Question

The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe L metre long. The length of the open pipe will be
(A) 4L
(B) L
(C) 2L
(D) $\dfrac{L}{2}$

Answer 1

Hint
Overtone and the frequency of organ pipes are closely related. Overtones are high-frequency standing waves and depend linearly on the number of harmonics.
For a closed organ pipe ${f_c} = (2n - 1)\dfrac{v}{{4L}}$ , and for an open organ pipe ${f_o} = \dfrac{{nv}}{{2L}}$, where f is the frequency, L is the length of the pipe, v is the speed and n is the number of harmonics.

Complete step by step answer
We are given two organ pipes: closed and open. We are asked to find the length of the open organ pipe when we know the overtone of the closed pipe. We know that the overtone is related to the number of harmonics as:
$\Rightarrow x = n - 1$ where x is the number of overtone and n is the number of harmonics.
So, let us consider the closed organ pipe. We know that the frequency is given as:
$\Rightarrow {f_c} = (2n - 1)\dfrac{v}{{4L}}$
We also know that
$\Rightarrow x = 1 \Rightarrow n = x + 1 = 2$
Thus, the frequency will be:
$\Rightarrow {f_c} = (2 \times 2 - 1)\dfrac{v}{{4L}} = \dfrac{{3v}}{{4L}}$ [Eq. 1]
Now, for the open organ pipe, the frequency is given as:
$\Rightarrow {f_o} = \dfrac{{nv}}{{2L}}$
And we are provided with the 2nd overtone, hence:
$\Rightarrow x = 2 \Rightarrow n = x + 1 = 3$
Putting the values in the frequency equation, we get:
$\Rightarrow {f_o} = \dfrac{{3v}}{{2{L_o}}}$ [Eq. 2]
Comparing Eq. 1 and Eq. 2, gives us:
$\Rightarrow \dfrac{{3v}}{{2{L_o}}} = \dfrac{{3v}}{{4L}}$
$\Rightarrow {L_o} = 2L$
Hence, the correct answer is option (C); 2L.

Note
Organ pipes are musical instruments that produce a sound when air is blown through them. Stationary waves are created in the pipe due to the air that travel from one end and are reflected on the other. Due to the interference of these waves, harmonics and overtones are produced at some frequency.