An organ pipe $P_1$, closed at one end and containing a gas of density $\rho_1$ is vibrating in its first…

An organ pipe $P_1$, closed at one end and containing a gas of density $\rho_1$ is vibrating in its first harmonic. Another organ pipe $P_2$, open at both ends and containing a gas of density $\rho_2$ is vibrating in its third harmonic. Both the pipes are in resonance with a given tuning fork. If the compressibility of gases is equal in both pipes, the ratio of the lengths of $P_1$ and $P_2$ is (assume the given gases to be monoatomic)
  1. $\frac{1}{3}$
  2. 3
  3. $\frac{1}{6} \sqrt{\frac{\rho_1}{\rho_2}}$
  4. $\frac{1}{6} \sqrt{\frac{\rho_2}{\rho_1}}$

Solution

Frequency of closed organ pipe for first harmonic $n_1=\frac{v_1}{4 l_1}$. Frequency of open organ pipe for third harmonic $n_3=\frac{3 v_2}{2 l_2}$ At resonance, $\quad n_1=n_3$ or $\quad \frac{v_1}{4 l_1}=\frac{3 v_2}{2 l_2}$ or $\quad \frac{l_1}{l_2}=\frac{1}{6}\left(\frac{v_1}{v_2}\right)$ $\frac{l_1}{l_2}=\frac{1}{6} \sqrt{\frac{B}{\rho_1}} \times \sqrt{\frac{\rho_2}{B}}$ $=\frac{1}{6} \sqrt{\frac{\rho_2}{\rho_1}}$

Asked in: MHT CET Full Test 7

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