Question

A resonance tube is resonated with a tuning fork of frequency \[256\,{\text{Hz}}\]. If the length of first and second resonating air columns are \[32\,{\text{cm}}\] and \[100\,{\text{cm}}\], then end correction will be
A. \[1\,{\text{cm}}\]
B. \[2\,{\text{cm}}\]
C. \[4\,{\text{cm}}\]
D. \[6\,{\text{cm}}\]

Answer 1

Hint: Use the formulae given by Rayleigh for the first and second resonating columns. This formula will give the relation between the wavelength, length of the first and second resonating columns and the end correction. Substitute the values of first and second resonating columns and solve these two equations to determine the value of the end correction.

Formulae used:
The Rayleigh equation for first resonating column is
\[\dfrac{\lambda }{4} = {l_1} + e\] …… (1)
Here, \[\lambda \] is the wavelength of the wave, \[{l_1}\] is the length of the first resonating column and \[e\] is the end correction.
The Rayleigh equation for second resonating column is
\[\dfrac{{3\lambda }}{4} = {l_2} + e\] …… (2)
Here, \[\lambda \] is the wavelength of the wave, \[{l_2}\] is the length of the second resonating column and \[e\] is the end correction.

Complete step by step answer:
We have given that the frequency of the tuning fork is \[256\,{\text{Hz}}\].
The lengths of the first and second resonating columns are \[32\,{\text{cm}}\] and \[100\,{\text{cm}}\] respectively.
\[{l_1} = 32\,{\text{cm}}\]
\[{l_2} = 100\,{\text{cm}}\]
We can determine the end correction using equations (1) and (2).
Substitute \[32\,{\text{cm}}\] for \[{l_1}\] in equation (1).
\[\dfrac{\lambda }{4} = \left( {32\,{\text{cm}}} \right) + e\]
\[ \Rightarrow \dfrac{\lambda }{4} = 32 + e\]
Substitute \[100\,{\text{cm}}\] for \[{l_2}\] in equation (1).
\[\dfrac{{3\lambda }}{4} = \left( {100\,{\text{cm}}} \right) + e\]
\[ \Rightarrow \dfrac{{3\lambda }}{4} = 100 + e\]
Substitute \[32 + e\] for \[\dfrac{\lambda }{4}\] in the above equation.
\[ \Rightarrow 3\left( {32 + e} \right) = 100 + e\]
\[ \Rightarrow 96 + 3e = 100 + e\]
\[ \Rightarrow 3e - e = 100 - 96\]
\[ \Rightarrow 2e = 4\]
\[ \Rightarrow e = \dfrac{4}{2}\]
\[ \therefore e = 2\,{\text{cm}}\]

Therefore, the end correction will be \[2\,{\text{cm}}\].Hence, the correct option is B.

Additional information:
In the resonating tube, the end correction occurs because the reflected wave goes a little above the mouth of the resonating wave.Hence, the length of the resonating tube becomes the original length of the resonating tube plus the end correction.

Note: One can also solve the same question by using the formula for end correction in terms of the lengths of first and second resonating air columns. This equation can be obtained by eliminating the wavelength term in these two equations. This equation for end correctly obtained is \[\dfrac{{{l_2} - 3{l_1}}}{2}\].