The ratio of frequencies of the first overtone produced by an open pipe to that of a closed pipe having the same length is

Question

TestHub · Accepted Answer

For an open pipe, the fundamental frequency is $n_{o} = \frac{v}{2 L}$ . The frequency of the first overtone is $n_{o 1} = 2 n_{o} = 2 (\frac{v}{2 L}) = \frac{v}{L}$ .

For a closed pipe, the fundamental frequency is $n_{c} = \frac{v}{4 L}$ . The frequency of the first overtone is $n_{c 1} = 3 n_{c} = 3 (\frac{v}{4 L}) = \frac{3 v}{4 L}$ .

The ratio of the frequency of the first overtone of the open pipe to that of the closed pipe is:

$\frac{n _{o 1}}{n _{c 1}} = \frac{\frac{v}{L}}{\frac{3 v}{4 L}} = \frac{4}{3}$

Thus, the required ratio is $4 : 3$ .

Answer: $4 : 3$

Physics - Wave on String Question with Solution | TestHub

Options:

Doubts & Discussion