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Question: An open pipe is suddenly closed with the result that, the second overtone of the closed pipe is foun...

An open pipe is suddenly closed with the result that, the second overtone of the closed pipe is found to be higher in frequency by 100Hz100Hz than the first overtone of the original pipe. Then, the fundamental frequency of open pipe is
(A) 200s−1200{s^{ - 1}}
(B) 100s−1100{s^{ - 1}}
(C) 300s−1300{s^{ - 1}}
(D) 250s−1250{s^{ - 1}}
(E) 150s−1150{s^{ - 1}}

Explanation

Solution

In order to solve this question, we will first find the frequency of the first overtone for open pipe and then of the second overtone for closed pipe and then we will solve for the fundamental frequency using the given frequency difference.

Formula Used:
For an organ pipe of length ll and velocity of sound be vv the nth{n^{th}} overtone frequency is given by fopen=(n+1)v2l{f_{open}} = \dfrac{{(n + 1)v}}{{2l}}
For an closed organ pipe, the nth{n^{th}} overtone frequency is given by fclosed=(2n+1)v4l{f_{closed}} = \dfrac{{(2n + 1)v}}{{4l}}

Complete answer:
We have given that, the second overtone of the closed pipe is found to be higher in frequency by 100Hz100Hz than the first overtone of the original pipe.
So, first overtone frequency for an open organ pipe is determined by using the formula fopen=(n+1)v2l{f_{open}} = \dfrac{{(n + 1)v}}{{2l}} where n=1n = 1 we get,
fopem=vl→(i){f_{opem}} = \dfrac{v}{l} \to (i)

Now, the second overtone of closed organ pipe is calculated using the formula fclosed=(2n+1)v4l{f_{closed}} = \dfrac{{(2n +1)v}}{{4l}} where n=2n = 2 we get,
fclosed=5v4l→(ii){f_{closed}} = \dfrac{{5v}}{{4l}} \to (ii) so their difference is given as
fclosed−fopen=100{f_{closed}} - {f_{open}} = 100 using equations (i) and (ii) we get,
5v4l−vl=100 v4l=100 vl=400→(iii)  \dfrac{{5v}}{{4l}} - \dfrac{v}{l} = 100 \\\ \dfrac{v}{{4l}} = 100 \\\ \dfrac{v}{l} = 400 \to (iii) \\\

And we know that the fundamental frequency of the organ pipe is v2l\dfrac{v}{{2l}} for an open organ pipe so, on dividing equation (iii) by 22 we get,
v2l=4002 v2l=200s−1  \dfrac{v}{{2l}} = \dfrac{{400}}{2} \\\ \dfrac{v}{{2l}} = 200{s^{ - 1}} \\\

So, the fundamental frequency of the pipe is 200s−1200{s^{ - 1}}

Hence, the correct option is (A) 200s−1200{s^{ - 1}}

Note: It should be remembered that here we have to use the fundamental frequency formula for open organ pipe because given organ pipe was originally opened and it was closed later and also the units of frequency s−1{s^{ - 1}} is also known as Hertz denoted by Hz.Hz.