A pipe resonates at successive frequencies of 540 Hz, 450 Hz, and 350Hz. Is this an open or a closed pipe?
L = (nλ)/2 or L = ((2n-1)/4)λ
v = fλ
The Attempt at a Solution
The difference between the first two frequencies (540 & 450) is 90Hz, and the difference between the last two frequencies (450 & 350) is 100Hz.
I have no idea as to how to solve this equation. I tried asking my physics teacher, and we could not come up with an answer, but I am still curious about the solution.
No temperature is given, therefore I cannot determine the speed of sound and rearrange for lambda. Unless I assume that the temperature is room temperature.
I also don't understand how to determine the difference between an open & closed pipe since the equation: L = (nλ)/2 is applicable to both open-open & closed-closed pipes.