In the resonance tube experiment, first resonant length is $l_1$ and second resonant length is $l_2$ then the third resonant length will be
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for first resonance
$\ell_{1}+\varepsilon=\frac{v}{4 \mathrm{f}_{0}}$
for second resonance
$\ell_{2}+\varepsilon=\frac{3 \mathrm{v}}{4 \mathrm{f}_{0}}$
$\ell_{3}+\varepsilon=\frac{5 \mathrm{v}}{4 \mathrm{f}_{0}}$
Solving get $\ell_{3}=2 \ell_{2}-\ell_{1}$
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