Question

An open pipe is suddenly closed at one end. As a result, the frequency of the third harmonic of the closed pipe is found to be higher by $100Hz$ than the frequency of the open pipe. The fundamental frequency of the open pipe is
A. $200Hz$
B. $300Hz$
C. $240Hz$
D. $480Hz$

Answer 1

Before solving this question, we need to know about the concepts for the harmonics for the open as well as closed pipe. According to the question, the open is suddenly closed from one end. We need to apply the formula for harmonics for both closed and open pipe. And hence we got the answer.

Complete step by step answer:
The fundamental (or known as first harmonic) for an open pipe have antinodes at both the ends. Whereas a closed pipe has a node at the closed end and antinode at the other end.
Now, writing frequency for the ${m^{th}}$ harmonic for the open pipe is
${v_m} = \dfrac{{mv}}{{2L}}$ $\ldots \left( 1 \right)$
where $L$ is the length of the pipe.
Frequency of the ${n^{th}}$ harmonic of the closed pipe is,
${v_n}^\prime = \dfrac{{nv}}{{4L}}$ $\ldots \left( 2 \right)$
We had used $\left( {v'} \right)$for the closed pipe and $\left( v \right)$ for the open pipe.Now, reading the question again provides us that the third harmonic of the closed pipe is $100Hz$ higher than the frequency of the open pipe which means the difference of the two frequencies is $100$.

Writing the third harmonic of closed pipe from $\left( 2 \right)$
${v_3}^\prime = \dfrac{{3v}}{{4L}}$
Frequency of the open pipe is given by ${v_1} = \dfrac{v}{{2L}}$
Now, ${v_3}^\prime - {v_1} = 100$
Putting the values,
\dfrac{{3v}}{{4L}} - \dfrac{v}{{2L}} = 100 \\\ \Rightarrow\dfrac{v}{L} = 400 \\\
Fundamental frequency of the open pipe be given by,
$\therefore{v_1} = \dfrac{v}{{2L}} = \dfrac{{400}}{2} = 200Hz$

The correct option is A.

Note: The frequency of the ${p^{th}}$ overtone of the closed pipe is given by $\left( {2p + 1} \right){n_1}$ where ${n_1}$ is the fundamental frequency. The frequency of harmonics is in the ratio of $1:3:5 \ldots$.The frequency of the ${p^{th}}$ overtone of the open pipe is given by $\left( {p + 1} \right){n_1}$ where ${n_1}$ is the fundamental frequency. The frequency of harmonics is in the ratio of $1:2:3 \ldots$.