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
Solution and Explanation
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 \( \frac{20}{9} \, \text{cm} \).
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