The pipe open at both ends and pipe closed at one end have same length and both are vibrating in fundamental…
The pipe open at both ends and pipe closed at one end have same length and both are vibrating in fundamental mode. Air column vibrating in open pipe has resonance frequency $n_1$ and air column vibrating in closed pipe has resonance frequency $\mathrm{n}_2$, then
$\mathrm{n}_1=2 \mathrm{n}_2$
$\mathrm{n}_1=\mathrm{n}_2$
$2 \mathrm{n}_1=\mathrm{n}_2$
$3 \mathrm{n}_1=4 \mathrm{n}_2$
Solution
Resonant frequency in open pipe, $\mathrm{n}_1=\frac{\mathrm{V}}{2 \mathrm{~L}}$
Resonant frequency in closed pipe, $\mathrm{n}_2=\frac{\mathrm{V}}{4 \mathrm{~L}}$
$\begin{array}{rlrl}
& \therefore & \mathrm{n}_2 & =\frac{\mathrm{V}}{2 \times 2 \mathrm{~L}} \\
& \mathrm{n}_2 & =\frac{\mathrm{n}_1}{2} \\
& \therefore & \mathrm{n}_1 & =2 \mathrm{n}_2
\end{array}$
.