Open ended pipe Harmonics Mastering Physics Question

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SUMMARY

The discussion centers on calculating the highest harmonic frequency for a pipe closed at one end, specifically a 45.0 cm long pipe using the speed of sound at 344 m/s. The relevant formula is f_n = (2n-1) * (v / 4L), leading to the conclusion that for n = 52, the harmonic number is 103. The user initially miscalculated the harmonic number as 52, but through clarification, it was established that the correct harmonic number is indeed 103, as it is the first odd number below 20,000 Hz.

PREREQUISITES
  • Understanding of wave mechanics and harmonics
  • Familiarity with the speed of sound in air (344 m/s)
  • Knowledge of the formula for harmonics in closed pipes: f_n = (2n-1) * (v / 4L)
  • Basic algebra for rearranging equations
NEXT STEPS
  • Study the properties of harmonics in open and closed pipes
  • Learn about the speed of sound variations in different mediums
  • Explore the implications of frequency limits in sound perception
  • Investigate the mathematical derivation of harmonic frequencies
USEFUL FOR

Students studying physics, particularly those focusing on wave mechanics, acoustics, and harmonic frequencies in musical instruments or sound engineering.

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



f_n = (2n-1)\frac{v}{4L}

The Attempt at a Solution



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

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

then:

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

and finally:

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

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:

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

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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The harmonics of a pipe closed at one end are all odd. For n = 2 you have the 3rd harmonic. For n=52 what harmonic do you have?
 
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:

f_n = \frac{nv}{4L}

for n = 1, 3, 5,...
 
Using:

f_n = \frac{nv}{4L}

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

TFM

Yes n is the harmonic number.
 
Success! n = 103.

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

TFM
 
2n - 1 is just another way of saying n = 1, 3, 5, ... . So if you stick n = 52 into 2n - 1 you get 103.
 
Last edited:
That makes sense.

Thanks,

TFM
 
  • #10
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:
 

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