Question

The fundamental frequency of a closed end organ pipe is n. Its length is doubled and radius is halved. Its frequency will become nearly:
A. \[\dfrac{n}{2} \\ \]
B. \[\dfrac{n}{3} \\ \]
C. \[n\]
D. \[2n\]

Answer 1

Hint: In the closed organ pipe, the node is formed at the end. For this question we need to use the relation between the fundamental frequency, speed and wavelength with end correction. When the length or radius is changed, the speed of the sound remains unchanged.

Formula used:
\[{\nu _0} = \dfrac{v}{{4\left( {l + 0.6r} \right)}}\]
where v is the speed of sound wave, l is the length of the organ pipe and r is the radius of the organ pipe.

Complete step by step solution:
Taking end correction into consideration, the fundamental frequency of the closed end organ pipe is given as,
\[{\nu _0} = \dfrac{v}{{4\left( {l + 0.6r} \right)}}\]
where v is the speed of sound wave, l is the length of the organ pipe and r is the radius of the organ pipe.

Let the initial length of the organ pipe is \[{l_1}\] and the radius of the end is \[{r_1}\]. Then the initial fundamental frequency is,
\[{\nu _{01}} = \dfrac{v}{{4\left( {{l_1} + 0.6{r_1}} \right)}}\]
It is given that the length is doubled and the radius is halved. So, the final length of the organ pipe will be \[{l_2} = 2{l_1}\] and the final radius of the end of the pipe will be,
\[{r_2} = \dfrac{{{r_1}}}{2}\]

Then the final fundamental frequency is,
\[{\nu _{02}} = \dfrac{v}{{4\left( {{l_2} + 0.6{r_2}} \right)}}\]
Taking ratio of both the fundamental frequencies, we get
\[\dfrac{{{\nu _{01}}}}{{{\nu _{02}}}} = \dfrac{{\dfrac{v}{{4\left( {{l_1} + 0.6{r_1}} \right)}}}}{{\dfrac{v}{{4\left( {{l_2} + 0.6{r_2}} \right)}}}}\]

On simplifying, we get
\[\dfrac{{{\nu _{01}}}}{{{\nu _{02}}}} = \dfrac{{\left( {{l_2} + 0.6{r_2}} \right)}}{{\left( {{l_1} + 0.6{r_1}} \right)}} \\ \]
\[\Rightarrow \dfrac{{{\nu _{01}}}}{{{\nu _{02}}}} = \dfrac{{\left( {2{l_1} + 0.6 \times \left( {\dfrac{{{r_1}}}{2}} \right)} \right)}}{{\left( {{l_1} + 0.6{r_1}} \right)}} \\ \]
\[\Rightarrow \dfrac{{{\nu _{01}}}}{{{\nu _{02}}}} = \dfrac{{\left( {2 + 0.3\left( {\dfrac{{{r_1}}}{{{l_1}}}} \right)} \right)}}{{\left( {1 + 0.6\left( {\dfrac{{{r_1}}}{{{l_1}}}} \right)} \right)}} \\ \]
As for the organ pipe length is larger than the radius of the organ pipe,
\[\dfrac{{{\nu _{01}}}}{{{\nu _{02}}}} = 2 \\ \]
\[\therefore {\nu _{02}} = \dfrac{{{\nu _{01}}}}{2}\]
The initial fundamental frequency is given as n, so the final fundamental frequency will be,
\[{\nu _{02}} = \dfrac{n}{2}\].

Therefore, the correct option is A.

Note: If the radius of the organ pipe is comparable to the length of the organ pipe, then we can not ignore the value of the ratio of radius to the length. In that case we need to be given the relation between the radius and the length of the pipe.