is that light output from the sun because it radiates in all directions, then the sort of spherical area here. So the surface area is going to be the surface area of just a sphere. So in other words, it's perfectly fine for us to use the surface area four pi R squared over here. So intensity is just going to be P average divided by four pi R squared. And we're still looking for this power output over here, we know what the intensity is. So really what happens is we just have to figure out the other variable in this equation which is to figure out the distance. What is the R value? What is the distance between the sun and the earth? So that's all I need to figure out once I figure that out I can plug it into this equation over here to figure out the power output of the sun. So how do I do that? Well, if you take a look at this problem, the distance is not given to us. And while you could go look at this up in your textbook and look it up online, it's very easy to uh we're, we're basically gonna see a really easy way to sort of solve this because when we, we can relate the distance that the light travels and also the speed of light to get a really sort of quick calculation of the distance between the earth and the sun. All right. So what's this R distance here? You could look it up, but you could also relate it to V equals delta X over delta T. So distance uh sorry displacement over time. So really, in this case, what happens is that displacement is going to be your R value. So it's just really easy to solve sort of solve. But this is pretty straightforward, you just take this equation here and just sort of rearrange and put delta T on the other side. So in other words, uh the V here is actually going to be the V, the speed of light. So in other words, VC equals three times 10 to the eighth. So really R is equal to VC times delta T. So in other words, it's just equal to the speed of light, which is three times 10 to the eighth meters per second. And then you multiply it by 500 seconds, right? So now what happens is the seconds will cancel and you'll end up with an approximate for the distance between the earth and the sun, which is 1.5 times 10 to the uh and this is gonna be I think the 10th. All right. So let's take a look at this. This is gonna be, I'm sorry, this is actually gonna be the 11 1.5 times 10 of the 11. So this if you actually look this up in meters, this is actually very close to what the actual distance between the earth and the sun is on average. So you take this number and you plug it back into this equation over here to figure out the power output of the sun. All right. So really what happens is once we rearrange this equation and we're just gonna bring this down here, we're gonna end up with the power output of the song on average is gonna be the intensity which we already know times. And then this is gonna be four pi times R squared. So this is gonna be 1.5 times 10 to the 11th. You're gonna take that whole number and you have to square that. So don't forget that. So you have to square this on the outside and then you have to multiply it by the intensity which we know is actually just equal to 1360. All right. So really, that's what this calculation ends up being once you rearrange. So the average power output of the sun when you do all these calculations is actually gonna be something like 3.85 times 10 to the 26 in watts. And if you actually look this N up for the power output of the sun, this is actually going to be very close to the real number over here. All right. So that's actually how we can take some real information and real data to figure out how much power our, our sun actually produces. And remember this is the amount of energy produces every single second. So it's an insane amount of power. Anyway, thanks for watching. Hopefully this made sense. Let's move on.

4

concept

The Inverse-Square Law for Intensity

Video duration:

5m

Play a video:

Hey, everyone. So in the last couple of videos, we saw the equation for the wave intensity of a wave, which is this equation over here. Now, in some problems, you have to compare the intensity of a wave between different distances. The example, we're going to work out down here has a siren that's producing waves in all directions at some distance. We're given the intensity of this value here. And we're going to have to calculate the distance at which the intensity falls to a different value. Now, in order to do this, we're going to use an equation called the inverse square law for intensity, I'm going to show you how it works. It's very straightforward. So in order to compare distances or to compare the intensity of different distances, we have to understand what happens as waves travel outwards from the source. So I've got this diagram here and let's say this, there's some power source like a loudspeaker or siren or something like that. So what happens is waves travel outwards, well, the distance outwards from the source is going to increase. So for example, if my power source is here and I have some distance. I'm gonna call this R one. There's a surface area here and I can calculate an intensity like this right here at some surface. So what happens as we travel outwards later on, later on the wave is going to travel some a bigger distance. I'm gonna call this R two, this one in blue here and R two is bigger than R one. So what happens here is that the power of the source is actually going to remain constant. So we have to take a look what happens in our variables inside of this equation here, the power is going to remain constant because as you travel outwards, as waves travel outwards, the thing that's actually producing the waves doesn't change. If the loudspeaker is producing 100 watts, it always is going to remain as 100 watts. So this P remains the same. What happens to the surface area though? Well, hopefully you guys realize that it's some greater distance R two. You're going to have a surface area. A two that is bigger than a one because the sphere, the area that encloses is going to be much bigger than the one that was enclosed by the red sphere, this R one distance. So your surface area is going to increase. And therefore this denominator of this equation is going to go up. So what do you think happens to the intensity? Hopefully, you guys realize that the intensity is going to decrease because the power remains constant, but your denominator gets bigger. So that's exactly what happens. The wave intensity is going to decrease here. So I two is going to be I less than I one. Basically, the power is going to get spread out or dissipated over a larger area. So the intensity is going to decrease. All right. So that's really all there is to it guys. So we can do is we can actually go ahead and rearrange this equation here and we want to get it in terms of P. So what happens here is that the intensity times the surface area equals P but this P here is going to remain the same no matter where you look, I remember how far you look away from that source. So we can do to compare two different distances. We can actually just set these ratios or set a ratio equal to each other. So we can do here is we can say that I one times four pi R one squared is equal to I two times four pi R two squared. And right, because both of these things are actually equal to the power. And if they're, if the power always remains the same, we can just set them equal to each other. Now, if we go and rearrange for this, cancel out some variables, we end up with this relationship here, which is the I one over I two is equal to R two over R one and they're both squared. So this equation right here is sometimes called the inverse square law for intensity. It says that the ratio of the intensities has to do with the ratio of the distances. The most important thing you have to remember is that these letters, the I one and the numbers I one and I two are going to be flipped from R two and R one I one over I two is R two squared over R one squared. All right, that's really all there is to it guys. Let's take a look at our equation or a problem here. So we have a siren that's producing sound waves radial outwards. Notice that we don't have the power, we have the distance. The first distance is 15. This is R one and the intensity at that distance I one is going to be 0.25. So we're going to calculate a distance. This is going to be R two. So I'm gonna calculate this in which the intensity I two falls to 0.01. So this is we're going to calculate here R two. Now, before we begin, I want to ask you what happens, what do you think this R two is going to be? Do you think it's gonna be bigger or smaller than this R one? Well, hopefully you guys realize that if the intensity from I one to I two is getting smaller is decreasing. If I two is less than I one. Then that means that R two has to be greater than R one. We're going farther. And so the intensity is decreasing. Let's go ahead and verify this by using our equation. So we have our intensity equation. I, our inver square law, this is I one over I two equals R two squared over R one squared. We want to figure out this R two here. So what I want to do is I'm going to rearrange this, I'm going to move this to the other side. So I've got the R one squared times. The ratio of I one over I two is equal to R two squared. Now, let's just go ahead and plug in some numbers. So I've got my R one is going to be 15. So this is going to be 15 squared. My I one is 0.25 and my I two is 0.01. What you're going to get here is R two squared and this is equal to 5625. So if you go ahead and take the square root, what you're just gonna get is you're gonna get 75 m. So just as we predicted, this R two is greater than the R one, we've gone farther. And so therefore, the intensity has decreased. That's really all there is to it guys, let me know if you have any questions

5

Problem

Problem

You measure the intensity from a sound source to be 0.3 W/m^{2} at a distance of 3.4 m. What will the intensity be if you walk closer to the source, to a distance of 2.5 m?