An organ pipe P_1, closed at one end and cont | 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, $n_1 = n_3$.Therefore, $\frac{v_1}{4 l_1} = \frac{3 v_2}{2 l_2}$, which implies $\frac{l_1}{l_2} = \frac{1}{6} \left( \frac{v_1}{v_2} \right)$.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.Since the compressibility is equal, $B_1 = B_2 = B$.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, we get $\frac{l_1}{l_2} = \frac{1}{6} \sqrt{\frac{\rho_2}{\rho_1}}$.

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