Question

An open and closed organ pipe have the same length. The ratio of the ${{p}^{th}}$ mode of frequency of vibration of air in the two pipes is:
$A)\text{ }p\left( 2p-1 \right)$
$B)\text{ }\dfrac{2p}{2p-1}$
$C)\text{ }p$
$D)\text{ }1$

Answer 1

int: This problem can be solved by using the direct formula for the ${{n}^{th}}$ mode of frequency for vibration in a closed and open organ pipe. Using these two values we can get the two required values and hence, find the required ratio from that.

Formula used:
${{f}_{closed,n}}=\dfrac{v}{4L}\left( 2n-1 \right)$
${{f}_{open,n}}=\dfrac{v}{2L}\left( n \right)$

Complete step by step answer:
We will use the direct formulae for the frequency of vibration of the ${{n}^{th}}$ mode of frequency for an open and closed organ pipe.
For a closed organ pipe of length $L$ , the ${{n}^{th}}$ mode of frequency ${{f}_{closed,n}}$ or harmonic is given by
${{f}_{closed,n}}=\dfrac{v}{4L}\left( 2n-1 \right)$ --(1)
Where, $v$ is the speed of sound in air inside the pipe.
Similarly, For an open organ pipe of length $L$ , the ${{n}^{th}}$ mode of frequency ${{f}_{open,n}}$ or harmonic is given by
${{f}_{open,n}}=\dfrac{v}{2L}n$ --(2)
Where, $v$ is the speed of sound in air in the pipe.
Now, according to the question the length of the closed and open organ pipes is the same.
Therefore, using (1) and (2), we get the ratio of the frequency of vibration of the ${{p}^{th}}$ mode in the open pipe to that in the closed pipe as
$\dfrac{{{f}_{open,p}}}{{{f}_{closed,p}}}=\dfrac{\dfrac{pv}{2L}}{\dfrac{v}{4L}\left( 2p-1 \right)}=\dfrac{2p}{2p-1}$
Hence, the required ratio is $\dfrac{2p}{2p-1}$.
Therefore, the correct answer is $B)\text{ }\dfrac{2p}{2p-1}$.

Note: The ${{n}^{th}}$ mode of frequency can also be termed as the ${{n}^{th}}$ harmonic or the ${{\left( n-1 \right)}^{th}}$ overtone. Students often get confused at this point and mix up the fact that the number of the overtone is one less than the number of the mode of frequency or harmonic. Students can remember that the word overtone has the word ‘over’ in it and thus can keep in mind that the ${{n}^{th}}$ overtone will always be one mode of frequency ‘over’ than the ${{n}^{th}}$ mode of frequency or ${{n}^{th}}$ harmonic.