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Resonance - class-XI
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A resonance tube is resonated with tuning fork of frequency 256 Hz. If the length of first and second resonating air columns are 32 cm and 100 cm, then end correction will be
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A
$2 cm$
💡 Explanation:
For a resonance tube, the resonance condition is L + e = (2n-1) * lambda / 4. Using the first two resonances, L1 + e = lambda / 4 and L2 + e = 3 * lambda / 4, we find 2 * (L2 - L1) = lambda. Substituting the values, lambda = 2 * (100 - 32) = 136 cm, so L1 + e = 136 / 4 = 34 cm; thus, e = 34 - 32 = 2 cm.