A piece of pipe, 2.0 m long and open at both ends, is subjected to sound waves of all audible frequencies (white noise). What is the fundamental frequency at which the pipe will resonate. Assume that the speed of the sound is 343 m/s.
What are the next two audible resonant frequencies of this same pipe?
Since the pipe is open at both ends it means there will be maximum of oscillation at both ends. This is equivalent to having inside the tube for the fundamental frequency only $\lambda/2$
($\lambda =$ wavelength) of the entire wave.
$\lambda/2= 2$ and thus $\lambda = 4$
but $\lambda = speed*time =speed/Frequency$
$Frequency = speed/lambda = 343/4 = 85.75 Hz$
The next resonant frequency will be when into the tube one will have $\lambda$ (there are 3 maximum inside the tube).
$\lambda = 2$
$F = speed/\lambda = 343/2 =171.5 Hz$
The third harmonic will be when into the tube one will has 4 maximum
($\lambda +\lambda/2 =3*\lambda/2)$
$3*\lambda/2 = 2$
$\lambda = 4/3$
$F = speed/\lambda = 343*3/4 =257.25 Hz$