The fundamental frequency of a pipe that is open at both ends is ...

Question

The fundamental frequency of a pipe that is open at both ends is 524 Hz. If one end is now closed, find (b) the wavelength

Accepted Answer

Everyone in this problem, we have an open ended tube and it's producing a second harmonic frequency of 380 hertz. Okay. And it gets closed at one end. And what we want to find is the first harmonic wavelength when the tube gets closed. Alright, so let's think about what we have with this tube. When the tube is completely open. Okay, recall that we're going to have a node node situation And when the tube gets closed at one end we're gonna end up with a node anti node situation. All right now we're told the second harmonic frequency is 380 hertz. Okay, so that means that F two Is equal to Hz. Alright, so let's start with what we know, we know what's going on with the open ended tube. Okay, so when the tube is open we have a no note situation. So recall we have no note. We have the wavelength lambda end is given by two L over N. So in this case we know the frequency of the second harmonic. So let's deal with land and equals two. Sorry. So we're gonna have lambda two, it's gonna be two L over two. Which is just going to equal, Oh, all right. Now, when we're talking about frequency with a node node, we have FN the frequency is equal to V. Or relaying the N. Okay, so the frequency of the second harmonic, which is what we know is equal to V over lambda two. Alright, so filling in the information we know too, we're told is 380 hertz. Okay. V here we're talking about an instrument or a tube producing sound. Okay, so that's going to be um the speed of sound. So this is going to be 343 m per second. Okay. And then we have lambda two, which is L. So we have divided by L. Here, let me move. So we have a little bit more space to write. Okay? And this is gonna give us times. L. Is equal to 343 camera. Gonna have L. Is equal to 0.90 to six m. Okay. Alright. So we found the length of our tube. Now, let's remember what we're trying to find. We're trying to find the first harmonic wave length of the tube once we close it. Okay, so we know the length of our tube now. So let's move over to the closed tube. Okay? Or one end closed. Now this is a node anti node situation. So instead of having the wavelength given by two L over N. We have the wavelength given by four L over end. Okay. And this is what we want to find. Okay, so the wavelength we want to find the first harmonic wavelength. Okay, so lambda one. We're gonna call it lambda one new because that's what the new one. When we close the tube. Okay, This is what we're trying to find. Well, if we substitute any close one, we're just going to end up with four L. Okay. And we know the value of L. From before we found the value of bell here so we can use it To find our wavelength. So we're gonna have four Time 0. - six m. This is going to give us a value of 3.6 m. And so our first harmonic wavelength when we close the tube is going to be lambda equals 3.6 m. Answer D. Thanks everyone for watching. See you in the next video.

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