An open organ pipe of length ' $l$ ' is sounded together with another open organ pipe of length…
An open organ pipe of length ' $l$ ' is sounded together with another open organ pipe of length $\left(l+l_1\right)$ in their fundamental modes. Speed of sound in air is ' $V$ '. The beat frequency heard will be ( $l_1 \ll l$ )
$\frac{\mathrm{V} l_1^2}{2 l}$
$\frac{\mathrm{V} l_1}{2 l^2}$
$\frac{\mathrm{V} l_1}{4 l^2}$
$\frac{\mathrm{V} l^2}{2 l_1}$
Solution
For a pipe open at both ends, $\mathrm{f}=\frac{\mathrm{V}}{2 l}$
$\therefore \quad \mathrm{f}_1=\frac{\mathrm{V}}{2 l} \quad ; \quad \mathrm{f}_2=\frac{\mathrm{V}}{2\left(l+l_1\right)}$
$\therefore \quad$ beat frequency, $\mathrm{f}_{\mathrm{b}}=\mathrm{f}_1-\mathrm{f}_2=\frac{\mathrm{V}}{2 l}-\frac{\mathrm{V}}{2\left(l+l_1\right)}$
$\therefore \quad \mathrm{f}_{\mathrm{b}}=\mathrm{V}\left[\frac{2\left(l+l_1\right)-2 l}{4 l\left(l+l_1\right)}\right]=\mathrm{V} \frac{2 l_1}{4 l\left(l+l_1\right)}$
$\therefore \quad \mathrm{f}_{\mathrm{b}}=\frac{\mathrm{V} l_1}{2 l\left(l+l_1\right)}$
Since $l_1 \ll l, l_1$ in the denominator can be taken as zero.
$\therefore \quad \mathrm{f}_{\mathrm{b}}=\frac{\mathrm{V} l_1}{2 l^2}$