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TFM
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[SOLVED] Open ended pipe Harmonics Mastering Physics Question
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?
[tex] f_n = (2n-1)\frac{v}{4L} [/tex]
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
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