The third overtone of a closed pipe of length ' L c ' has the same frequency as the third overtone of the open pipe of length ' L 0 '. Both the pipes have same diameters. The r

Question

The third overtone of a closed pipe of length ' L c ​ ' has the same frequency as the third overtone of the open pipe of length ' L 0 ​ '. Both the pipes have same diameters. The ratio L c ​ : L 0 ​ is equal to

TestHub · Accepted Answer

For a closed pipe, the frequency of the $n$ -th overtone is given by $f_{c} = (2 n + 1) \frac{v}{4 L _{c}}$ .

For the third overtone ( $n = 3$ ), the frequency is $f_{c} = 7 \frac{v}{4 L _{c}}$ .

For an open pipe, the frequency of the $n$ -th overtone is given by $f_{o} = (n + 1) \frac{v}{2 L _{0}}$ .

For the third overtone ( $n = 3$ ), the frequency is $f_{o} = 4 \frac{v}{2 L _{0}} = \frac{2 v}{L _{0}}$ .

Given that the frequencies of the third overtones are equal:

$7 \frac{v}{4 L _{c}} = \frac{2 v}{L _{0}}$

$\frac{L _{c}}{L _{0}} = \frac{7}{8}$

The ratio $L_{c} : L_{0}$ is $7 : 8$ .

Answer: $7 : 8$

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