1

concept

## Standing Waves

8m

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Hey, guys. In this video, we're gonna talk about standing waves, which are types of waves that occur due to interference between overlapping waves. All right, let's get to it. Remember that waves along a string fixed their bounty at a boundary are gonna be reflected backwards as inverted waves. Okay, so these waves right here were a bunch of waves that had been produced by this hand sent forward and have already reflected their inverted. And remember that these inverted waves will have the same frequency as the incident waves as the initial waves that hit the boundary. Okay, now imagine holding the end of a fixed string. Oh, sorry. The end of the string fixed will wall and vibrating it right. We're going to go up and down. We're just gonna shake it, okay? We're going to produce a bunch of pulses that are going to travel from the free end towards the fixed end, and they're going to reflect backwards. Okay? These inverted reflections, they're gonna have the same frequency as they reflect off, and then they're gonna encounter new pulses that are being produced. Okay, forward pulses and the reflections since they're gonna be in the same space at the same time are going to interfere. Okay, Now, normally, what's going to be produced is some sort of uncoordinated vibration. Okay, Essentially nothing. Just a string that's doing a whole bunch of this. But it's specific frequencies. You are going to produce what are known as standing waves. Okay. I drew one instance of a standing wave here. In another instance of a standing wave here, the first standing wave that I drew is a wave that essentially looks like this. Okay. And then if you go faster, one part is gonna go down as the other is going to go up like this. That's the second one that I drew. Okay. Thes air called standing waves. Because they appear to be standing still. Instead of individual wave pulses that are moving forward, we now have just a pulse that's going up and down or going up and down, alternating like this. I don't have three hands, so I can't really do another standing wave which would be formed of three pulses. Okay. Two types of standing waves exist. Node node, standing waves and node anti node standing waves and different scenarios produce either node node or nodes. Anti node. What a node is is It's a point of no displacement. Okay, I highlight. I bolted and underscored this right there. So that will help you remember what a node is, Node. No displacement. Okay, An anti node is the opposite of a node is just a point of maximum displacement. So if we look at the standing waves drawn, this is a node. And this is a note. This is what we would call a node node standing wave because both ends of the string are notes, and then this point is an anti node. Okay, here we have three notes. A node, a node, another node. This is still a node node standing wave because both ends of the stringer at nodes and we have to anti notes. Okay, Now, let's look at the differences between node node and node. Anti notes, standing ways. Okay for node node, standing waves. As I said, Both ends or notes for node anti notes, standing waves, one end is a node. The other is an anti note. Okay, so I drew pictures off the three largest wavelength or lowest frequency of each. If you see here on either end, the wave returns back to the horizontal axis. It returns back to a node. Here on one end is a node, but on the other end, it's always gonna be at the amplitude. You can see here. This is the amplitude. It's the point of maximum displacement. Okay, Because of that, wavelengths and frequencies have to abide by equations. Okay, for node node standing waves. Theologian wavelengths are given by this equation. Where n is any integer, Okay. And the allowed frequencies are given by this equation. Where in is any integer? Okay, as long as your wavelength or your frequency agrees with those equations, it's a standing wave. If you have a frequency, for instance, if you whip your hand at a frequency that does not follow this equation, you will not produce a standing way. If you're just going to produce a bunch of uncoordinated vibrations, that don't mean anything. All right, I'm gonna minimize myself for node anti notes, standing waves. The allowed wavelengths obey this equation, but in has to be odd. This is crazy important to remember that in which is called the harmonic number we call in the harmonic number is any integer for node node standing waves. But it has to be odd. You can only have odd harmonic numbers for node anti node standing waves. Okay. And the allowed frequencies for note anti notes. Standing waves are given by this equation Where once again, the harmonic number has to be odd. Okay, that's crazy important that you remember that the harmonic number must be odd. All right, let's do a quick example to finish this off. You want to produce standing waves on the string shown in the figure below By vibrating the end, you're grasping. What is the third largest frequency? Produce a ble on the string. If it has a massive 0.5 kg and attention of Newtons, assume there's no friction between the ring and the end of the string. Sorry. The ring at the end of the string and the pole. Okay. What kind of standing waves do you think these are gonna be, guys? Your hand is gonna be a node. But this end, which is free to move, is gonna be an anti node. Okay, so this is a node. Anti node. If we want to alert the third largest frequency. We have to use this equation. What's the third largest harmonic number for a note? Anti node, Remember, In has to be odd. So it's 13 etcetera. The third largest is five. Okay, It's not three because two doesn't count. The third largest is five. Okay, So the fifth harmonic, which is the third largest harmonic in this case, is five V over four l and all we have to do is figure out what V is the speed of this wave on this string. We know that the string has a mass of 0.5 kg and attention of 150 Newtons and we know it's length is 15 centimeters. So the speed is gonna be the spirit of tea over meal which is the square root of 150 Newtons. The tension divided by 05 the mass over the length 15 right. And that's going to be 21 to meters per second. Now that we know the speed we can solve for the frequency, this is just gonna be five 21. over four. The length is 0.15 and so that is 177 hurts. Alright, guys, that wraps up this introduction into standing waves. Thanks for watching

2

Problem

In the following figure, what is the harmonic number of the standing wave? The wavelength of the standing wave? If the frequency of the standing wave is 30 Hz, what is the speed of the waves producing the standing wave?

A

n=5, λ=0.1 m, f=3 m/s

B

n=5, λ=10 m, f=300 m/s

C

n=6, λ=0.083 m, f= 2.5 m/s

D

n=6, λ=8.3 m, f= 250 m/s

3

example

## Unknown Harmonic Frequency

4m

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Hey, guys, let's do an example. A string fixed at both ends produces a harmonic frequency of hertz, with the next highest harmonic frequency being 780 hertz. If the same, the same string can produce a harmonic frequency of hertz. What harmonic frequency is the next highest. Okay, so we have two different pairs of harmonic frequencies. Initially were saying some harmonic frequency, which I will call F n, which is in V over to El right. It's fixed that both ends. So this is node node, right? So I'm using envy over too well, which is our note note equation. This is 650 hertz. Now I'm saying the next highest, which is in plus one, right? If in was 650 then the next highest is in plus one that's going to be in plus one the over to L, which is 780 hertz. But notice something I Kenbrell up that in plus one into two separate fractions. I can say that in the over to l plus the over to l equals 780 hertz. What's key here? is that this term right here? V over to El. That is how much the frequency grows every time you go up a new harmonic number. So if I'm at 17 and I go up to 18 it's gonna grow by V over to El. So V over 12 is exactly what I need to know to solve the problem. And luckily, I know what envy over to l is right. It's just 650 hertz. So this tells me Hertz plus V over to L, which is our step in the frequency. It's how much the frequency increases by when we increase our harmonic number by one. That's 780 hertz. So V over to l is 7. 80 minus 6 50 which is 130 hertz. Okay, so this is sort of the linchpin to solving that problem. You need to know the step. We don't know anything about the length of the string, the tension, the mass in the string, the speed. We don't know any of that. So we can't actually solve for V over to el. But we found it by comparing one arbitrary harmonic frequency to the next highest. Now we're at some other harmonic frequency. Some other unknown. It really doesn't matter. We'll call it M. And we're saying f of M, which is M V over to l is 390 hertz. Now, the next one above that is in plus one. So this is F M plus one, which is m plus one V over to L. And once again, we can split up this fraction M v over to l plus V over to l. So you see there we've exposed this. Is that increase in frequency whenever you increase the harmonic number by one. This was our original harmonic frequency at M, which was 390 hertz. So this becomes 390 hertz plus that step frequency that we found, plus 130 hertz, which is 520 hurts. And you see, we didn't know anything about this problem. We didn't know how long the string was. We didn't know the speed of the string. We didn't even know what these harmonic numbers were. But all of that is irrelevant because there was enough information to solve for that step that increase in harmonic frequency every time you increase the harmonic number by one. All right, guys, Thanks for watching

4

example

## Standing Wave On A Guitar

3m

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Hey, guys, let's do an example. Ah, guitars played by plucking the string, which emits a sound at the frequency it vibrates to play different tunes. The string is held down by your finger at what are called fret locations. Each subsequent fret, shortening the length of the string by about two centimeters. If a guitar string has a full length of 64 centimeters and attention of 70 Newton's in a mass sprint length of one times 10 to the negative kg per meter. What is the frequency emitted by the guitar if held at the 10th Fret location? Okay, so remember, the string starts out at 64 centimeters, and each fret shortens the length of the string by about two centimeters. So if you're at the 10th fret, that means that you're at minus 20 centimeters for the length. So the new effective length, which all called El Prime, is 64 centimeters minus centimeters, which is 44 centimeters. So you're actually playing a string that's Onley 44 centimeters long because you are holding one end of the string down to shorten the total length of the string. Now the string is fixed at both ends. So this is a node node standing wave right. Which means that the frequency is going to be in times V over two times our new length, not the 64 centimeters, because when you're holding out of the 10 Threat, you've effectively shorten the length of the guitar. The question is, what mode are we in? What harmonic number? Well, if the guitar string is fixed at both ends and you pluck it, what happens is it vibrates like this up and down. Okay, this is in equals one. Alright. And that's all we need. So the frequency it's gonna be v over two times the new length. Alright, that speed, we don't quite know, but we know enough to solve for it, right? The speed is the square root of the tension which is 70 divided by the mass per unit length which is one times 10 to the negative three, which is Sorry, 265 meters per second. Couldn't read my notes for a second 265 m per second. So this is going to be 265 m per second, divided by two times our new length, which is 44 centimeters or 44 which is going to be about 301 hurts. That's gonna be about the frequency emitted by the guitar. Alright, guys, that wraps up this problem. Thanks for watching.

5

Problem

An unknown mass hangs on the end of a 2 m rope anchored to the ceiling when a strong wind causes the rope to vibrate and hum at its fundamental frequency of 100 Hz. If the rope has a mass of 0.15 kg, what is the unknown mass?

A

12.24 kg

B

1224 kg

C

24.49 kg

D

2449 kg

E

48.98 kg

F

4898 kg

Additional resources for Standing Waves

PRACTICE PROBLEMS AND ACTIVITIES (10)

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- The wave function of a standing wave is y(x, t) = 4.44 mm sin[(32.5 rad/m)x] sin[(754 rad/s)t]. For the two tr...
- The wave function of a standing wave is y(x, t) = 4.44 mm sin[(32.5 rad/m)x] sin[(754 rad/s)t]. For the two tr...
- The wave function of a standing wave is y(x, t) = 4.44 mm sin[(32.5 rad/m)x] sin[(754 rad/s)t]. For the two tr...
- A piano tuner stretches a steel piano wire with a tension of 800 N. The steel wire is 0.400 m long and has a m...
- A wire with mass 40.0 g is stretched so that its ends are tied down at points 80.0 cm apart. The wire vibrates...
- A 1.50-m-long rope is stretched between two supports with a tension that makes the speed of transverse waves 6...
- A 1.50-m-long rope is stretched between two supports with a tension that makes the speed of transverse waves 6...
- A 1.50-m-long rope is stretched between two supports with a tension that makes the speed of transverse waves 6...