Velocity of Transverse Waves - Video Tutorials & Practice Problems
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1
concept
Velocity of Waves on a String
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3m
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Video transcript
Hey guys. In the last couple of videos, I introduced you to waves and I also showed you the wave speed equation V equals lambda F, which applies to all kinds of waves, whether they are transverse or longitudinal. Now, in some problems, we're gonna be dealing with transverse waves on strings. You're pulling some string with some tension, you're whipping it up and down. But this equation isn't going to be enough to solve our problems. So we're going to need another equation to solve. And that's what I'm gonna show you in this video. I'm going to show you a special equation that we use to calculate the velocity for waves specifically went on strings. So let's go ahead and check this out here. Take a look at a problem. We got a uh tension of 100 newtons and we're flicking it up and down to create a transverse wave. So basically, you're doing this, you've got some little string like this, it's attached to the wall at some points and you're just basically whipping it up and down to create a transverse wave. So let's go ahead and take a look here. We have the string has a mass of 0.5 kg. So first, the tension is 100 our mass is 0.5 and the length of the string L is going to be 1.2 m. So we want to figure out what the frequency of this wave is. So you're flicking this thing up and down, what's the frequency? But we're told that the wavelength is 15 centimeters. So the last piece of information we know is that Lambda is equal to 0.15 once you do the conversion here. So what's F? Well, the only equation we've seen so far is the wave speed equation V equals lambda F. So let's start there. So we've got V equals lambda F. Now I'm gonna solve for F. So I've got this F here is equal to V divided by lambda. So if I want to figure out the frequency, I just need both of these things. Now, the problem is, I actually am told what the wavelength is. But unfortunately, I don't know what this wave speed is, what is V and given the other variables that I have here, there's no way to calculate V. So it turns out that this equation isn't enough to solve this problem. And so we're gonna need another equation. That's the point of this video. So we vehicles lambda F that applies to all waves. But wave speed is also determined by the physical properties of the medium itself. What does. That mean it means that the velocity for waves on a string depends on the physical properties of the string. Those are things that you can measure things like tension and mass and length. So this new equation here that relates to speed with these physical properties is gonna be the square root of the tension divided by this Greek letter mu. This Greek letter mu is just the mass divided by the length of the string. Most textbooks refer to this as the mass density of the string. So we have V equals lambda F but specifically for waves on strings, we also have this other equation. So some problems we're gonna have to use both. So let's go ahead and get back to our problem here. We want to figure out the wave speed. Now we're just going to use this new equation here. So we have V equals the square roots of the tension divided by mu. Now we're not given me directly, right? We are given the tension so we can go ahead and pop that in. But we're not given the mu but we actually do have mass and length so we can figure it out. So we, one thing we can do here is we can actually just go ahead and rewrite this entire expression. So we're gonna have the tension on top. And remember, mu is really just mass divided by the length. So really we're just gonna have this fraction dividing by another fraction here. This is gonna be M over L. Now, I'm just gonna plug in all the numbers. So this is gonna be the square roots of 100 divided by and this is gonna be 0.5 divided by 1.2. So if you got to work this out, you're gonna get a speed of 15.4 m per second. Are we done yet? Not quite because remember this is just the v this is just the wave speed we're supposed to find the frequency. So we got to pop this thing back into this equation over here and then we'll solve. So we got, now, this wave speed here is 15.4. Now we're divided by the wavelength which is 0.15. When you go to work this out here, you're gonna get 100 and three Hertz. So that's how to calculate the frequency. Sometimes you just have to use a combination of these equations. All right guys. So that's it for this one. Let me know if you have any questions.
2
Problem
Problem
A rope with length 2.50 m and mass 0.10 kg is stretched and pulled to create transverse waves of frequency 40.0 Hz and wavelength of 0.750 m. How much tension is exerted on the rope?
A
FT = 36 N
B
FT = 6 N
C
FT = 1.2 N
D
FT = 22,500 N
3
example
Example 1
Video duration:
3m
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Video transcript
Hey, everyone. So I want you to imagine this problem that you and your friend have sort of a rope that is in between you. So you've got you like this, you're holding this little bit of rope and you've got it connected to your friend who's over here. And basically, the idea is that you're gonna whip it up and down to create a wave pulse. So you're gonna whip this, this string of rope up and down and that wave is gonna travel along until it reaches your friend. If this wave travels with some speed V, then how long does it take for the wave pulse to travel? So how long is gonna be a delta T? What we're really looking for is how much time does it take for that wave pulse to travel along the rope and reach the other side and reach your friend? So how do we go ahead and solve for DEL D for delta T? Well, what happens here is we're gonna use an old relationship between the velocity delta T and the length of the rope, which is delta X. This delta X here is gonna be the length of the rope uh between the two sides. And basically, if the wave pulse is gonna be traveling at a fixed speed, it doesn't accelerate, then the relationship is just that V equals delta X over delta T right displacement over time, we saw that a long time ago in physics. So to calculate this delta T, we're really just gonna rearrange and this delta T is gonna be the delta X the displacement divided by the velocity. Now, what's the displacement? Well, if you could just assume that the rope of the rope which is 7 m long, 7.5 m long doesn't change a whole lot as you whip it up and down. Then we can just see this delta X is equal to L which is 7.5. So we need to do is we have the delta X. Now we just need to find the velocity in order to figure out the time. So how do we calculate then the velocity? What we're just gonna use now are velocity equations for waves on string. Remember we can always use this equation that's always works the frequency and wavelength, but we're not given any of that information. Instead, we're going to use this one which is the um it was just the velocity for, for waves on strings. So we can only use this for strings. This V here equals the square roots of the tension force divided by the mass per unit length. So in other words, I'm just gonna expand that, that's the square root of the tension force divided by this is going to be mass per length. All right. So that's basically what we need because we have the tension. We're told that as you pull this between your friend, there is 30 newtons of tension on the rope and we've got the mass of it, which is 0.5 and we've also got the length of it. All right. So we're basically just gonna go ahead and plug in a bunch of numbers to calculate this. This is going to be the square roots of 30 divided by. And this is gonna be 0.5 divided by 7.5. Now, when you work this out, we remember what you're gonna get here is 21.2 m per second. Now, remember this is not our final answer because we need the velocity in order to plug it back into this equation here to find the time. So this delta T is just gonna be the delta X, which is, in other words, this is the 7.5 divided by uh this is gonna be 7.5 divided by the 21.2 m per second and which you'll get here is 0.35 seconds. So that's how long it takes for that wave pulse to travel between you and your friend. That's it for this one, folks.
4
concept
Solving Problems with Waves on Strings
Video duration:
5m
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Video transcript
Hey guys. So in the last couple of videos, we took a look at the equation for waves speed for waves on strings, which is really just this guy right here. Well, in some problems, they're going to give you initial setup of a problem. So for example, we have an oscillating blade that's creating transverse waves on a stretched string. And what what they're gonna do in different parts of the problems is they're going to change some of the variables. So for example, in part A, we're gonna call drupal the tension in part B, we're going to double the frequency and they're going to ask you to calculate what happens as a result of this change. So we're going to calculate the new wave speed and the new wavelength. Now in these problems, it's often difficult to understand which variables affect others inside of this equation here. So I'm going to show you how this works. I'm gonna give you some very simple rules for solving these problems. Let's go ahead and check this out here. We're going to start off with part A. So our setup here tells us that we have an oscillating blade which is really just a little blade that's attached to the string that's on a motor that's, that's vibrating up and down. It's creating waves at a frequency of 35 Hertz. So it's basically spinning up and down at a frequency of 35. We're told the tension on this string here is 98. We're also told that the mass density of this is true is just two. And initially, we calculate the wave speed. So this whole entire wave is moving to the right with 7 m per second. And we're also told that the wavelength this lambda here is equal to 0.2. Now you can just trust me on that or you can actually plug in all the values inside of this equation and you actually get those numbers. So in part A, we're going to quadruple the tension on the string and then we want to calculate the new wave speed. So basically what they're saying here is I'm going to calculate this uh this, this new tension in which I'm gonna call ft prime. It's just gonna be quadruple what the initial tension was. So it's gonna be four times 98 which is equal 392. And now we want to calculate the new wave speed. So I'm gonna call this va and that brings us to the first rule. The first rule to solve these problems is that the velocity, remember depends on the mass the, the, the sorry, the tension, the mass and the length, it depends on the physical properties of the medium. So anytime you change any one of these values, tension mass or length, you're always going to change the velocity. So remember to change any one of these values here, like the tension or the mass and the l it's going directly impact the speed of the wave that's on the string. So what happens is to calculate our new velocity? All we have to do is just use this same equation here. But now we just have to use our updated or a new value for tension. So we're just going to use the square roots of FT prime divided by the mu. Now FT prime is the only one that's changed. Now, we just have 392 and our new value hasn't, it's just two, right? The mass and length hasn't changed. So what you end up getting here is you get getting 4 14 m per second, which is exactly twice what the initial wave speed was. I'm gonna call this V knot here. And that's because we quadruple the value that's inside this numerator. So effectively, we are doubling the wave speed. All right. So that's the, that's the first rule. Let's take a look. Now at the second part of the problem. Now we're going to double the frequency of the oscillator. So what happens here is remember that we have this little spinning blade that's oscillating, vibrating up and down. And all that happens is that now we're going to increase that, we're gonna double it. So what happens is we're gonna move this, this vibrating blade and it's gonna go faster. It's gonna create these little bunched up waves like this. We want to do the same thing here. We want to calculate the new wave speed and also the new wavelength. So now what happens is we have our new frequency F prime is going to be twice the original frequency. So I'm going to call this original frequency 35 Hertz. So we just have two times 35 which equals 70. And now we want to calculate the new wave speed. I'm gonna call this VB and I'm gonna call this lambda B. So now what happens is we're going to take a look at our second rule. Our second rule says that the frequency of a wave depends only on the oscillator frequency. So what happens here is that this oscillator frequency is directly going to impact what this F is. So I'm going to call this F oscillator, which is really what this is and these things are equal to each other. So whatever this is, that's going to be the F that's inside of your problems. Now remember that the frequency of the wave speed depends only on the physical properties of the medium tension mass and length. So because F depends only on the oscillator frequency changing, it does not affect the velocity, changing the oscillator frequency only affects F and LAMBDA. So what happens is any change that you make to, this is going to directly impact the F, it's not going to impact the V string. So what happens as a result? Well, basically, if F has to increase or decrease, then that means that your wavelength is going to have to decrease or increase proportionally because this V has to remain the same. That's what the rule is. So what this means here is that by changing the oscillator frequency, your wave speed actually doesn't change at all. So VB is just the initial wave speed which is going to be 7 m per second. And that's the answer your wave speed does not change at all. So what happens to the wavelength as a result? We can actually just go ahead and calculate this by using this equation here string equals lambda. F. If you solve this lambda, which you're gonna get is VV initial divided by the new frequency which you're going to get here is the 7 m per second divided by the new frequency of 70 you're gonna get a wavelength that's 0.1 m. So this actually makes some sense. If you're doubling the frequency of the oscillator, you're going to create these or bunched up waves like this. And that just means that the wavelength is gonna decrease because now you're gonna be creating more bunched up waves like this, the wave speed remains the same, your frequency increases, but your wavelength is going to decrease as a result. Now it's 0.1 whereas initially it was 0.2. All right. So that's how these problems work. Let me know if you guys have any questions.
5
Problem
Problem
An oscillating blade creates waves on a string. If the amplitude of the wave doubles, what happens to the wavelength λ and v?
A
λ and v remain the same
B
λ doubles, v remains the same
C
λ remains the same, v doubles
6
example
Example 2
Video duration:
2m
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Video transcript
Hey guys. So let's work this problem out together. So in this problem, we're told that the wave speed for a certain string under tension is 20 m per second. So I'm gonna call this V equals 20. I'm gonna draw this out real quick. That's my sort of wave like this. And this is sort of like my axis right. So now we're asked to calculate, well, if you cut the tension in half, half, so the tension is half of its original value, then what happens to the new wave speed? So let's take a look here because obviously, we're going to be dealing with some kind of a velocity or a wave speed equation. And we know this is going to be a string. So because we have a string, we can, we can always just use this formula over here. So we have that V equals the square root of the tension divided by the mass per length. Now notice how this problem has really no information other than the 20 m per second. And this is a classic kind of proportional reason question because they're asking what happens to something, if something else gets cut in half, right? If you change something, how does another variable respond? So here's what I'm gonna do here. I wanna say that V New is really just gonna be the square root of FT New over the mass per unit length. The only thing we're, we're told here is that the tension is cut in half. So we can assume that the mass per unit length will also stay the same. So here's what happens this FT new here, what they're asking or what they're saying here is that it's gonna be one half of the original tension. So what happens is this equation here, this V equals squared of FT over mu is equal to 20. We actually know that already. But now what happens is V new is gonna be the new tension divided by Mu. OK. And what we can do here is we can basically just replace this FT new with this expression, the one half FT. So it's gonna be the square roots of one half of the original tension divided by the mass per unit length. And what we can do here is we can basically just pull out the one half that's inside of the square root. And this just becomes the square root of one half times the square root of FT over mu, right? That's what happens when you sort of extract it, basically just splitting it up the the square roots. So notice how this piece right here. This equation. This FFT over U is what shows up in our new equation. But now we just have an extra factor of the square root of one half that's on the outside. So that's what usually happens with these proportional reasoning type questions is that you can basically just extract out the same expression in the before case, but with just an extra constant that's out in front. So remember this expression here is equal to 20. We don't know what all the variables are, but we just know that the whole thing equals 20. So really what happens is this just becomes the square root of one half times 20 right? Because this thing is equal to 20. So in other words, the velocity, the new wave speed is actually just gonna be the square root of one half times 20 which is equal to uh this is gonna be 14.1 m per second. Now, this should make some sense that the new wave speed is gonna be less because again, if you look at the equation here, if you reduce the tension, then that means the speed is going to be reduced as well. It's not gonna be half because of the square root that's inside here, but it is going to be reduced. That's it for this one guys.
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