The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe of length…
The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe of length ' $L$ '. The length of the open pipe will be
$\frac{L}{2}$
$\mathrm{L}$
$\mathrm{2L}$
$\mathrm{4L}$
Solution
The length of closed pipe is denoted using $\mathrm{L}$. Let $l$ be the length of open pipe and $\mathrm{v}$ be the velocity.
Frequency of second overtone of an open organ pipe is $\mathrm{n}_{\mathrm{o}}=\frac{3 \mathrm{v}}{2 l}$
Frequency of first overtone of a closed pipe is $\mathrm{n}_{\mathrm{c}}=\frac{3 \mathrm{v}}{4 \mathrm{~L}}$
Given: $\mathrm{n}_0=\mathrm{n}_{\mathrm{c}}$
$\frac{3 \mathrm{v}}{2 l}=\frac{3 \mathrm{v}}{4 \mathrm{~L}}$
$\mathrm{L}=\frac{l}{2}$
$\therefore \quad l=2 \mathrm{~L}$