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## Physics

Learn the toughest concepts covered in Physics with step-by-step video tutorials and practice problems by world-class tutors

18. Waves & Sound

# Velocity of Transverse Waves (Strings)

1
concept

## Waves On A String

3m
Play a video:
Hey, guys, in this video, we're gonna talk about waves on strings. Okay? If you were to fix a wave at one end against something like a wall, grab the other end and just whip it up and down, you would see that you produce a bunch of waves traveling down the length of that string. All right, so this is what we want to talk about. Now, as we've said before, a wave on a string is a very common example of a trans verse wave. One who's oscillations are perpendicular to its propagation. Okay, we explain why this was trance first before, but if you whip this string up and down, you're gonna produce these wave pulses that travel towards the wall. Okay, These waves that travel towards the wall now a wave on a string is no different, really than any other wave. In a lot of respects, it has the same mathematical description. It has the same characteristics on amplitude of propagation, speed, wavelength, frequency, etcetera. But there are a couple things that are unique to each wave. The specific energy carried by that wave is going to be unique to the wave and the specific propagation. Speed of that wave is gonna be unique to the wave. Remember that propagation speed depends upon two things. The type of wave and the characteristics of the medium that the wave is in. Okay, so what we're gonna focus ourselves on in this video is the speed of this wave. Okay? And of course, we're talking about the propagation speed. All right. The propagation speed of the wave on a string is given by the following equation. Spirit of tea over mu. Okay, where t is the tension in the string and mu? Is the mass per unit length of that strength? Okay, let's do a quick example. Ah, 1.2 m string as attention of 100 Newtons. If it has a massive 5 kg, what is the frequency of a wave with a 12 centimeter wavelength on this street? Okay, well, we know we want to find the frequency of the wave, okay? And we're told the wavelength. Since we are dealing with both frequency and wavelength, we're probably going to be using the equation. V equals Lambda F. And we can easily solve for F by dividing Lambda over. So the only question is What's the the speed on the string? Well, we know that for waves on a string, the speed is just the tension divided by the mass per unit length. Okay, we know the tension. We know the length and we know the mass. So we know everything we need to solve this problem. This is gonna be 100 Newtons of tension divided by the mass per unit length. So it's 5 kg and mass and it's 1.2 m in length. So that whole thing is going to be about 4.9 meters per second. Okay, This means that the frequency is 012 m. It was given in centimeters, but I need to convert 2 m divided by 4.9 m per second, which equals hurts. All right. No big deal. That wraps up our discussion on our introduction. Toe waves on the strength. Thanks for watching guys
2
example

## Unknown Mass Of A String

2m
Play a video:
Hey, guys, let's do a quick example. An easy way to find the mass of a string is to produce a wave on it. A five centimeter string off unknown mass produces a wave with a centimeter wavelength that has a 70 hertz frequency. When it's tension is 10 Newtons Given this, what must the mass of the spring be? Sorry. The string. Be okay? Well, we're the only equation that we have so far that deals with the mass. On a string is the speed equation. Speed of waves on a string are given by the spirit of the tension divided by the mass per unit length. Okay, this is the tension divided by the mass per unit length. What we want is toe isolate that mass. Okay, so we can just rewrite This is the square root. Sorry. This squared of the speed is t over em. Over. L take em over. Ellen, multiply into the numerator. Take V squared divided into the denominator. This is M over. L is t over v squared And then lastly take l multiply it into the numerator and we have em. Is t l over b squared. So what? We need to know is V squared. Well, the relationship for speed off any wave is that it's equal to the wavelength times the frequency. We know that it's a 12 centimeter wavelength and a 70 hertz frequency. So this is gonna be a 0.12 You wanna in meters times 70 which is 84 m per second. Now we console for the mass, so the mass is gonna be t l over V squared. The tension is 10 Newtons. The length is five centimeters or m. And the speed we found was 84 and we that squared. And that comes out to 0007 kg. Or if I reduce that by a factor of 1000 that's 7 g. Either of these answers are perfectly fine. Alright, guys, that wraps up this problem. Thanks for watching
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Problem

A 2 m long string is arranged as shown in the figure below. If the mass of the string is 0.5 kg, what would the frequency be of a wave with a 2 cm wavelength?

4
example

## Speed Of A Wave Traveling Up A String

2m
Play a video:
Hey, guys, let's do a quick example. A string is hung vertically from an anchor point. If you were to grasp the string at the bottom and shake it to produce a wave traveling up the string, what happens to the speed of the wave as it moves up the string? Okay, so we have some anger point here, a rope hanging. We grab that rope and we produce a wave on it by whipping it. So this is traveling upwards. What happens to the speed of that wave as it travels up the rope? Okay. A lot of people are probably gonna want to say it decreases right off the bat, but don't associate this with freefall motion. We know the speed of a wave on a string equals attention. Divided by the mass screen at length. The mass screen length is not changing, but the tension and the rope is okay. Tension and rope is always increasing the higher up the rope you get. Okay. The simple reason is this Imagine the rope was built up, was broken up into a bunch of chains. Chain links, right. How much tension is in the chain link? Second from the bottom. Well, all that's holding all that it's supporting is one little mg, which I'll say is the weight of a single link. So the tension is just one little mg. But what if we go? One, two, three, four, five up? Then there's four chains below it. So what's the weight? It's supporting four little MGI, so it's tension is four little MGI. Let's go. We have toe to the top of this chain link. It has one, two, three, four, five, six, seven, eight, nine. So the way that it's supporting is nine little mg. So the tension is nine little mg. So clearly the tension in the rope increases the higher you go up because there's mawr rope being supported, the higher you go up. Okay. So as the tension goes up, the speed goes up, so the speed of the wave actually increases as it goes up the rope. Okay, This is contrary to our initial sort of visceral guests that it should actually decrease on its way up like objects in free fall. Alright, guys, that wraps up this question. Thanks for watching
5
concept

9m
Play a video: