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Question: The fundamental frequency of a closed pipe is equal to the frequency of the second harmonic of an op...

The fundamental frequency of a closed pipe is equal to the frequency of the second harmonic of an open pipe. The ratio of their length is :
(A) 1:21:2
(B) 1:41:4
(C) 1:81:8
(D) 1:161:16

Explanation

Solution

We know that fundamental frequency of closed pipe,
υ1=V4L1{\upsilon _1} = \dfrac{V}{{4{L_1}}}.
We can use the idea that sound in air is constant.
Frequency of second harmonic of open pipe,
υ2=V2L2{\upsilon _2} = \dfrac{V}{{2{L_2}}}

Complete step by step answer:
Given, v1=v2{v_1} = {v_2} then the fundamental frequency of a closed pipe is equal to the frequency of the second harmonic of an open pipe we are calculating the ovation by length of closed pipe is that of open pipe is fundamental frequency of closed pipe,
υ1=V4L1{\upsilon _1} = \dfrac{V}{{4{L_1}}}.
Where V is the velocity of sound in air.
L1{L_1}is the length of the closed pipe.
Frequency of second harmonic of open pipe,
υ2=V2L2{\upsilon _2} = \dfrac{V}{{2{L_2}}}
L2{L_2}is the length of the open pipe.
Given,

υ1=υ2 L1 {\upsilon _1} = {\upsilon _2} \\\ {L_1} \\\

V4L1=V2L2 \Rightarrow \dfrac{V}{{4{L_1}}} = \dfrac{V}{{2{L_2}}}
14L1=12L2\Rightarrow \dfrac{1}{{4{L_1}}} = \dfrac{1}{{2{L_2}}}
L1L2=24=12\Rightarrow \dfrac{{{L_1}}}{{{L_2}}} = \dfrac{2}{4} = \dfrac{1}{2}
Then the correct answer is (A) 1:21:2.

Note: We should note the idea of question, the pipes are different. One is closed and another one is open pipe.