The standing waves in air in a pipe of length L that is open at both ends have a speed v.The frequencies of the three lowest harmonics are
A)v/L,2v/L,and 3v/L
B)v/2L,v/L,and 3v/2L
C)$\lambda$/2,$\lambda$,and 3$\lambda$/2
D)L/v,2L/v,and 3L/v
E)$\lambda$/3,2$\lambda$/3,and 3$\lambda$/3

Question

Quizplus · Accepted Answer

For a pipe open at both ends, the possible standing waves are those for which both ends are nodes. This means that the wavelength of the wave must be equal to twice the length of the pipe (since one half wavelength fits between each end of the pipe). 
So, we have λ = 2L for all harmonics.
The speed of sound in air is v = fλ, where f is the frequency of the wave.
For the lowest harmonic (the fundamental), we have v = f1λ = f1(2L). Solving for f1, we get f1 = v/(2L).
For the second harmonic, we have one full wavelength fitting into the length of the pipe, so we have λ = L. Thus, f2 = v/λ = v/L.
For the third harmonic, we have three half wavelengths fitting into the length of the pipe, so we have λ = 2L/3. Thus, f3 = v/λ = 3v/2L.
Therefore, the frequencies of the three lowest harmonics are v/2L, v/L, and 3v/2L, which corresponds to choice B.

The Standing Waves in Air in a Pipe of Length λ\lambdaλ

The Standing Waves in Air in a Pipe of Length $\lambda$