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Question

Show that all harmonics are present in case of an air column vibrating in a pipe open at both ends.

Solution
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For a tube open at both ends the antinodes should be formed at the open ends as shown below: [Ref. image]

In the fundamental mode or case 1 as shown above There are one node at the center and two antinodes at the open ends.

We can see that $$\dfrac{\lambda_1}{2} = L$$, where $$L$$ is the length of pipe.
Let the velocity of the sound wave be '$$v$$'

Then frequency $$\nu_1 = \dfrac{v}{\lambda_1} = \dfrac{v}{2L}$$

For second case similarly we see
$$\lambda_2 = L$$
So, $$\nu_2 = \dfrac{v}{\lambda_2} = \dfrac{v}{L} = \dfrac{2v}{2L}$$

For third case
$$\dfrac{3\lambda_3}{2}=L$$
So, $$\nu_3 = \dfrac{v}{\lambda_3} = \dfrac{3v}{2L}$$

Similarly if we take for $$n^{th}$$ harmonic we will get
$$\nu_n = \dfrac{v}{\lambda_n} = \dfrac{nv}{2L}$$

We can see from above that all the harmonics
$$\dfrac{v}{2L} , \dfrac{2v}{2L}, \dfrac{3v}{2L}.......\dfrac{nv}{2L}$$

is present in case of air column vibrating in a pipe open at both ends and it can be explained because '$$n$$' can take any positive integer value from $$1$$ to infinity.

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