18. Waves & Sound
Wave Intensity
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Hey, everyone. So in the last couple of videos, we saw the equation for the wave intensity of a wave, which was this equation over here. Now in some problems, you have to compare the intensity of a wave between different distances. The example we're gonna 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. We're going to have to calculate the distance in 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 as 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 going to 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 later on, the wave is gonna travel some bigger distance. I'm gonna call this r 2, this one in blue here. And r 2 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 gonna 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 a 100 watts, it always is gonna remain as a 100 watts. So this PEA remains the same. What happens to the surface area though? Well, hopefully you guys realize that at some greater distance distance r 2. You're gonna have a surface area a 2 that is bigger than a one because the sphere, the area that encloses is gonna 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 gonna go So what do you think happens to the intensity? Hopefully, you guys realize that the intensity is going to decrease because their power remains constant, but your denominator gets bigger. So So that's exactly what happens. The wave intensity is going to decrease here. So I 2 is gonna be I less than I 1. Basically, the power is gonna get spread out or dissipated over a larger area, so the intensity is gonna decrease. Alright. So that's really all there is to it guys. So what we can do is we can actually go ahead and rearrange this equation here, and we wanna 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 gonna remain the same no matter where you look or no matter how far you look away from that source. So what we can do to compare 2 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 4pi r one squared is equal to I 2 times 4pi r 2 squared. And right because both of these things are actually equal to the power, and if they're both if the power always remains the same, we can just set them equal to each other. Now if we go ahead and rearrange for this, cancel out some variables, we'll 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 gonna be flipped from r two and r one. I one over I two is r two squared over r one squared. Alright? That's really all there is to it, guys. Let's take a look at our equation or our problem here. So we have a siren that's producing sound waves radially 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 gonna be 0.25. So we're gonna calculate a distance. This is gonna be r 2. So I'm gonna calculate this, at which the intensity I 2 falls to 0.01. So this is we're gonna calculate here, r 2. Now before we begin, I wanna ask you, what happens what do you think this r 2 is going to be? 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 2 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 inverse square law. This is I one over I two equals r two squared over r one squared. We wanna figure out this r two here. So what I wanna do is I'm gonna rearrange this. I'm gonna 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 gonna be 15. So this is gonna be 15 squared. My I one is 0.25, and my I two is 0.01. What you're gonna get here is r two squared, and this is equal to 5,625. So if you go ahead and take the square root, what you're just gonna get is you're gonna get 75 meters. So just as we predicted, this r two is greater than the r one. We've gone farther and so, therefore, the intensity is decreased. That's really all there is to it, guys. Let me know if you have any questions.
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