Sound WavesMediumBloom L3
Question
If l1 and l2 are the lengths of air column for the first and second resonance when a tuning fork of frequency n is sounded on a resonance tube, then the distance of the displacement antinode from the top end of the resonance tube is:
Options
A.2(l2 – l1)
B.$\frac{1}{2}$ (2l1 – l2)
C.$\frac{l_{2} - 3l_{1}}{2}$
D.$\frac{l_{2} - l_{1}}{2}$
Solution
Sol. f = $\frac{\nu}{4(l_{1} + e)}$ (considering end correction)
Also f = $\frac{3\nu}{4(l_{2} + e)}$
⇒ $\frac{4(l_{2} + e)}{3}$ = 4(l1 + e)
⇒ l2 + e = 3l1 + 3e
⇒ l2–3l1 =2e
⇒ e = $\frac{l_{2} - 3l_{1}}{2}$
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