Question

The successive resonance frequencies in an open organ pipe are 1944 Hz and 2600 Hz. The length of the pipe if the speed of sound in air is \[328{\text{ m/s}}\], is:
A. 0.40 m
B. 0.04 m
C. 0.50 m
D. 0.25 m

Answer 1

Hint: In this question, we need to determine the length of the pipe such that the speed of sound in air is 328 m/s. As the frequencies of the two successive resonance frequencies are given so by comparing the frequencies of the successive frequencies we will find the length of the organ pipe.

Formula used:
\[f = \dfrac{{nv}}{{2l}}\] where, ‘f’ is the frequency, ‘v’ is the velocity of the sound in air and ‘l’ is the length of the pipe.

Complete step by step answer:
The speed of sound in air is \[v = 328m{s^{ - 1}}\]
Let the successive resonance frequencies in an open organ pipe n and (n+1), hence we can say
\[{f_n} = 1944Hz\]
\[{f_{n + 1}} = 2600Hz\]
We know the length of an organ pipe is \[l = n\left( {\dfrac{\lambda }{2}} \right)\]and since we know \[\lambda = \dfrac{v}{f}\], hence we can say the frequency of sound in an open organ pipe is
\[f = \dfrac{{nv}}{{2l}}\]
So the frequency for the \[{n^{th}}\]node will be
\[{f_n} = \dfrac{{nv}}{{2l}} - - (i)\]
And the frequency for the \[n + {1^{th}}\]node will be
\[{f_{n + 1}} = \dfrac{{\left( {n + 1} \right)v}}{{2l}} - - (ii)\]
Now since we have to find the length of the open organ pipe for the successive frequency so we will subtract equation (i) from the equation (ii), so we get
\[{f_{n + 1}} - {f_n} = \dfrac{{\left( {n + 1} \right)v}}{{2l}} - \dfrac{{\left( n \right)v}}{{2l}}\]
Now we will substitute the values of the frequency in the above equation, hence we get
\[2600 - 1944 = \dfrac{v}{{2l}}\]
Since, \[v = 328m{s^{ - 1}}\], so we get the length of the pipe as
\[
  656 = \dfrac{{328}}{{2l}} \\
   \Rightarrow 2l = \dfrac{{328}}{{656}} \\
   \Rightarrow l = \dfrac{1}{4} \\
   \therefore l = 0.25m \\
 \]
Therefore the length of the successive resonance frequencies in an open organ pipe \[ = 0.25m\]

So, the correct answer is “Option D”.

Note:
In an open organ pipe both the sides are open and the length between the two successive progressions is \[l = n\dfrac{\lambda }{2}\].
Students must note that in an open organ pipe it’s both ends are open so when air is blown to it the wave reaches to the other end from where it gets reflected.