Open ended pipe Harmonics Mastering Physics Question

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TFM
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[SOLVED] Open ended pipe Harmonics Mastering Physics Question

Homework Statement



Consider a pipe 45.0cm long if the pipe is open at both ends. Use v = 344m/s.
Now pipe is closed at one end.

What is the number of the highest harmonic that may be heard by a person who can hear frequencies from 20 Hz to 20000 Hz?

Homework Equations



[tex]f_n = (2n-1)\frac{v}{4L}[/tex]

The Attempt at a Solution



I have an answer that works, but masteringphysics doesn't accept. I first rearranged the equation to give me:

[tex](2n-1) = \frac{f_n * 4L}{v}[/tex]

then:

[tex]2n = (\frac{f_n * 4L}{v})+1[/tex]

and finally:

[tex]n = ((\frac{f_n * 4L}{v})+1)/2[/tex]

inserting the values gives 52.5 so I inserted 52 as the answer. wrong, I have tried 51-54, all wrong. so I thought tpo go backwards, using:

[tex](2n-1) = \frac{f_n * 4L}{v}[/tex]

and inserting values, to find the value which is the closest to 20000, buit under it - guess what, the value that came out:

52!

Any ideas

TFM
 
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It will be the 53rd Harmonic. The trouble is, I have put 53 in, and it says its the wrong answer!
 
TFM said:
It will be the 53rd Harmonic. The trouble is, I have put 53 in, and it says its the wrong answer!

Sorry that third harmonic was a bad example. The harmonics are given by 2n-1. So if n is 52 what is the harmonic. An easier way to have thought about it would to have solved for:

[tex]f_n = \frac{nv}{4L}[/tex]

for n = 1, 3, 5,...
 
Using:

[tex]f_n = \frac{nv}{4L}[/tex]

and using n = 103,

I get a frequency of 19684, which is the first odd number below 20000. would this be the harmonic number?

TFM
 
Success! n = 103.

IOne thing does bother me slightly - where does my orginal answer of 52 fit in?

TFM
 
That makes sense.

Thanks,

TFM
 
TFM said:
That makes sense.

Thanks,

TFM

What I was originally aiming at was for you to put the n = 52 into that equation and get 103 but I used a stupid example which probably mislead you slightly. :smile: