Question

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:

• A. \( \frac{20}{7} \, \text{cm} \)
• B. \( \frac{15}{7} \, \text{cm} \)
• C. \( \frac{20}{9} \, \text{cm} \)
• D. \(\frac{15}{9} \, \text{cm} \)

Hint:

For tubes with different densities, the length of the tube is inversely proportional to the square root of the density ratio.

Answer 1

The Correct Option is C

To determine the length of an open tube, we equate the ninth harmonic frequency of a closed tube to the fourth harmonic frequency of an open tube.

For a closed organ pipe (closed at one end), the harmonic frequency formula is:

\( f_n = \frac{n v_1}{4L_1} \)

where \( n \) is an odd integer, \( v_1 \) is the speed of sound in the closed tube's gas, and \( L_1 \) is its length.

For an open organ pipe, the harmonic frequency formula is:

\( f_n = \frac{n v_2}{2L_2} \)

where \( n \) is any integer, \( v_2 \) is the speed of sound in the open tube's gas, and \( L_2 \) is its length.

Given that 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 the equation:

\( \frac{9 v_1}{40} = \frac{4 v_2}{2L_2} \)

\( \frac{9 v_1}{40} = \frac{2 v_2}{L_2} \)

Rearranging to solve for \( L_2 \):

\( L_2 = \frac{80 v_2}{9 v_1} \)

The speed of sound \( v \) in a medium is defined as:

\( v = \sqrt{\frac{B}{\rho}} \)

where \( B \) is the bulk modulus and \( \rho \) is the gas density. Assuming identical bulk moduli \( B \) for both gases, the speed ratio is:

\( \frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}} \)

With the given density ratio \( \rho_1 : \rho_2 = 1 : 16 \):

\( \frac{v_1}{v_2} = \sqrt{\frac{16}{1}} = 4 \)

Substituting the speed ratio \( \frac{v_1}{v_2} = 4 \) into the equation for \( L_2 \):

\( L_2 = \frac{80 \times v_2}{9 \times 4 \times v_2} = \frac{80}{36} = \frac{20}{9} \, \text{cm} \)

Therefore, the length of the open tube is \( \frac{20}{9} \, \text{cm} \).