Question

A pipe of length \[{l_1}\] closed at one end is kept in a chamber of gas of density \[{\rho _1}\]. A second pipe open at both ends is placed in the second chamber of gas of density \[{\rho _2}\]. The compressibility of both the gases is equal. Calculate the length of the second pipe if the frequency of the first overtone in both the cases is equal.
A. \[\dfrac{4}{3}{l_1}\sqrt {\dfrac{{{\rho _2}}}{{{\rho _1}}}} \\ \]
B. \[\dfrac{4}{3}{l_1}\sqrt {\dfrac{{{\rho _1}}}{{{\rho _2}}}} \\ \]
C. \[{l_1}\sqrt {\dfrac{{{\rho _2}}}{{{\rho _1}}}} \\ \]
D. \[{l_1}\sqrt {\dfrac{{{\rho _1}}}{{{\rho _2}}}} \]

Answer 1

Hint: The first case is when the pipe is closed at one end, while the second case is when the pipe is open at both the ends. Thus by comparing the frequency of the first overtone in both the cases we can find the length of the pipe.

Formula used:
Frequency of first overtone when a pipe is closed at one end is given as,
\[{\upsilon _1} = \dfrac{3}{4}\dfrac{v}{l}\]
\[{\upsilon _1} = \dfrac{3}{{4l}}\sqrt {\dfrac{Y}{\rho }} \]
Where l is the length of the pipe, \[\rho \] is the density and Y is the compressibility of the gas.
Frequency of first overtone when a pipe is open at both end is given as,
\[{\upsilon _1} = \dfrac{1}{l}\sqrt {\dfrac{Y}{\rho }} \]

Complete step by step solution:
In the case when a pipe of length \[{l_1}\] closed at one end,
Frequency of first overtone,
\[{\upsilon _1} = \dfrac{3}{{4{l_1}}}\sqrt {\dfrac{Y}{{{\rho _1}}}} \]
In the second case when a pipe of length \[{l_2}\] open at both ends,
Frequency of first overtone,
\[{\upsilon _1} = \dfrac{1}{{{l_2}}}\sqrt {\dfrac{Y}{{{\rho _2}}}} \]

Now it is given that the frequency of the first overtone in both the cases is equal. So, we can write that;
Frequency of first overtone of closed pipe = Frequency of first overtone of open pipe
\[\dfrac{3}{{4{l_1}}}\sqrt {\dfrac{Y}{{{\rho _1}}}} = \dfrac{1}{{{l_2}}}\sqrt {\dfrac{Y}{{{\rho _2}}}} \\ \]
\[\Rightarrow \dfrac{{3{l_2}}}{{4{l_1}}} = \sqrt {\dfrac{{{\rho _1}}}{{{\rho _2}}}} \\ \]
\[\Rightarrow {l_2} = \dfrac{4}{3}{l_1}\sqrt {\dfrac{{{\rho _1}}}{{{\rho _2}}}} \\ \]
Therefore, the length of the second pipe is,
\[{l_2} = \dfrac{4}{3}{l_1}\sqrt {\dfrac{{{\rho _1}}}{{{\rho _2}}}} \]

Hence option B is the correct answer

Note: Open organ pipe is the one in which both the ends are opened and then the sound is passed through it. Whereas the closed organ pipe is the one in which only one end is open and the other is closed and then sound is passed. The instrument is called an open-end air column if both ends of the pipe are open.