Question

In a resonance pipe the first and second resonances are obtained at depths $22.7cm$ and $70.2cm$ respectively. What will be the end correction (in $cm$)?

Answer 1

Let $l_1$ be the depth of first resonance of the pipe, $l_1=22.7cm$

Let $l_2$ be the depth of second resonance of the pipe, $l_2=70.2cm$

For end correction,

$\Rightarrow \frac{l_2+x}{l_1+x}=\frac{3\frac{\lambda }{4}}{\frac{\lambda }{4}}$

$\Rightarrow \frac{l_2+x}{l_1+x}=3$

$\Rightarrow x=\frac{l_2-3l_1}{2}$

Substituting the values, we get

$\Rightarrow x=\frac{70.2-3\times 22.7}{2}$

$\Rightarrow x=\frac{70.2-68.1}{2}$

$\Rightarrow x=\frac{2.1}{2}$

$\Rightarrow x=1.05cm$

Final Answer: The end correction is $1.05cm$.