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(a) Draw a diagram for a pipe that is closed on both ends and have 6 loops of standing ` waves and derive an equation for nth frequency of vibration in terms of spe...

Question

(a) Draw a diagram for a pipe that is closed on both ends and have 6 loops of standing ` waves and derive an equation for nth frequency of vibration in terms of speed of sound and length of the pipe: (6) If the length of the pipe is 120 cm; what is fundamental frequency and first overtone produced by the pipe if speed of sound in air is 343 m/s.

(a) Draw a diagram for a pipe that is closed on both ends and have 6 loops of standing ` waves and derive an equation for nth frequency of vibration in terms of speed of sound and length of the pipe: (6) If the length of the pipe is 120 cm; what is fundamental frequency and first overtone produced by the pipe if speed of sound in air is 343 m/s.



Answers

Find the fundamental frequency and the frequency of the first three overtones of a pipe 45.0 $\mathrm{cm}$ long (a) if the pipe is open at both ends and (b) if the pipe is closed at one end. Use
$v=344 \mathrm{m} / \mathrm{s} .$ (c) For each of these cases, what is the number of the highest harmonic that may be heard by a person who can hear frequencies from 20 $\mathrm{Hz}$ to $20,000 \mathrm{Hz}$ ?

So here we know the length equals 45 centimeters, or we can say the length is equal to a 450.45 meters now for part A. They want the fundamental frequency such that the pipe is open at both ends, So this will be the velocity divided by two l hear it the velocities that's going to be the speed of sound with an air so 344 meters per second, divided by two times point for five. And this is giving us 382 hertz. Remember that this is for a pipe that is open at both ends. The the 1st 3 overtones will be the second harmonic. This will simply be two times 3 82 so 7 64 hurts. The third harmonic will be three times the fundamental frequency, or 1,146 hertz, and then the fourth harmonic or the third overtone will be equal to four times the fundamental frequency. So 1,528 hurts now for part B. They want the fundamental frequency such that it is closed at one end. So if it's closed at one end That means that the freak that the wavelength is equal to four times the length so it will be the fundamental frequency of the pipe that has closed at one end will be equal to the speed of sound, with an air divided by four times the length or the wavelength. And this will equal 344 divided by four times 40.45 And we're getting 191 hurts. Note that this is exactly half of this of 3 82 And so we can say that the frequency of the first overtone the pipe when the pipe is closed at one end and is not goingto end can only take on odd and treasures not even and odd. So the first overtone is going to be F sub three instead of F sub, too, and this is simply going to be three times the fundamental frequency. So 573 hurts, and then f sub five the second, the second overtone will be equal to five times 191. So 955 hertz and then the fourth overtone well, second through first rather than third overtone will be f sub seven and this will be seven times 191. So 1,337 hertz. So again, 1st 2nd and third overtone now for part. See, they're asking us if if someone can here a frequency of 20,000 hertz. What harmonic is that? So for the open pipe, the harmonic is going to be equal to the frequency of that harmonic divided by the fundamental frequency. So this will be 20,000 hertz divided by 382 hertz, and this is equal in 52. So the highest harmonic that can be heard is the 52nd harmonic and then here closed that one end. This means that and that can only equal f sub and divided by f f the fundamental frequency. However, here it's going to be 20,000 divided by 1 91 and this will be 104 now notice if it's closed at one end and can only take odd integers. So this is an even into jher. Therefore, on equals one less than this. So it becomes alright, Esso and equals 103. So that would be the highest that the Ah, listen, er can hear if the pipe is closed at one end. This would be the highest the listener can hear. If the pipe is open at both ends, that is the end of the solution. Thank you for watching.

Western. We have given the pipe is open board games. So basically, this is something would you look like in this scenario? You already know that You kind of some sort off the thing is that over there? Uh hmm. What will be good? Yeah. So this is has sounds like this one. Okay, basically over there, I think Lambda two equals to the land of r two equals toe the length off the five millions language request tutorial. Andi, over there. We have given some quantity over there about the question, which is you. First of all, they have given velocity of the sound. So basically, in the person aereo velocity of the sandwiches 2 33 31 m once again on board frequency, Fundamental frequency 3. 20 hearts. And so when you find the length so basically, we know that the fundamental frequency over there will be nothing but f d upon to l. So basically toe l so l over there. That will be just the upon to f. So be over the 3. 31 upon, uh, to have two times off, which each two times of 3. 20 which is your 200.52 m, little 0.52 m. And in the next scenario, we have to find the In the next scenario, we have to find the other frequency as well. So basically, we will find us again. Harmonic. So we just multiply by two. So two times off, 0.52 which will be nothing. But what will be that? A 0.5, not first harmonic, which is 3 20. So 3. 20. So was willing 6. 40. How did you And in the 6. 40 and 6 40 F three. The rule of three times off 3. 20 which will be nine 60 heart knowing where the next part of this question See in this question? Yeah. We have to find the frequency Wyndham velocity off. The sound changes to 33 7 m per second. So and we know that frequency off the sound is nothing but al frequency or the velocity is nothing but a lender. Times of the frequency frequency over there that will be received upon number 337 upon prevalence evidence. And you'll find five to our velocity is nothing but a million times of frequency. Yeah, right. So if you can see where there will be that and that is anything. But when we do, we need todo. We just need to calculate the bylaws over there, which is a utility seven. 33 7 on developed by 0.52 which will be just comes out of the 6. 48 point you know, seven our yoga. So this will be the frequency and the window. That is when the spirit different and this is the fundament.

So a discussion. I want to find the difference between length of pipes A and pipe be or l A minus L b since in this question, tubes have only one and open. So the fundamental frequency is given by one equal two the over four times l So from here can say l is vey over four times of one. Um so l a would be equal to Ah, the over four F one of A and L B is, um V over four times that one off. I have the I have one of a but we don't have one of beets to find. I wanna be I'm going to look at the beat frequency beat frequency is 12 hurts. So, uh, should we say if one ah b minus F one a is 12 hurts or if one a minus one B is 12 hertz. Which one? So to find that out Ah, you're going to look at again if one is a cultural V over four em. So here we can see that f and l are inversely proportional. So if f and so if l is larger ah, if should be smaller. So here we know that l A is bigger than it will be. Therefore, we can say if one of a is smaller than if one oh, feet. Okay, So if if one b is larger than if one, eh? So we can say f one B minus if one A is equal to 12 hurts. So you know, if one A is given as 256 6 So if one B is a culture 12 plus 256 and this is a call to 260 eight hertz Okay, so now I have one. Be so I can say from here l A minus. L b is equal to 343 meter per second, divided by four times F one A, which is 200 56 hertz minus. It'll be 340 three meter per second, divided by four times, 268 hertz. And this is equal to 0.0 15 meters

So for party we have an open open pipe or an open pipe by the pipe. Open at both ends. And so with this we can we know that. Then the displacement of the node for the fundamental frequency is equal to half of the length of the pipe. Given that the anti nodes are at the ends of the pipe. So X would be equaling two L over two. This would be 1.20 divided by 2.60 m. Now for the first overtone We have that X is equal and then all over four. three Yell over four And then acts given 1.20 m divided by 4.30 meters. three times 1.2 3.60 divided by four 90 m. Again, this has given that L. Is equaling 1.20 m. So given that this would be the displacement for of the note for the first overtone. Now, for the second overtone there are three nodes. There are three notes where X is then equaling out over six comma three L over six. Well L over to essentially And five L over six. And so solving this would be 1.2 x six, So .20 m Half, so .60 m on five Times 1.20 m six m, about by 6, 1 m Or 1.00 m for part B. We have a we know that a closed end is always a node. So here if a closed it's a closed and is always a node. We know that. Then the displacement for a note or displacement of the note for the fundamental frequency must be zero. So X would be equal to 0m. And for the first overtone we can say that the node displacement of the node X would be equal to zero and two L over three, So X. is equaling zero m And then two times 1.20, x 3.80 meters. And then finally for the second overtone we have the X. Is going to be equal to zero 2/5 And four L over five. Again, L is equaling 1.20. So we have zero m point For eight m and we have .96 m. And these would be again the displacement of the nodes for the second overtone. That is the end of the solution. Thank you for watching.


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