Calculating the 'Wavelength' Of the first harmonic

In summary, the open ended tube has a resonant frequency of 261.6Hz and the speed of sound is 343m/s. The length of the tube is calculated to be .5λ, 1λ, 1.5λ, 2λ and so on.
  • #1
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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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  • #2
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
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
(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
Oh, so when calculating the wavelength of the first harmonic, I'm still calculating one full wavelength? simple enough!

Thanks!
 

1. What is the first harmonic?

The first harmonic is the fundamental frequency of a standing wave pattern. It is the lowest frequency at which a standing wave can occur in a given system.

2. How do you calculate the wavelength of the first harmonic?

To calculate the wavelength of the first harmonic, you divide the speed of the wave by the frequency. The speed of the wave is typically given in meters per second, and the frequency is given in hertz (Hz).

3. What is the relationship between wavelength and frequency in the first harmonic?

In the first harmonic, the wavelength and frequency are inversely proportional. This means that as the frequency increases, the wavelength decreases, and vice versa.

4. Can the first harmonic exist in any system?

No, the first harmonic can only exist in systems that have boundaries or fixed ends, such as a string or air column. These boundaries create the standing wave pattern necessary for the first harmonic to occur.

5. How is the first harmonic different from higher harmonics?

The first harmonic is the lowest frequency and has the longest wavelength compared to the higher harmonics. Higher harmonics are integer multiples of the first harmonic and have shorter wavelengths and higher frequencies.

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