91Ó°ÊÓ

The frequency of wave in an open pipe is given as$\mathbf{f}\mathbf{=}\mathbf{n}\frac{\mathbf{v}}{\mathbf{2}\mathbf{L}}$ were, were,f is the frequency of nth harmonic, v is the velocity of the wave,$\mathbf{n}\mathbf{→}\mathbf{nth}$harmonic (n — 1, 3, 5, ...), L is the length of the pipe.

The distance between two nodes equals $\frac{\mathrm{λ}}{2}$at any standing wave both ends are closed, so the molecules at this end can't move Therefore, each end is considered a node.

When the length of the pipe =$\frac{{\mathrm{λ}}_1}{2}$ we have two nodes at the ends of the pipe

and to add additional node, we add$\frac{{\mathrm{λ}}_2}{2}$

$\mathrm{L}=\frac{{\mathrm{λ}}_1}{2}=\frac{{\mathrm{λ}}_2}{2}=\frac{{\mathrm{λ}}_2}{2}=\frac{2{\mathrm{λ}}_2}{2}={\mathrm{λ}}_2$and so on…

$\mathrm{L}=\frac{{\mathrm{²Ôλ}}_{\mathrm{n}}}{2}$hence, proved

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{λ}}{\mathrm{T}}={\mathrm{λ´Ú}}_1 \\ \mathrm{f}=\frac{\mathrm{v}}{\mathrm{λ}} \\ {\mathrm{f}}_{\mathrm{n}}=\frac{\mathrm{v}}{{\mathrm{λ}}_{\mathrm{n}}}=\frac{\mathrm{v}}{2\mathrm{L}/\mathrm{n}}=\frac{\mathrm{²Ôλ}}{2\mathrm{L}} \end{gathered}$$

Hence proved.

The fundamental frequency IS n=1, n=2, n=3

Use the formula,${\mathrm{f}}_{\mathrm{n}}=\frac{\mathrm{²Ôλ}}{2\mathrm{L}}$

$$\begin{gathered} {\mathrm{f}}_1=\frac{\mathrm{²Ôλ}}{2\mathrm{L}}=\frac{1×344\mathrm{m}/\mathrm{s}}{2\left(2.5\mathrm{m}\right)}=68.8\mathrm{Hz} \\ {\mathrm{f}}_2=\frac{\mathrm{²Ôλ}}{2\mathrm{L}}=\frac{2×344\mathrm{m}/\mathrm{s}}{2\left(2.5\mathrm{m}\right)}=137.6\mathrm{Hz} \\ {\mathrm{f}}_3=\frac{\mathrm{²Ôλ}}{2\mathrm{L}}=\frac{3×344\mathrm{m}/\mathrm{s}}{2\left(2.5\mathrm{m}\right)}=206.4\mathrm{Hz} \end{gathered}$$

Therefore, the three fundamental frequencies are 68.8 Hz, 137.6Hz, 206.4Hz