Question

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 $100Hz$ than the first overtone of the original pipe. Then, the fundamental frequency of open pipe is
(A) $200{s^{ - 1}}$
(B) $100{s^{ - 1}}$
(C) $300{s^{ - 1}}$
(D) $250{s^{ - 1}}$
(E) $150{s^{ - 1}}$

Answer 1

Hint: 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 $l$ and velocity of sound be $v$ the ${n^{th}}$ overtone frequency is given by ${f_{open}} = \dfrac{{(n + 1)v}}{{2l}}$
For an closed organ pipe, the ${n^{th}}$ overtone frequency is given by ${f_{closed}} = \dfrac{{(2n + 1)v}}{{4l}}$

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

Now, the second overtone of closed organ pipe is calculated using the formula ${f_{closed}} = \dfrac{{(2n +1)v}}{{4l}}$ where $n = 2$ we get,
${f_{closed}} = \dfrac{{5v}}{{4l}} \to (ii)$ so their difference is given as
${f_{closed}} - {f_{open}} = 100$ using equations (i) and (ii) we get,
$
  \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 $\dfrac{v}{{2l}}$ for an open organ pipe so, on dividing equation (iii) by $2$ we get,
$
  \dfrac{v}{{2l}} = \dfrac{{400}}{2} \\
  \dfrac{v}{{2l}} = 200{s^{ - 1}} \\
 $

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

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

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 ${s^{ - 1}}$ is also known as Hertz denoted by $Hz.$