Wave Intensity - Online Tutor, Practice Problems & Exam Prep

Wave intensity measures the energy a wave produces over a distance, defined as power divided by surface area. For three-dimensional waves, intensity decreases with distance due to the inverse square law, where I = P4πr2. As waves radiate outward, the energy spreads over a larger area, leading to lower intensity at greater distances. Understanding this relationship is crucial for analyzing sound and other wave phenomena.

Wave Intensity

Guys, 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 right, and they also carry some amount of energy. Right? So one-dimensional waves just move along to the right and carry energy from point A to point B. 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 in 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 three-dimensional waves that travel outwards like the sound source here at the center of our diagram, it's going to radiate energy outwards in all dimensions like this. And at a certain distance, at any distance of r away from that source, you can sort of imagine the sphere, the surface area is \( 4\pi r^2 \). And the intensity of the wave is going to be \( p \) divided by that surface area. So the equation for this is \( \frac{p}{4 \pi r^2} \), and the units for this are going to be in watts per meter squared. Alright. So let's go ahead 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 meters. What this means here is that \( r = 10 \). 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 \( I = \frac{p}{a} \). So it's going to be \( p \)divided by so sorry. I'm actually going to use \( 4 \pi r^2 \) here. So \( 4 \pi r^2 \). So we're going to use \( \frac{500}{4 \pi \times 10^2} \). And if you go and work this out, what you're going to get is 0.4, and that's watts per meter squared. So watts per meter squared. Alright. That's that's it for this one, guys. Very straightforward example, and let's go ahead and take a look at some practice problems.

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/m². 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

Example 1

Welcome back, everyone. So this is kind of a cool problem because it actually has to do with real-life information, real data here about the wave intensity of sunlight. We're told that wave intensity also applies to sunlight, which is an electromagnetic wave. 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 meter. We're asked to find what's the total power output or the average power output of our sun. Alright, so, basically, what happens is we're looking for \(P_{\text{average}}\). Alright. we're going to use our wave intensity equation because we're also told some information about the wave intensity. Alright? So let's go ahead and figure this out here. I know that I'm just going to start off with my equation. \(I = \frac{P_{\text{average}}}{A}\), 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. The sun 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 or 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 \(4 \pi r^2\) over here. So intensity is just going to be \(P_{\text{average}}\) divided by \(4 \pi r^2\), 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. We just need 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 look it up in your textbook and look it up online, it's very easy to, We're basically going to see a really easy way to solve this because when we can relate the distance that the light travels and also the speed of light to get a really quick calculation of the distance between the earth and the sun. Alright? So what's this \(r\) distance here? You could look it up, but you could also relate it to \(v = \frac{\delta x}{\delta t}\). So, distance 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 sort of solve for this. It's 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, the \(v\) here is actually going to be the \(v\) speed of light. So in other words, \(v_c = 3 \times 10^8\). So really, \(r\) is equal to \(v_c \times \delta t\). So in other words, it's just equal to the speed of light, which is \(3 \times 10^8\) 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 \(1.5 \times 10^{11}\). 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. Alright, so really what happens is, once we rearrange this equation, we're just going to bring this down here, we're going to end up with the power output of the sun on average is going to be the intensity, which we already know, times, and then this is going to be \(4 \pi\) times \(r^2\). So this is going to be \(1.5 \times 10^{11}\). You're going to 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. Alright? 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 going to be something like \(3.85 \times 10^{26}\) watts. And if you actually look this number up for the power output of the sun, that is actually going to be very close to the real number over here. Alright? So that's actually how we can take some real information and real data to figure out how much power our sun actually produces. And remember, this is the amount of energy it produces every single second, so it's an insane amount of power. Anyway, thanks for watching. Hopefully, this made sense. Let's move on.

The Inverse-Square Law for Intensity

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 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. 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 there's some power source like a loudspeaker or siren or something like that. 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 r₁. 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, the wave is going to travel some bigger distance. I'm going to call this r₂, this one in blue here. And r₂ is bigger than r₁. So what happens here is that the power of the source is actually going to remain constant. So we have to take a look at 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 PEA remains the same. What happens to the surface area though? Well, hopefully you guys realize that at some greater distance r₂ you're going to have a surface area a₂ that is bigger than a₁ because the sphere, the area that encloses, is going to be much bigger than the one that was enclosed by the red sphere, this r₁ distance. So your surface area is going to increase and therefore, this denominator of this equation is going to 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 I₂ is going to be I<I₁ . Basically, the power is going to get spread out or dissipated over a larger area, so the intensity is going to 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 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 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 I₁ times 4πr₁² is equal to I₂ times 4πr₂². I1 I2 = r 2 2 r 1 2 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 I₁ over I₂ is equal to r₂ over r₁ 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 I₁ and the numbers I₂ are gonna be flipped from r₂ and r₁. I₁ over I₂ is r₂ squared over r₁ 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 r₁, and the intensity at that distance, I₁, is going to be 0.25. So we're going to calculate a distance. This is going to be r₂. So I'm going to calculate this, at which the intensity I₂ falls to 0.01. So this is we're going to calculate here, r₂. Now before we begin, I wanna ask you, what happens what do you think this r₂ is going to be? Do you think it's going to be bigger or smaller than this r₁? Well, hopefully, you guys realize that if the intensity from I₁ to I₂ is getting smaller, is decreasing, if I₂ is less than I₁, then that means that r₂ has to be greater than r₁. 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 I₁ over I₂ equals r₂ squared over r₁ squared. We wanna figure out this r₂ 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 r₁ squared times the ratio of I₁ over I₂ is equal to r₂ squared. Now let's just go ahead and plug in some numbers. So I've got my r₁ is going to be 15. So this is going to be 15 squared. My I₁ is 0.25, and my I₂ is 0.01. What you're going to get here is r₂ squared, and this is equal to 5,625. So if you go ahead and take the square root, what you're just going to get is you're going to get 75 meters. So just as we predicted, this r₂ is greater than the r₁. 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.

You measure the intensity from a sound source to be 0.3 W/m² 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?

• A. 0.41 W/m²
• B. 0.16 W/m²
• C. 0.55 W/m²
• D. 0.22 W/m²

Here’s what students ask on this topic:

What is wave intensity and how is it calculated?

Wave intensity measures the energy a wave produces over a certain distance. It is defined as the power (P) of the wave divided by the surface area (A) over which the energy is spread. For three-dimensional waves, the surface area is a sphere, so the formula for intensity (I) is:

$I=\frac{P}{4\pi r^2}$

where r is the distance from the source. The units for intensity are watts per meter squared (W/m²).

How does wave intensity change with distance?

Wave intensity decreases with distance due to the inverse square law. As waves radiate outward from a source, the energy spreads over a larger surface area. The intensity (I) at a distance (r) from the source is given by:

$I=\frac{P}{4\pi r^2}$

As the distance (r) increases, the surface area (4πr²) increases, causing the intensity to decrease. This relationship is crucial for understanding phenomena like sound waves and light intensity.

What is the inverse square law for wave intensity?

The inverse square law for wave intensity states that the intensity (I) of a wave is inversely proportional to the square of the distance (r) from the source. Mathematically, it is expressed as:

$I=\frac{P}{4\pi r^2}$

This means that if the distance from the source is doubled, the intensity is reduced to one-fourth. This law is essential for understanding how wave energy dissipates over distance.

How do you calculate the distance at which wave intensity falls to a certain value?

To calculate the distance at which wave intensity falls to a certain value, you can use the inverse square law. If you know the initial intensity (I₁) at a distance (r₁) and the desired intensity (I₂), you can find the new distance (r₂) using:

$\frac{I1}{I2}=\frac{r^2}{r^1}$

Rearrange to solve for r₂:

$r^2=\sqrt{r^1\times \frac{I}{1}\times \frac{1}{I}}$

This allows you to find the distance at which the intensity falls to the desired value.

What are the units of wave intensity?

The units of wave intensity are watts per meter squared (W/m²). This unit measures the power (in watts) distributed over a surface area (in square meters). Intensity quantifies how much energy a wave transmits through a given area per unit time.