Standing waves and length of tube

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SUMMARY

The discussion focuses on determining the length of a tube that supports standing waves at frequencies of 390 Hz, 520 Hz, and 650 Hz. The fundamental frequency is identified as 130 Hz, derived from the relationship between the given frequencies and their harmonic modes. The speed of sound is noted to be either 344 m/s or 340 m/s, affecting the calculated tube length. The consensus is that the tube is open at both ends, which is essential for the observed frequencies to resonate correctly.

PREREQUISITES
  • Understanding of wave mechanics and standing waves
  • Familiarity with the equation for wave frequency: f = m(v/(2L))
  • Knowledge of harmonic series and their relationship to frequency
  • Basic principles of sound propagation in air
NEXT STEPS
  • Calculate the length of a tube using the fundamental frequency and speed of sound
  • Explore the concept of harmonics in open and closed tubes
  • Investigate the effects of varying the speed of sound on wave frequency calculations
  • Learn about resonance and its applications in musical instruments
USEFUL FOR

Students studying physics, particularly those focusing on acoustics and wave phenomena, as well as educators seeking to explain the principles of standing waves in tubes.

aliaze1
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Homework Statement



A narrow column of air is found to have standing waves at frequencies of 390 Hz, 520 Hz, and 650 Hz and at no frequencies in between these. The behavior of the tube at frequencies less than 390 Hz or greater than 650 Hz is not known.

How long is the tube?


Homework Equations



f = m(v/(2L)) = mf
m = 1,2,3,4...


The Attempt at a Solution



I keep getting .4358 or .4410 (depending on if I use 344 or 340 for the speed of sound, respectivly)
 
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one-end open tube or both ends open tube?
btw, your answers seem wrong either
 
Last edited:
mjsd said:
one-end open tube or both ends open tube?
btw, your answers seem wrong either

open on both ends, basically like a pipe..
 
aliaze1 said:

Homework Equations



f = m(v/(2L)) = mf
m = 1,2,3,4...

did u work out the corresponding m's for your frequencies? indeed, different wave speed will give different L.
 
huh?

mjsd said:
did u work out the corresponding m's for your frequencies? indeed, different wave speed will give different L.

so each frequency will give a different length? so what length would be correct?
 
mjsd said:
one-end open tube or both ends open tube?
btw, your answers seem wrong either

well it said 'narrow column of air' so i assumed open on both ends
 
aliaze1 said:
so each frequency will give a different length? so what length would be correct?

no, what I meant was, you should work out the value of your m's corresponding to each frequency:
say m_{f1} = k, m_{f2}=k+1, m_{f3}=k+2 where f1, f2 and f3 are the 390, 520, 650 Hz.

and you would only get different answer if your speed of sound is different.
it appears that this can only be the open-both-ends case for it to work. (just from the wavelength to m relations)
 
Last edited:
Um, I'm not sure on this question either. Can someone guide me through it? It's also been giving me problems.
 
Hate to revive an old thread but I can't figure out this one for the life of me...I've tried entering the length for every frequency (with the accompanying mode) and none of them seem to give me the correct length.
 
  • #10
Three frequencies 390, 520, and 650 Hz can be wrightten as 3x130, 4x130 and 5x130. The same open tube can resonate in these three modes. Hence fundamental resonant frequency of the tube must be 130 Hz. Using this frequency and the velocity of sound, you can calculate the length of the tube.
 

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