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

Question

Aicharya · Accepted Answer

For end correction $x$, $\frac{l_2 + x}{l_1 + x} = \frac{3\lambda / 4}{\lambda / 4} = 3$ $\Rightarrow x = \frac{l_2 - 3l_1}{2} = \frac{70.2 - 3 	imes 22.7}{2} = 1.05	ext{ cm}$

In a resonance pipe, the first and second resonances are obtained at depths 22.7 cm and 70.2 cm,…

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