Calculating the 'Wavelength' Of the first harmonic

  • Thread starter mHo2
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  • #1
mHo2
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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!
 

Answers and Replies

  • #2
Delphi51
Homework Helper
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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.
 
  • #3
mHo2
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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!
 
  • #4
Delphi51
Homework Helper
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(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).
 
  • #5
mHo2
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Oh, so when calculating the wavelength of the first harmonic, i'm still calculating one full wavelength? simple enough!

Thanks!
 

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