If ' $l$ ' is the length of pipe, ' $r$ ' is the internal radius of the pipe and ' $v$ ' is the velocity of…
If ' $l$ ' is the length of pipe, ' $r$ ' is the internal radius of the pipe and ' $v$ ' is the velocity of sound in air then fundamental frequency of open pipe is
$\frac{\mathrm{V}}{2(l+1 \cdot 2 \mathrm{r})}$
$\frac{\mathrm{V}}{(l+1 \cdot 2 \mathrm{r})}$
$\frac{\mathrm{V}}{(l+0 \cdot 3 \mathrm{r})}$
$\frac{\mathrm{V}}{(l+0 \cdot 6 \mathrm{r})}$
Solution
For an open organ pipe, the length of the pipe with end correction is given as:
$\begin{aligned}
& \mathrm{L}=l+2 \mathrm{e}=l+2 \times 0.6 \mathrm{r} \ldots(\because \mathrm{e}=0.6 \mathrm{r} \text { for open pipe }) \\
& \mathrm{L}=l+1.2 \mathrm{r}
\end{aligned}$
$\therefore \quad$ The fundamental frequency of open pipe is:
$\mathrm{f}=\frac{\mathrm{v}}{2 \mathrm{~L}}=\frac{\mathrm{v}}{2(l+1.2 \mathrm{r})}$
~