Question

A closed organ pipe and an open organ pipe of same length produce 4 beats when they are set onto vibrations simultaneously. If the length of the organ pipes is doubled, the number of beats produced will be
A. 2
B. 4
C. 1
D. 8

Answer 1

In this question, the concept of difference in frequencies will be applied. The magnitude of difference in frequencies gives the number of beats. We shall use the expressions relating the length of the organ pipe and wavelength of the wave for both the organ pipes. Further we will use the relation$\lambda = \dfrac{v}{\upsilon }$ to express the length in terms of frequency and then take their difference to equate with the number of beats.

Complete step by step answer:
For an open organ pipe, the length of the organ pipe and the wavelength are related as
$L = \dfrac{{n\lambda }}{2}$
where L is the length of the organ pipe and $\lambda$ is the wavelength of the sound wave.
We know that the wavelength is given as the ratio of the velocity of the wave to the frequency of the wave.Hence,
$\lambda = \dfrac{v}{{{\upsilon _o}}}$
where v is the velocity of the wave and ${\upsilon _o}$ is the frequency of the wave in the open organ pipe.
Substituting in the equation, we get,
$L = \dfrac{{n \times \dfrac{v}{{{\upsilon _o}}}}}{2}$
$\Rightarrow L = \dfrac{{nv}}{{2{\upsilon _o}}}$
This can be rewritten as
${\upsilon _o} = \dfrac{{nv}}{{2L}}\,\,\,\,\,\,\,\,\,\,.........(1)$
For the closed organ pipe, the length of the organ pipe and the wavelength are related as
$L = \dfrac{{(2n + 1)\lambda }}{4}$
where $L$ is the length of the organ pipe and $\lambda$ is the wavelength of the sound wave.
We know that the wavelength is given as the ratio of the velocity of the wave to the frequency of the wave.Hence,
$\lambda = \dfrac{v}{{{\upsilon _c}}}$
where $v$ is the velocity of the wave and ${\upsilon _c}$ is the frequency of the wave in the closed organ pipe.

Substituting in the equation, we get,
$L = \dfrac{{(2n + 1) \times \dfrac{v}{{{\upsilon _c}}}}}{4}$
$\Rightarrow L = \dfrac{{(2n + 1)v}}{{4{\upsilon _c}}}$
This can be rewritten as
${\upsilon _c} = \dfrac{{(2n + 1)v}}{{4L}}\,\,\,\,\,\,\,\,\,\,.........(2)$
The difference of frequencies gives the number of beats.
Given the number of beats is 4.
Hence, ${\upsilon _c} - {\upsilon _0} = 4$

Substituting the values in the equation, we get,
$\dfrac{{(2n + 1)v}}{{4L}}\, - \dfrac{{nv}}{{2L}}\, = 4$
Further solving the equation,
$\dfrac{{2nv + v - 2nv}}{{4L}}\, = 4$
$\Rightarrow \dfrac{v}{{4L}}\, = 4$
Now dividing both sides by 2
$\dfrac{v}{{4(2L)}}\, = \dfrac{4}{2}$
The above expression is for the organ pipes with twice the original length.
$\therefore \dfrac{v}{{4(2L)}}\, = 2$

Hence, option A is the correct answer.

Note: The expressions relating the length of the organ pipe and the wavelength are different for different organ pipes. The difference is due to the fact that nodes and antinodes in closed and open organ pipes are located at different distances. And hence the expression changes. However, the expression $\lambda = \dfrac{v}{\upsilon }$ is universal and can be applied to both organ pipes.