Velocity of Transverse Waves - Online Tutor, Practice Problems & Exam Prep

Waves on strings can be analyzed using the wave speed equation, v=λf, and a specific equation for wave speed, v = Tμ, where T is tension and μ is mass density. Changes in tension or mass directly affect wave speed, while frequency changes impact wavelength without altering speed. Understanding these relationships is crucial for solving wave problems effectively.

Velocity of Waves on a String

Guys, in the last couple of videos, I introduced you to waves and I also showed you the wave speed equation, \( v = \lambda f \), which applies to all kinds of waves whether they are transverse or longitudinal. Now in some problems, we're going to 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 going to show you in this video. I'm going to show you a special equation that we use to calculate the velocity for waves, specifically well on strings. So let's go ahead and check this out here. Let's take a look at a problem. We got a tension of 100 Newtons and we're flicking it up and down to create a transverse wave. Basically, you're doing this. You've got some little string like this. It's attached to the wall at some point 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. The string has a mass of 0.5 kilograms. So first, the tension is 100. Our mass is 0.5, and the length of the string \( l \) is going to be 1.2 meters. So we want to figure out what the frequency of this wave is.

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, \( v = \lambda f \). So let's start there. We've got \( v = \lambda f \). Now I'm going to 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 going to need another equation. That's the point of this video. So \( v = \lambda f \), 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 the things that you can measure. Things like tension and mass and length. So this new equation here that relates the speed with these physical properties is going to 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 \( v = \lambda f \), but specifically for waves on strings, we also have this other equation. Since some problems, we're going to 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 \( v = \sqrt{\frac{\text{tension}}{\mu}} \). Now we're not given \( \mu \) directly. Right? We are given the tension, so we can go ahead and pop that in. We're not given the \( \mu \), but we actually do have mass and length so we can figure it out. So one thing we can do here is we can actually just go ahead and rewrite this entire expression. So we're going to have the tension on top and remember, \( \mu \) is really just mass divided by the length. So really we're just going to have this fraction dividing by another fraction here. This is going to be \( m \) over \( l \). Now I'm just going to plug in all the numbers. So this is going to be the square roots of 100 divided by and and this is going to be 0.5 divided by 1.2. So if you go to work this out, you're going to get a speed of 15.4 meters 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 gotta 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 going to divide it by the wavelength, which is 0.15. When you go ahead and work this out here, you're going to get 103 Hertz. So that's how to calculate the frequency. Sometimes you just have to use a combination of these equations.

Alright, guys. So that's it for this one. Let me know if you have any questions.

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. F_T = 36 N
• B. F_T = 6 N
• C. F_T = 1.2 N
• D. F_T = 22,500 N

Example 1

Hey, everyone. So I want you to imagine in this problem that you and your friend have sort of a rope that is 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 going to whip it up and down to create a wave pulse. You're going to whip this string or rope up and down, and that wave is going to 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 going to 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 \( \Delta t \)? Well, what happens here is we're going to use an old relationship between the velocity, \( \Delta t \), and the length of the rope, which is \( \Delta x \). This \( \Delta x \) here is going to be the length of the rope, between the two sides, and basically, if the wave pulse is going to be traveling at a fixed speed, it doesn't accelerate, then the relationship is just that \( v = \frac{\Delta x}{\Delta t} \). Right? Displacement over time. We saw that a long time ago in physics. So to calculate this \( \Delta t \), we're really just going to rearrange, and this \( \Delta t \) is going to be the \( \Delta x \), the displacement divided by the velocity. Now what's the displacement? Well, if you could just assume that the rope, which is 7 meters long 7 and a half meters long, doesn't change a whole lot as you whip it up and down, then we can just say this \( \Delta x \) is equal to \( l \), which is 7.5. So what we need to do is have the \( \Delta x \). Now we just need to find the velocity in order to figure out the time. So how do we calculate the velocity? Well, we're just going to use now our velocity equations for waves on strings. Remember, we can always use this equation, that 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 velocity force for waves on strings. So we can only use this for strings. This \( v \) here equals the square root of the tension force divided by the mass per unit length. So in other words, I'm just going to expand that. That's the square root of the tension force divided by this is going to be mass per length. Alright? 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. Alright? So we're basically just going to go ahead and plug in a bunch of numbers to calculate this. This is going to be the square root of 30 divided by and this is going to be 0.5 divided by 7.5. Now when you work this out, we remember what you're going to get here is 21.2 meters 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 going to be the \( \Delta x \), which is, in other words, the 7.5 divided by, this is going to be 7.5 divided by the 21.2 meters per second, and what 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.

Solving Problems with Waves on Strings

Guys, in the last couple of videos, we took a look at the equation for wave speed for waves on strings, which is really just this guy right here. Well, in some problems, they're going to give you the initial setup of a problem. For example, we have an oscillating blade that's creating transverse waves on a stretched string. What they're going to do in different parts of the problems is they're going to change some of the variables. For example, in part a, we're going to quadruple the tension. In part b, we're going to double the frequency. 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 going to give you some very simple rules for solving these problems. Let's go ahead and check this out here. We're going to start with part a. 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 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 just 2. And initially, we calculate the wave speed, so this whole entire wave is moving to the right with 7 meters per second. And we also thought 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'll 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 new tension, which I'm going to call \(f_t'\). It's just going to be quadruple what the initial tension was. So it's going to be \(4 \times 98\), which is going to equal 392, and now we want to calculate the new wave speed, so I'm going to call this \(v_A\). 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 tension, the length. It depends on the physical properties of the medium. So any time 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 length, it's going to 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 our new value for tension. So we're just going to use the square roots of \( \frac{f_t'}{\mu} \). Now \(f_t'\) is the only one that's changed, now we just have 392, and our \(\mu\) value hasn't, it's just 2, right? The mass and length hasn't changed. So, what you end up getting here is you end up getting 14 meters per second, which is exactly twice what the initial wave speed was. I'm going to call this \(v_0\) here. And that's because we quadrupled the value that's inside this numerator, so effectively you are doubling the wave speed. Alright? So 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 now we're going to increase that and we're going to double it. So what happens is we're going to move this vibrating blade and it's going to go faster. It's going to 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 \(f'\) is going to be twice the original frequency, so I'm going to call this original frequency 35 hertz. So, we just have \(2 \times 35\), which equals 70, and now we want to calculate the new wave speed. I'm going to call this \(v_B\) and I'm going to call this \(\lambda_B\). So now what happens is we're going to take a look at our second rule. Oursecond 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 \(f\) that's inside 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 their 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 \(f\). 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 now 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 \(v_B\) is just the initial wave speed, which is going to be 7 meters 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, \(v_{string} = \lambda f\). If you solve for \(\lambda\), what you're going to get is \(v_v\), the initial divided by the new frequency, what you're going to get here is the 7 meters per second divided by the new frequency of 70. You're going to get a wavelength that's 0.1 meters. So, this actually makes some sense. If you're doubling the frequency of the oscillator, you're going to create these more bunched-up waves like this, and that just means that the wavelength is going to decrease because now you're going to create 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. Alright? So that's how these problems work. Let me know if you guys have any questions.

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

Example 2

Hey, guys. So let's work this problem out together. In this problem, we're told that the wave speed for a certain string under tension is 20 meters per second. I'm going to call this v=20. I'm going to 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, so the tension is half of its original value, then what happens to the new wave speed? Let's take a look here because, obviously, we're going to be dealing with some kind of velocity or a wave speed equation, and we know this is going to be a string. So, because we have a string, we can always just use this formula over here. So we have that v=FTμ . Now notice how this problem has really no information other than 20 meters per second. And this is a classic kind of proportional reasoning 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? Here's what I'm going to do here. I'm going to say that vnew is really just going to be the square root of FTnew over the mass per unit length. The only thing 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 FTnew here, what they're asking or saying here, is that it's going to be one-half of the original tension. So what happens is this equation here, v=FTμ, is equal to 20. We actually know that already. But now what happens is vnew is going to be the new tension divided by μ. Okay? And what we can do here is we can basically just replace this FTnew with this expression, the one-half FT. So it's going to be the square root 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 μ, right? That's what happens when you sort of extract it, basically just splitting up the square roots. So notice how this piece right here, this equation, this FT over μ, 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. 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 going to be the square root of one-half times 20, which is equal to 14.1 meters per second. Now this should make some sense that the new wave speed is going to 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 going to 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.

Here’s what students ask on this topic:

What is the equation for the velocity of transverse waves on a string?

The velocity of transverse waves on a string is given by the equation:

$v=\sqrt{\frac{T}{\mu }}$

where $v$ is the wave speed, $T$ is the tension in the string, and $\mu$ is the mass per unit length (mass density) of the string.

How does tension affect the speed of a transverse wave on a string?

The speed of a transverse wave on a string is directly related to the square root of the tension in the string. According to the equation:

$v=\sqrt{\frac{T}{\mu }}$

if the tension $T$ increases, the wave speed $v$ also increases. Specifically, if the tension is quadrupled, the wave speed will double.

What is the relationship between wave speed, frequency, and wavelength for transverse waves?

The relationship between wave speed $v$, frequency $f$, and wavelength $\lambda$ for transverse waves is given by the equation:

$v=\lambda f$

This means that the wave speed is the product of the wavelength and the frequency. If either the frequency or the wavelength changes, the other must adjust to keep the wave speed constant, assuming the medium does not change.

How do you calculate the mass density of a string?

The mass density $\mu$ of a string is calculated by dividing the mass $m$ of the string by its length $l$:

$\mu =\frac{m}{l}$

This value is essential for determining the wave speed on the string using the wave speed equation.

What happens to the wavelength if the frequency of a transverse wave is doubled?

If the frequency $f$ of a transverse wave is doubled, the wavelength $\lambda$ is halved, assuming the wave speed $v$ remains constant. This relationship is described by the equation:

$v=\lambda f$

Since the wave speed is the product of wavelength and frequency, if the frequency increases, the wavelength must decrease proportionally to maintain the same wave speed.