A closed organ and an open organ tube filled by two different gases having the same bulk modulus but different densities \( \rho_1 \) and \( \rho_2 \), respectively. The frequency of the 9th harmonic of the closed tube is identical with the 4th harmonic of the open tube. If the length of the closed tube is 10 cm and the density ratio of the gases is \( \rho_1 : \rho_2 = 1 : 16 \), then the length of the open tube is:
Show Hint
- \( \frac{20}{7} \, \text{cm} \)
- \( \frac{15}{7} \, \text{cm} \)
- \( \frac{20}{9} \, \text{cm} \)
- \(\frac{15}{9} \, \text{cm} \)
The Correct Option is C
Approach Solution - 1
To solve this problem, we need to find the length of an open tube when the ninth harmonic of a closed tube matches the fourth harmonic of the open tube.
For a closed organ pipe (closed at one end), the formula for the frequency of the nth harmonic is:
\( f_n = \frac{n v_1}{4L_1} \)
where \( n \) is an odd integer, \( v_1 \) is the speed of sound in the gas inside the closed tube, and \( L_1 \) is the length of the closed tube.
For an open organ pipe, the formula for the frequency of the nth harmonic is:
\( f_n = \frac{n v_2}{2L_2} \)
where \( n \) is any integer, \( v_2 \) is the speed of sound in the gas inside the open tube, and \( L_2 \) is the length of the open tube.
Given: The ninth harmonic of the closed tube (\( n=9 \)) matches the fourth harmonic of the open tube (\( n=4 \)):
\( \frac{9 v_1}{4 \times 10} = \frac{4 v_2}{2L_2} \)
Simplifying that:
\( \frac{9 v_1}{40} = \frac{4 v_2}{2L_2} \)
\( \frac{9 v_1}{40} = \frac{2 v_2}{L_2} \)
\( L_2 = \frac{80 v_2}{9 v_1} \)
The speed of sound \( v \) in a medium is given by:
\( v = \sqrt{\frac{B}{\rho}} \)
where \( B \) is the bulk modulus, and \( \rho \) is the density of the gas. Given that the bulk modulus \( B \) is the same for both gases, the speed ratio is:
\( \frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}} \)
Given the density ratio \( \rho_1 : \rho_2 = 1 : 16 \):
\( \frac{v_1}{v_2} = \sqrt{\frac{16}{1}} = 4 \)
Substituting \( \frac{v_1}{v_2} = 4 \) into the equation for \( L_2 \):
\( L_2 = \frac{80 \times v_2}{9 \times 4 \times v_2} = \frac{80}{36} = \frac{20}{9} \, \text{cm} \)
Thus, the length of the open tube is \( \frac{20}{9} \, \text{cm} \).
Approach Solution -2
Speed of sound in a gas: \(v=\sqrt{B/\rho}\). With the same bulk modulus \(B\), the ratio of speeds is \[ \frac{v_1}{v_2}=\sqrt{\frac{\rho_2}{\rho_1}}=\sqrt{16}=4 \] so \(v_1=4v_2\).
For a closed tube (one end closed) the fundamental frequency is \(f_{c1}=\dfrac{v_1}{4L_c}\). If we interpret the “9th harmonic” as the 9th multiple of the fundamental (9th partial, which is allowed because 9 is odd), its frequency is \[ f_{c9}=\frac{9\,v_1}{4L_c}. \]
For an open tube the \(n\)-th harmonic is \[ f_{o n}=\frac{n\,v_2}{2L_o}. \] For the 4th harmonic \[ f_{o4}=\frac{4\,v_2}{2L_o}=\frac{2v_2}{L_o}. \]
Equate the two frequencies: \[ \frac{9v_1}{4L_c}=\frac{2v_2}{L_o}. \] Solve for \(L_o\): \[ L_o=\frac{8L_c\,v_2}{9v_1}. \] Substitute \(v_2/v_1=\tfrac{1}{4}\) and \(L_c=10\ \text{cm}\): \[ L_o=\frac{8\times10\ \text{cm}\times\frac{1}{4}}{9}=\frac{2\times10}{9}\ \text{cm}=\frac{20}{9}\ \text{cm}. \]
Answer
\(\displaystyle L_o=\frac{20}{9}\ \text{cm}.\) (Option 3)
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