The fundamental frequency of a closed organ pipe of length $20\text{ cm}$ is equal to the second overtone of…
The fundamental frequency of a closed organ pipe of length $20\text{ cm}$ is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is [CBSE AIPMT 2015]
80 cm
100 cm
120 cm
140 cm
Solution
The fundamental frequencies of closed and open organ pipe are given as $\nu_c = \frac{v}{4l}$ and $\nu_o = \frac{v}{2l'}$, respectively.
Given, the second overtone (i.e. third harmonic) of open pipe is equal to the fundamental frequency of closed pipe,
i.e. $3 \nu_o = \nu_c \Rightarrow 3 \frac{v}{2l'} = \frac{v}{4l}$
$\Rightarrow l' = 6l = 6 \times 20 = 120\text{ cm}$