For a certain organ pipe, three successive resonance frequencies are observed at 425,595 and $765…
For a certain organ pipe, three successive resonance frequencies are observed at 425,595 and $765 \mathrm{~Hz}$, respectively. The length of the pipe is (speed of sound in air $=340 \mathrm{~ms}^{-1}$ )
$0.5\ m$
$1\ m$
$1.5\ m$
$2\ m$
Solution
For closed organ pipe only odd harmonies are possible.
$\therefore$ Fundamental frequency
Fundamental frequency for a closed organ pipe
$\begin{aligned}
& v_0=\frac{v_0}{4 l} \Rightarrow 85=\frac{340}{4 \times l} \\
& {\left[\because v_0=\frac{595-425}{2}=\frac{765-595}{2}=85 \mathrm{~Hz}\right]} \\
& l=\frac{340}{4 \times 85}=1 \mathrm{~m}
\end{aligned}$