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. (c) the frequency of the new fundamental.

Accepted Answer

Hey everyone in this problem we have a musical note and we're told that it has a fundamental frequency of 300 hertz. Okay. And it's played by blowing air into an open ended bamboo instrument. Okay, what we're asked to do is to find the new fundamental frequency if one end is closed. Okay, So you have this instrument, you're going to close one end with your hand. What's the new frequency going to be? Alright. So what we wanna do is think about the instrument. So when the instrument is open, okay, it's open on both ends, recall that this is going to be a node node situation. Okay. And when we close one end, it's going to become a node anti node scenario. Okay. All right. But we also know we know F1K. We're told the fundamental frequency. So that's F one. The frequency of the first harmonic is 300 Hz. Alright, so let's consider when the instrument is open. Okay, now recall we have a no note situation. So we have an equation for the wavelength lambda end given by two L. Over N. Okay, in this case we're talking about fundamental frequency. So the frequency or the end value is one. So we have lambda. One is equal to two L. Alright, similarly to wavelength, we have an equation for the frequency F. N. Which is given by V over land N. Again, F is one. So we have F one is equal to V over two L. Now we're talking about speed here. The speed of sound. Sorry, when we're talking about V we're talking about the speed and we're having a musical instrument. So we're talking about the speed of sound. And so we could plug in the value V. Of the speed of sound. 343 m per second and work like that. We're not going to do that because in this problem we don't need to and I'll show you why. So we have 300 is equal to V over two L. Okay, so let's stop there before we find or before we use the value of V. Now let's think about when we go to the closed and instrument. Okay, now when we're closed and we have node anti note. So our equation for the wavelength and the frequency are a bit different. Okay, so when we're talking about the wavelength, we have four L. Over N. Okay, so instead of two L. Over and we have four and L over in and we want to find the fundamental frequency. Okay, so we're still talking about And this one. Okay, so we have λ. one new. It's going to be four. Just substituting the value of n equals one. We get this lambda value and similarly to the open ended case we have the frequency in this case again, V over lambda. N. Okay, N is one. So the frequency one new is going to be V over four L. Well, thinking about this V over four. Oh we can just write this as one half V over two L. Okay so we pull a factor of one half out. We have one half V over two L. Well, we know that V over two L. Is this the over two l. Is just equal to 300. So this value here can be replaced by 300. So this is going to be 1/2. Let me just move a little bit. So we have some more space. Right? 1/2 times 300. Okay, so this is gonna be 100 50 hertz. Alright so we could have plugged in the value of V. The speed of sound. Found the length L. And then use that in our calculation. But in this case we didn't need to. Okay. And we find our new fundamental frequency when we close the end of our instrument as 150 hertz. Okay. That's going to correspond with answer. C. Thanks everyone for watching. See you in the next video.

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