A pipe P_{C} closed at one end and a pipe P_{ | vedclass

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(B) For a pipe $P_{O}$ open at both ends,the frequency of the $n$-th harmonic is $f_n = \frac{n v}{2 L_O}$. The second overtone corresponds to the $3^

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$\frac{5}{6}$

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(B) For a pipe $P_{O}$ open at both ends,the frequency of the $n$-th harmonic is $f_n = \frac{n v}{2 L_O}$. The second overtone corresponds to the $3^{rd}$ harmonic $(n=3)$,so $f_3 = \frac{3 v}{2 L_O}$.For a pipe $P_{C}$ closed at one end,the frequency of the $n$-th harmonic is $f_n = \frac{(2n-1) v}{4 L_C}$. The second overtone corresponds to the $3^{rd}$ overtone mode,which is the $5^{th}$ harmonic $(n=3)$,so $f_5 = \frac{5 v}{4 L_C}$.Since both pipes are in resonance with the same tuning fork,their frequencies are equal: $f_3 = f_5$.$\frac{3 v}{2 L_O} = \frac{5 v}{4 L_C}$$\frac{3}{2 L_O} = \frac{5}{4 L_C}$$\frac{L_C}{L_O} = \frac{5 	imes 2}{4 	imes 3} = \frac{10}{12} = \frac{5}{6}$.

$A$ pipe $P_{C}$ closed at one end and a pipe $P_{O}$ open at both ends are vibrating in their second overtone. They are in resonance with a given tuning fork. The ratio of the length of pipe $P_{C}$ to $P_{O}$ is (Neglect end correction):

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