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Last updated date: 30th Nov 2023
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# An open pipe is in resonance in $2^{nd}$ harmonic with frequency ${f_1}$ . Now one end of the tube is closed and frequency is increased ${f_2}$ such that the resonance again occurs in ${n^{th}}$ harmonic. Choose the correct option .A. $n = 3;{f_2} = \dfrac{{3{f_1}}}{4}$B. $n = 3;{f_2} = \dfrac{{5{f_1}}}{4}$C. $n = 5;{f_2} = \dfrac{{3{f_1}}}{4}$D. $n = 5;{f_2} = \dfrac{{5{f_1}}}{4}$

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Hint: Here we are given two types of pipe, open and closed. We are given at first that the pipe is open and after that as its face is closed, it becomes a closed pipe. We need to know the formula to find the frequency of the open and closed pipe. It is seen that resonance occurs at a harmonic, thus we can find the frequency and match with answers.

Let the speed of sound in air be $v$ and the length of the pipe be $l$. Thus, for an open pipe we know the frequency of second harmonic is given by:
$\dfrac{{nv}}{{2l}}$ , here n = 2 , hence ${f_1} = \dfrac{v}{l} - - - - (1)$ .
Now as its one end is closed it becomes a closed pipe, thus the formula for the ${n^{th}}$ frequency of a closed pipe ${f_2} = \dfrac{{(2n - 1)v}}{{4l}} . - - - - (2)$
${f_2} = \dfrac{{(2n - 1){f_1}}}{4} - - - - (3)$ .
Thus we now have the required value, now we just have to place the value as indicated in the question and get the required values . Let us put n = $3$ in the equation, thus we get ${f_2} = \dfrac{{(2n - 1){f_1}}}{4}$ , thus ${f_2} = \dfrac{{5{f_1}}}{4}$.
Now let us put $n = 5$ in the equation, thus we get ${f_2} = \dfrac{{9{f_1}}}{4}$.