Velocity of Longitudinal Waves - Video Tutorials & Practice Problems
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concept
Speed of Longitudinal Waves (Fluids & Solids)
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5m
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Hey guys. So up until now, we've been dealing with the velocity equation for transverse waves on strings, which was this guy right here. But sometimes in problems you're not gonna have transverse waves and strings, you're gonna have longitudinal waves that are propagating through other typess of mediums like a liquid or a solid or something like that. So I'm gonna show you the equation for the speed of longitudinal waves and fluids and solids. What we're going to see is that it's very similar to this equation right here. It's just a couple of letters that are different. And then we're gonna check out some examples very straightforward. Let's get to it. So basically, your longitudinal wave speed equations break down to two different categories. Whether your wave is traveling through a fluid, uh like water or you know, some other liquid or if you're traveling through a solid, like a metal rod or something like that. If you've ever like gone up to a metal pole and you've knocked on it, you've probably heard the sound wave travel up uh and sort of echo along the way. So sound waves can travel through fluids and solids now, your textbooks are gonna do long derivations for this. I'm just gonna give you the equations because your professors won't have you memorize this. So the speed for fluids is gonna be beta divided by a row where beta is the bulk modulus of a fluid. This is a number that's always gonna give, be given to you if you need it divided by the density of the fluid. So these are both uh letters that were probably uh sort of given to you in um or talked about when you did the Elasticity chapter. Now in solids, it's gonna look very similar, except it's gonna be instead at young's modulus because we're talking about solids, rods, uh solids and rods divided by the density of the material. Now remember that all waves, regardless of whether they are transverse or longitudinal, always are equal to lambda times frequency. So this relationship still holds for longitudinal waves. That's it. So that's all there is to it. Let's go ahead and take a look at some uh and some examples here. So we have a liquid. So we know we're gonna be dealing with this equation right here and we're told the density of this liquid is 1200. We know it's a longitudinal wave and we're told that the frequency of these waves are 400 Hertz. And we're told that the wavelengths are measured to be 8 m. And we want to do is we wanna calculate the bulk modulus of this liquid here. So basically, that's gonna be that beta term. So I'm gonna use this equation right here. Let's go ahead and set it up. So I've got my V is equal to the square roots of beta, divided by row is equal to lambda times the frequency. So let's see, I don't have the beta. In fact, that's actually what I'm looking for here, but I do have the density of the fluid. I also have the wavelength and the frequency of that wave. So mirror this is basically just um it's just giving you four, you have four variable variables and I have three out of the four. So I can go ahead and figure this unknown. Now, this one that's unknown. So I'm just going to square both sides. I'm actually just gonna go ahead and start plugging in some numbers. So I've got the square roots of this bulk modulus divided by 1200 divided by four times the frequency, which is 800 here. So when I square both sides, I'm gonna get beta divided by 1200 is equal to, this is gonna be 3200 and I'm gonna square that. And so finally, once you go ahead and plug everything in, you're gonna get that beta is equal to. Uh let's see, I got 1.23 times 10 to the 10th. Um And that is the units for that are in pascals. Let's look at, look at the second example, the second example, we're gonna strike a 60 m long brass rod at one end. So let me go ahead and draw this. I've got this little like long brass pipe like this and I've got the distance here. This delta X is equal to 60 m. So how long does it take for a person on the other end of the rod to hear the sound? What they're actually asking us for is delta T. So the idea here is that you're gonna sh strike went into the brass rod and the sound wave is gonna propagate, it's going to travel down through that solid and eventually hit the person on the other side. So this person is standing over here like this. So we wanna calculate what's the delta T between you hitting it and then hearing the sound here. All right. So we're given all the constants that we need the young modulus and the density and stuff like that. So how do we figure this out? Well, if I'm looking for a delta T and I have velocity and delta X, I'm actually gonna set up an old mo one dimensional motion equation. Remember that V is equal to delta X over delta T, right? So it's just displacement over time. So what I want is I want to figure out this delta T. So I'm actually just gonna trade places between both these variables. So delta T is equal to delta X divided by the velocity. So I have what the delta X is. And if I want to figure out delta T, I just need to figure out what is the velocity. So how do I do that? Well, now I have the equation to do that, the velocity of a longitudinal wave given young's modulus and the density here. So that's all I have to do. I just have to basically go over here and figure out what is this, what is this V here by using that equation? So remember this V is just gonna be the square roots and this is gonna be Young's modulus divided by row. So this is gonna be the square roots of uh let's see, I get um I got nine times 10 to the 10th divided by uh 8600. And what you should get is the velocity is gonna be uh 3000, yeah, 3200 m per second. So that's the speed at which a sound wave travels through brass. So this is actually about 10 times faster than the speed of sound through air. And that makes sense. So sound travels much faster through metals and solids because the density of material is much higher there. So this is 3200 m per second. Now, we just plug this back into this equation here. So to finish this off, we have delta T is equal to, this is gonna be 60 divided by 3200 and you should get a delta T that's equal to 0.02 seconds. So it takes almost no time for that sound to reach them. All right. So that's it. This one guys, let me know if you have any questions.
2
Problem
Problem
A metal bar has a length of 1.5 m and a density of 6400 kg/m3. Sound waves take 3.9 × 10-4 s to travel along the length of the bar. What is Young's modulus for this metal?
A
2,311 Pa
B
9.47 × 1010 Pa
C
2.46 × 107 Pa
D
3.744 Pa
3
example
Example 1
Video duration:
2m
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Welcome back. Everyone. In this problem, we're focused on a container full of oxygen gas, 32 g of oxygen occupies a volume of 0.0224. The speed of sound in this container is 317 m per second. We want to figure out the bulk modulus for oxygen gas. So in other words, we have a mass of 0.032 kg. I'm just gonna convert that to kilograms a volume that it occupies which is 0.0224 that's going to be meters cubed. We're told the speed which is a little v of sound in this container is 317. And ultimately, we don't want to figure out the bulk modulus, which is the capital letter beta here. So that's really what we're looking for. Remember in these types of problems here, we're looking with waves of longitudinal uh of waves or sorry speed of longitudinal waves. There's two forms one with beta over row and one with young over row. One is for liquids. This actually sort of should say fluids because in this case, we're dealing with a gas, but it behaves like a liquid. Um And then one for solids. So we're really just looking at this equation over here our velocity equation. All right. So let's sort of put all this stuff together. And we're just going to use our velocity equation for longitudinal waves and solids or sorry in fluids, which is going to be the bulk modulus divided by row. We're looking for this bulk modulus here. So we just need those other two variables. I need to figure out the density and and I also need the speed of sound. Now, we actually have what the speed of sound is. We're just given that we're just 370 m per second. What we really need to do is find the density here because all we have is mass and volume. So that leads us to how do we find density? How do we figure out this row here inside of this equation? Well, we're really just gonna come over here and remember the definition of density is always just mass divided by volume. Now, remember mass over volume has to be in kilograms per meters cubes. So you have to have things in the right units. So that's why we converted this thing to 0.032. So this really just becomes 0.032 that's in kilograms divided by a volume of 0.0224. And that's gonna be in meters cubed. Now, if you go try to figure this out. What you're gonna get actually here um Is you are going to get 1.43 and that's gonna be kilograms per meters cubed. So this is the density that you're gonna plug into this equation over here. All right. So now I'm just going to go ahead and rearrange. So first we're gonna have to square both sides to get rid of the square roots. So this is just going to be beta over row. And then lastly, all we have to do is just multiply the row over. So in other words, beta is equal to V squared times row. So in other words, it's 317, that's the speed of sound in this container, squared times our density that we figured out was 1.43. Now, go ahead and work this out. What you're gonna get is you're going to get a bulk modulus of 1.44 times 10 to the fifth and bulk modulus are listed in pascals. So this is our final answer and looking through our answer choices, it looks like that's going to be answer choice. C, so that's the correct answer. Hopefully you got that one right. Uh Let me know if you have any questions if you didn't. Thanks for watching.
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