I have the answer to this problem, I just need to know how todo it. Thanks Here

Question

I have the answer to this problem, I just need to know how todo it. Thanks Here are the answers (a) lpipe=.46 m f overtone= 772 hz (b) f fundamental 193 Hz f overtone 579 Hz (c) n=51 (a) What length of pipe open at both ends has a fundamentalfrquency of 3.86X 102. Find the Firstovertone. (b) If one end of this pipe is now closed what is the newfundamenal frequency? Find the first overtone. (C) If the pipe is open at one end only how many harmonics arepossible in the normal hearing range from 20 to 20,000 hz? A previous question came before it, if you feel that you aremissing information it might have come from the firstquestion. Here is the info. A pipe is 2.34m long. Take 355 m/s as the speed of sound inair. I have the answer to this problem, I just need to know how todo it. Thanks Here are the answers (a) lpipe=.46 m f overtone= 772 hz (b) f fundamental 193 Hz f overtone 579 Hz (c) n=51 (a) What length of pipe open at both ends has a fundamentalfrquency of 3.86X 102. Find the Firstovertone. (b) If one end of this pipe is now closed what is the newfundamenal frequency? Find the first overtone. (C) If the pipe is open at one end only how many harmonics arepossible in the normal hearing range from 20 to 20,000 hz? A previous question came before it, if you feel that you aremissing information it might have come from the firstquestion. Here is the info. A pipe is 2.34m long. Take 355 m/s as the speed of sound inair.

Explanation / Answer

Fundamental frequency f= 386 Hz

We know f = v / 2L

From this length of the pipe L = f / 2v

Where v = speed of sound = 355 m / s

Plug the values wgete   L = 0.5436 m

First overtone frequency f ‘ =2f      = 772 Hz

Since it is a open pipe.So, it forms all harmonics.

(b). New fundamental frequency F = v / 4L

                                                       = 193 Hz

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