Calculating the 'Wavelength' Of the first harmonic

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Homework Help Overview

The discussion revolves around calculating the length of an open-ended tube based on its resonant frequency and the speed of sound. The original poster is trying to understand the relationship between wavelength and tube length in the context of harmonics.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the wavelength using the formula v = fλ and expresses confusion about why the calculated wavelength does not match the expected tube length. Other participants discuss the nature of resonant frequencies in open tubes and the implications of harmonics on wavelength and tube length.

Discussion Status

Participants are exploring the relationship between wavelength and tube length, particularly in the context of the first harmonic. Some guidance has been offered regarding the concept of resonant frequencies and the significance of the term "lowest" in the problem statement. There is an ongoing clarification about the distinction between full and half wavelengths.

Contextual Notes

The original poster initially omitted the term "lowest" in their problem statement, which led to confusion regarding the calculation of the wavelength associated with the first harmonic. The discussion highlights the need to consider multiple wavelengths that can fit within the tube.

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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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