An open pipe is in second harmonic with frequency f₁ . One end of the tube is closed and frequency is increased to f₂ and it resonates again with nth harmonic. For what value of n , the ratio of f₁ f₂ is 4 5 ?

Step-by-step solution

For an open pipe of length L , the frequency of the second harmonic is given by f₁ = 2 v 2L = v L , where v is the speed of sound. When one end is closed, the pipe becomes a closed pipe of length L . The frequencies of a closed pipe are given by f = m v 4L , where m is an odd integer ( 1, 3, 5, ). The n -th harmonic of a closed pipe corresponds to the n -th odd harmonic, so f₂ = n v 4L , where n is an odd integer. The ratio of the frequencies is given as f₁ f₂ = 4 5 . Substituting the expressions for f₁ and f₂ : v/