An organ pipe P_1,closed at one end and conta | vedclass

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(D) The frequency of a closed organ pipe vibrating in its first harmonic is $n_1 = \frac{v_1}{4 l_1}$.The frequency of an open organ pipe vibrating in

Vedclass · Accepted Answer

$\frac{1}{6} \sqrt{\frac{\rho_2}{\rho_1}}$

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(D) The frequency of a closed organ pipe vibrating in its first harmonic is $n_1 = \frac{v_1}{4 l_1}$.The frequency of an open organ pipe vibrating in its third harmonic is $n_3 = \frac{3 v_2}{2 l_2}$.Since both pipes are in resonance with the same tuning fork,their frequencies are equal: $n_1 = n_3$.Therefore,$\frac{v_1}{4 l_1} = \frac{3 v_2}{2 l_2}$,which simplifies to $\frac{l_1}{l_2} = \frac{1}{6} \left( \frac{v_1}{v_2} \right)$.The speed of sound in a gas is given by $v = \sqrt{\frac{B}{\rho}}$,where $B$ is the bulk modulus (reciprocal of compressibility) and $\rho$ is the density.Given that the compressibility is equal in both pipes,$B_1 = B_2 = B$.Thus,$\frac{v_1}{v_2} = \sqrt{\frac{B/\rho_1}{B/\rho_2}} = \sqrt{\frac{\rho_2}{\rho_1}}$.Substituting this into the ratio of lengths: $\frac{l_1}{l_2} = \frac{1}{6} \sqrt{\frac{\rho_2}{\rho_1}}$.

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