Nathanial Levine

2022-09-05

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.

goffaerothMotw1

Beginner2022-09-06Added 10 answers

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

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

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

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

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