# 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.

Mar 15, 2018

For an open ended tube, both the ends represent antinodes,so distance between two antinodes =$\frac{\lambda}{2}$ (where,$\lambda$ is the wavelength)

So,we can say $l = \frac{2 \lambda}{2}$ for $2$ nd harmonic,where $l$ is the length of the tube.

So,$\lambda = l$

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

Given, $v = 340 m {s}^{-} 1 , l = 4.8 m$

So,$\nu = \frac{v}{\lambda} = \frac{340}{4.8} = 70.82 H z$