A closed organ pipe of length L_c and an open | vedclass

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(A) For an open organ pipe,the fundamental frequency is $f_o = \frac{v_2}{2 L_o}$. The first overtone is $f_{o1} = 2 f_o = \frac{v_2}{L_o}$.For a clos

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$\frac{4 L_c}{3} \sqrt{\frac{\rho_1}{\rho_2}}$

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(A) For an open organ pipe,the fundamental frequency is $f_o = \frac{v_2}{2 L_o}$. The first overtone is $f_{o1} = 2 f_o = \frac{v_2}{L_o}$.For a closed organ pipe,the fundamental frequency is $f_c = \frac{v_1}{4 L_c}$. The first overtone is $f_{c1} = 3 f_c = \frac{3 v_1}{4 L_c}$.Given that the frequencies are equal,$f_{o1} = f_{c1}$,so $\frac{v_2}{L_o} = \frac{3 v_1}{4 L_c}$.Rearranging for $L_o$,we get $L_o = \frac{4 L_c}{3} \frac{v_2}{v_1}$.The speed of sound in a gas is given by $v = \sqrt{\frac{1}{\rho \beta}}$,where $\beta$ is the compressibility. Since $\beta$ is the same for both gases,$\frac{v_2}{v_1} = \sqrt{\frac{\rho_1}{\rho_2}}$.Substituting this into the expression for $L_o$,we get $L_o = \frac{4 L_c}{3} \sqrt{\frac{\rho_1}{\rho_2}}$.

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