Question

The fundamental frequency of an open organ pipe is $300Hz$. The first overtone of the pipe has the same frequency as the first overtone of a closed organ pipe. If the speed of sound is $330m/s$, then the length of the closed organ pipe is:
(a). $41cm$
(b). $37cm$
(c). $31cm$
(d). $80cm$

Answer 1

Hint: This problem can be solved by first finding out the length of the open organ pipe using the formula for its fundamental frequency. Then using this value, the first overtone frequency of the open organ pipe has to be equated with the formula for the first overtone of the closed organ pipe, thereby giving the value for the length of the closed organ pipe.

Formula used:
The ${{n}^{th}}$ harmonic frequency ${{f}_{n}}$ or ${{\left( n-1 \right)}^{th}}$ overtone is given by,
${{f}_{n}}=n\dfrac{v}{2l}$, $n=1,2,3.....$ [For open organ pipe]

${{f}_{n}}=\left( n-\dfrac{1}{2} \right)\dfrac{v}{2l}$, $n=0,1,2,3.....$ [For closed organ pipe]
where $v$ is the speed of sound in the pipe and $l$ is the length of the pipe.
${{f}_{1}}$ is the fundamental frequency.

Complete step by step answer:
For an open organ pipe, the ${{n}^{th}}$ harmonic frequency ${{f}_{n}}$ or ${{\left( n-1 \right)}^{th}}$ overtone is given by,
The ${{n}^{th}}$ harmonic frequency ${{f}_{n}}$ or ${{\left( n-1 \right)}^{th}}$ overtone is given by,
${{f}_{n}}=n\dfrac{v}{2l}$, $n=1,2,3.....$ [For open organ pipe] --(1)

${{f}_{n}}=\left( n-\dfrac{1}{2} \right)\dfrac{v}{2l}$, $n=0,1,2,3.....$ [For closed organ pipe] --(2)
where $v$ is the speed of sound in the pipe and $l$ is the length of the pipe.
${{f}_{1}}$ is the fundamental frequency.
Now, let us analyze the question.
The given fundamental frequency of the open organ pipe is ${{f}_{1,open}}=300Hz$
The given speed of sound in air (in the tube) is $v=330m/s$.
Let the length of the open organ pipe be $l$.
Now, plugging these values in equation (1), we get,
${{f}_{1,open}}=1\times \dfrac{v}{2l}$
$\therefore 300=\dfrac{330}{2l}$
$\therefore l=\dfrac{330}{2\times 300}=0.55m=55cm$ $\left( \because 1m=100cm \right)$ ---(3)

Now, let the length of the closed organ pipe be $L$.
Using (1), the first overtone (or second harmonic) of the open organ pipe is given by,
${{f}_{2,open}}=2\dfrac{v}{2l}$ --(4)
Now, using (2), the first overtone (or second harmonic) of the closed organ pipe is given by,
${{f}_{2,closed}}=\left( 2-\dfrac{1}{2} \right)\dfrac{v}{2L}=\dfrac{3v}{4L}$ --(5)
Now, according to the question, the first overtone of the closed organ pipe is equal to the first overtone of the open organ pipe. Hence, equating (4) and (5), we get,
${{f}_{2,open}}={{f}_{2,closed}}$
$\Rightarrow 2\dfrac{v}{2l}=3\dfrac{v}{4L}$
$\Rightarrow \dfrac{1}{l}=\dfrac{3}{4L}$
$\Rightarrow L=\dfrac{3l}{4}$
$\Rightarrow L=\dfrac{3\times 55}{4}=41.25cm\approx 41.25cm$
Hence, the length of the closed organ pipe is $41cm$.
Therefore, the correct option is (a) $41cm$.

Note: Students often get confused about the numbers of the harmonics and overtones, that is, the number of the overtone corresponds to the harmonic one less in number than that. For example, the second overtone means the first harmonic. An easy way to remember is to notice that the word overtone has ‘over’ in it, which can act as sort of a mnemonic or reminder that the number of the overtone is one number ‘over’ the number of the harmonic (that is one number more than the number of the harmonic). In this way there is a lot less chance of confusion and silly mistakes.