Question

The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both ends. The length of organ pipe open at both the ends is:
A. 120 cm
B. 140 cm
C. 80 cm
D. 100 cm

Answer 1

A transverse wave is the one which travels through a medium, making crests and troughs. Wave on a string is an example of a transverse wave. A longitudinal wave is the one which travels through the medium making compressions and rarefactions. Sound waves are an example of longitudinal waves. When two waves approach each other in the same place, on the same line, standing waves are formed. Organ pipe is an example of the formation of standing waves.

Formula used:
$\nu = \dfrac{v}{4L} (for\ closed\ organ\ pipe)\ and\ \nu= \dfrac{3v}{2L}(second\ overtone \ open\ organ\ pipe)$

Complete step-by-step answer:
Organ pipe is related to the experiments on sound waves. Open organ pipe is the one in which both ends are opened and then sound is passed through it. Closed organ pipe is the one in which only one end is open and the other is closed and then sound is passed.
Now, for a closed organ pipe, the fundamental frequency is given $\nu = \dfrac{v}{4L}$, where ‘v’ is the velocity of sound in the medium of organ pipe and ‘L’ being the length of pipe. For an open organ pipe, the second overtone frequency is given $\nu = \dfrac{3v}{2L}$, where ‘v’ is the velocity of sound in the medium of organ pipe and ‘L’ being the length of pipe.
Since both frequencies are equal, thus;
$\dfrac{3v}{2L_o} = \dfrac{v}{4L_c}$
$\implies L_o = {6L_c}$
$\implies L_o = 6\times 20 = 120\ cm$
Hence, option A. is correct.

So, the correct answer is “Option A”.

Note: Fundamental frequency is the frequency corresponding to the $1^{st}$ harmonic of the frequency of the pipe. One must know that sound waves create standing waves inside the tube so that we can analyze the situation and know the parameters like frequency and conduct the experiment. Students are advised to derive these expressions once by their own.