Calculating the 'Wavelength' Of the first harmonic

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SUMMARY

The discussion centers on calculating the length of an open-ended tube based on its resonant frequency of 261.6 Hz and the speed of sound at 343 m/s. The correct length of the tube is determined to be 0.65 m, which corresponds to half the wavelength of the first harmonic. The confusion arises from the distinction between the wavelength and the tube length, as the tube can resonate at multiples of half wavelengths. The fundamental frequency formula f = nv/(2L) is crucial for understanding the relationship between frequency, wavelength, and tube length.

PREREQUISITES
  • Understanding of wave equations, specifically v = fλ
  • Knowledge of harmonic frequencies in open tubes
  • Familiarity with the concept of resonant frequencies
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the relationship between frequency and wavelength in acoustics
  • Learn about harmonic series in open and closed tubes
  • Explore the implications of standing waves in different mediums
  • Investigate the derivation and applications of the formula f = nv/(2L)
USEFUL FOR

Students studying physics, particularly those focusing on wave mechanics and acoustics, as well as educators teaching concepts related to sound and resonance in tubes.

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Homework Statement


Hi, i have a problem that I'm trying to figure out:
An open ended tube has a resonant frequency of 261.6Hz, and the speed of sound is 343m/s. What is the tubes' length?

Homework Equations


v=(f)(λ)
v=(f)(λ/2)

The Attempt at a Solution


calculating with the equation v=(f)(λ)
v/f=λ
(343m/s)/(261.6hz)=λ
λ=1.3m

But the answer is half of that, (0.65m), why is this? Doesn't calculating the wavelength from the resonant frequency calculate the 'Half' Wavelength? equation: v=(f)(λ/2)

Thanks for all of your help!
 
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Your wavelength calc is correct. But the wavelength is not necessarily the length of the tube. The thing is the pressure wave in an open tube must have zero pressure at each end so it will resonate (just fit in the tube) when half a wavelength, a full wavelength, 1.5 wavelengths, etc. are equal to the length of the tube. See the pictures of the wave here:
http://en.wikipedia.org/wiki/Acoustic_resonance#Cylinders

The correct answer is a multiple one. The length of the tube could be
.5λ, 1λ, 1.5λ, 2λ and so on. There are an infinite number of possible correct answers.

In the link, note the shortcut formula f = nv/(2L) where n is any positive integer. If the question said the LOWEST resonant frequency was 261.6 Hz, you could set n =1, solve for L and get the .65 m answer. The word LOWEST has been left out of the question, ruining it.
 
Sorry, I had forgotten to add the 'lowest' part (first harmonic) i was writing this down from my memory.

The problem is, by calculating the wavelength in that problem, i thought i was calculating the wavelength of the first harmonic (so only half of a wave), because when i am calculating the actual frequency of the harmonic(when the wavelength&speed are known), i would use:
(v)/(.5λ)=f0
(notice the .5(λ).)

So i thought by reversing the problem, the half wavelength would be what i would be finding out, not a full wavelength. I'm not sure where I'm getting confused.

Thanks for the help so far!
 
(v)/(.5λ)=f0 is not correct. The wave equation is v =f*λ so f = v/λ.
For the fundamental, half a wavelength equals the length L = .5λ
Putting the two together gives f = v/(2L).
 
Oh, so when calculating the wavelength of the first harmonic, I'm still calculating one full wavelength? simple enough!

Thanks!
 

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