Organ Pipe and Fundamental Frequency

AI Thread Summary
The discussion revolves around calculating the wavelength of the fundamental frequency of a 2.0 m organ pipe open at both ends, which has a fundamental frequency of 128 Hz. The wavelength is determined using the formula lambda = 2L, resulting in a wavelength of 4.0 m. Additionally, the conversation addresses how to find the blockage position when the frequency shifts to 262 Hz (middle C). Participants note that the speed of sound is approximately 330 m/s, allowing for the calculation of the new wavelength and the corresponding length of the pipe. The discussion emphasizes the relationship between frequency, wavelength, and pipe length in acoustic physics.
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Homework Statement


The organ pipe is 2.0 m long, was open at both ends, and was originally tuned to a fundamental frequency of 128 Hz (C below middle C).
a) what is the wavelength of the fundamental?
b)if the note you now hear is closer to 262 Hz (middle C), where is the blockage with respect to the opening at the bottom of the pipe?


Homework Equations


for a pipe open on both ends: lambda=2L


The Attempt at a Solution


a) lambda=2L= 4.0 m
b) I know the pipe will have a displacement antinode at each end, and a pressure node at each end. I don't understand how to find L given only the frequency 262 Hz; I looked at all my equations, but couldn't find one that seemed to work.
Help please!
 
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v=f\lambda

Using that relationship you can change the frequency in wavelength (speed\ of\ sound=330 ms^{-1} if not specified.) So from there it's should be pretty straight forward to solve for L
 
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