Organ Pipe and Fundamental Frequency

[yellowgators]

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!

 

[Deadleg]

$$v=f\lambda$$

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 $$L$$

 

[baba89]

To find the location of the blockage, we can use the formula for the fundamental frequency of an open pipe: f = v/2L, where v is the speed of sound and L is the length of the pipe. Since the pipe is now producing a frequency of 262 Hz, we can rearrange the equation to solve for L: L = v/2f. The speed of sound in air is approximately 343 m/s, so L = 343 m/s / (2 * 262 Hz) = 0.655 m. This means that the blockage is located 0.655 m from the bottom opening of the pipe.