Question

In a resonance tube experiment, $3^{\text{rd }}$ resonance occurs at $111\,cm$ and $5^{\text{th }}$ resonance occurs at $200\,cm$ . Find the end correction (in $cm$ ) is $\alpha $ . Write the value of $100\alpha $ .

Answer 1

In the resonance tube, for $3^{rd}$ resonance,
$\frac{5 \lambda }{4}=l_{1}+e$ .....(i)
For $5^{\text{th}}$ resonance,
$\frac{9 \lambda }{4}=l_{2}+e$ .....(ii)
From (i) and (ii), $\lambda =l_{2}-l_{1}=200-111=89cm$
From (i), end correction $e=\frac{5 \lambda }{4}-l_{1}=\frac{5}{4}\times 89-111$
$\Rightarrow e=0.25cm=\alpha 100\alpha =25$