When Sequoia measured the sound from the tube open at one end and closed at the other, she only observed three, equally-spaced, peaks in the Fourier transform window, and thus recorded three frequencies (150 Hz, 450 Hz, and 750 Hz). If the length of the tube was 0.55 m, what would be the speed of sound that she would calculate using the fundamental frequency she measured?

v= 330m/s

Explanation: the length (l) of an open closed pipe is related to the wavelength (λ) as shown in the formulae below

l = λ/4

where l = 0.55m, hence we have that

0.55 = λ/4

λ = 0.55*4

λ = 2.2m

from the question, it can be seen that the fundamental frequency is 150Hz.

Recall that we have v=fλ, where v = speed of wave, f= frequency, λ= wavelength

v =2.2 * 150

v = 330 m/s

whichever one it is, it's definitely not one that you included on the list you provided.

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