Question

What is the frequency of the second harmonic sound wave in an open-ended tube that is 4.8 m long? The speed of sound in air is 340 m/s.

Answer 1

For an open ended tube, both the ends represent antinodes,so distance between two antinodes =`lambda/2` (where,`lambda` is the wavelength)

So,we can say `l=(2lambda)/2` for `2` nd harmonic,where `l` is the length of the tube.

So,`lambda=l`

Now,we know, `v=nulambda` where, `v` is the velocity of a wave,`nu` is the frequency and `lambda` is the wavelength.

Given, `v=340ms^-1,l=4.8m`

So,`nu=v/lambda=340/4.8=70.82 Hz`