(N/A) In an open pipe,antinodes are produced at both ends. The length $L$ of the pipe for the $n^{\text{th}}$ harmonic is given by $L = \frac{n \lambda_n}{2}$,where $n = 1, 2, 3, \ldots$ and $\lambda_n$ is the wavelength.
Therefore,the wavelength is $\lambda_n = \frac{2L}{n}$.
Using the relation $v = \nu_n \lambda_n$,where $v$ is the speed of sound,the frequency $\nu_n$ is:
$\nu_n = \frac{v}{\lambda_n} = \frac{n v}{2L} = n \left( \frac{v}{2L} \right) = n \nu_1$,where $\nu_1 = \frac{v}{2L}$ is the fundamental frequency.
For $n=1$,$\nu_1 = \frac{v}{2L}$ (First harmonic or fundamental frequency).
For $n=2$,$\nu_2 = 2 \left( \frac{v}{2L} \right) = 2 \nu_1$ (Second harmonic).
For $n=3$,$\nu_3 = 3 \left( \frac{v}{2L} \right) = 3 \nu_1$ (Third harmonic).
Since $n$ can take any integer value $(1, 2, 3, \ldots)$,all harmonics are possible in an open pipe.