Question

An organ pipe \[{P_1}\] closed at one end vibrating in its first harmonic and pipe \[{P_2}\] open at both ends vibrating in its third harmonic are in resonance with the same tuning fork. The ratio of their lengths is
A. \[3/8\]
B. \[3/4\]
C. \[1/8\]
D. \[1/6\]

Answer 1

Hint:Use the formula for velocity of the wave in terms of its frequency and wavelength. Use the formula for the lengths of first and third harmonic of organ pipes open at one end and closed at both ends to determine the frequencies of the wave. Equate these two frequencies and determine the ratio of lengths.

Formulae used:
The velocity of a wave is given by
\[v = n\lambda \] …… (1)
Here, \[v\] is the velocity of the wave, \[n\] is the frequency of the wave and \[\lambda \] is the wavelength of the wave.
The length \[{L_1}\] of the organ pipe closed at one end for the first harmonic is
\[{L_1} = \dfrac{\lambda }{4}\] …… (2)
Here, \[\lambda \] is the wavelength of the wave.
The length \[{L_3}\] of the organ pipe open at both ends for the third harmonic is
\[{L_3} = \dfrac{{3\lambda }}{2}\] …… (3)
Here, \[\lambda \] is the wavelength of the wave.

Complete step by step answer:
We have given that an organ pipe \[{P_1}\] closed at one end vibrating in its first harmonic and pipe \[{P_2}\] open at both ends vibrating in its third harmonic are in resonance with the same tuning fork. We can determine the frequency of the wave in an organ pipe \[{P_1}\] closed at one end vibrating in its first harmonic. Rewrite equation (1) for the frequency \[{n_1}\] of the wave in the organ pipe \[{P_1}\].
\[{n_1} = \dfrac{v}{\lambda }\]
Here, \[\lambda \] is the wavelength of the wave and \[v\] is the velocity of the wave in the organ pipe.

Rearrange equation (2) for the wavelength of the wave in the organ pipe \[{P_1}\].
\[\lambda = 4{L_1}\]
Substitute \[4{L_1}\] for \[v\] in the above equation for frequency.
\[{n_1} = \dfrac{{4{L_1}}}{\lambda }\]
This is the frequency of the wave in the organ pipe \[{P_1}\]. We can determine the frequency of the wave in an organ pipe \[{P_2}\] closed at both ends vibrating in its third harmonic. Rewrite equation (1) for the frequency \[{n_3}\] of the wave in the organ pipe \[{P_2}\].
\[{n_3} = \dfrac{v}{\lambda }\]
Here, \[\lambda \] is the wavelength of the wave and \[v\] is the velocity of the wave in the organ pipe.

Rearrange equation (2) for the wavelength of the wave in the organ pipe \[{P_2}\].
\[\lambda = \dfrac{{2{L_3}}}{3}\]
\[{P_2}\]
Substitute \[\dfrac{{2{L_3}}}{3}\] for \[v\] in the above equation for frequency.
\[{n_3} = \dfrac{{\dfrac{{2{L_3}}}{3}}}{\lambda }\]
\[ \Rightarrow {n_3} = \dfrac{{2{L_3}}}{{3\lambda }}\]
This is the frequency of the wave in the organ pipe \[{P_2}\]. Since the organ both the organ pipes \[{P_1}\] and are in resonance in their first and third harmonic respectively, the frequencies \[{n_1}\] and \[{n_3}\] must be equal.
\[{n_1} = {n_3}\]
Substitute \[\dfrac{{4{L_1}}}{\lambda }\] for \[{n_1}\] and \[\dfrac{{2{L_3}}}{{3\lambda }}\] for \[{n_3}\] in the above equation.
\[\dfrac{{4{L_1}}}{\lambda } = \dfrac{{2{L_3}}}{{3\lambda }}\]
\[ \Rightarrow \dfrac{{{L_1}}}{{{L_3}}} = \dfrac{2}{{12}}\]
\[ \therefore \dfrac{{{L_1}}}{{{L_3}}} = \dfrac{1}{6}\]
Therefore, the ratio of the lengths of the organ pipes is \[\dfrac{1}{6}\].

Hence, the correct option is D.

Note: If instead of the organ pipe the resonance is obtained in the air column or tube then also the same method is used for the solution. There are not two different waves in the two organ pipes. The wave is the same but the harmonics of the wave are different and the condition for the ends of the organ pipe is different. Hence, the velocity and wavelength of the wave is the same for both organ pipes.