Question

In a pipe open at both ends, ‘$n_1$’ and ‘$n_2$’ be the frequencies corresponding to vibrating lengths ‘$l_1$’ and ‘$l_2$’ respectively. The end correction is

• A. $\frac{n_{1}l_{1}-n_{2}l_{2}}{2\left(n_{1}-n_{2}\right)}$
• B. $\frac{n_{2}l_{2}-n_{1}l_{1}}{2\left(n_{2}-n_{1}\right)}$
• C. $\frac{n_{2}l_{2}-n_{1}l_{1}}{2\left(n_{1}-n_{2}\right)}$
• D. $\frac{n_{1}l_{1}-n_{2}l_{2}}{\left(n_{1}-n_{2}\right)}$

Answer 1

Answer: (C)

The frequency of the $p$th mode of vibration in an open organ pipe,
$v = \frac{pv}{2(L + 2e)}$
Where, $e$ is the end correction of one side and $L$ is the length of the pipe
Given : For $L = l_1, $ frequency of $p$th mode $= n_1$
For $L = l_2$, frequency of $p$th mode $= n_2$
$\Rightarrow n_1 = \frac{pv}{2(l_1 + 2e)} \,\dots(1)$
$\Rightarrow n_2 = \frac{pv}{2(l_2 + 2e)} \,\dots (2)$
Dividing (1) and (2), we get
$\frac{n_1}{n_2} = \frac{pv}{2(l_1 + 2e)} \times \frac{2(l_2 + 2e)}{pv}$
$\therefore \frac{n_1}{n_2} = \frac{l_2 + 2e}{l_1 + 2e}$
$\Rightarrow e = \frac{n_2l_2 - n_1l_1}{2(n_1 - n_2)}$