A tuning fork of frequency ' n ' is held near the open end of a tube which is dipped in water and length of the tube is adjusted until resonance occurs. If the two shortest lengths that prod

Question

A tuning fork of frequency ' n ' is held near the open end of a tube which is dipped in water and length of the tube is adjusted until resonance occurs. If the two shortest lengths that produce resonance are l 1 ​ and l 2 ​ , the speed of sound in air is (neglect end correction)

TestHub · Accepted Answer

For a resonance tube, the water level acts as a closed end, forming a closed organ pipe.

The first resonance occurs at length $l_{1} = \frac{λ}{4}$

The second resonance occurs at length $l_{2} = \frac{3 λ}{4}$

The difference between the two resonating lengths is:

$l_{2} - l_{1} = \frac{3 λ}{4} - \frac{λ}{4} = \frac{λ}{2}$

$\Rightarrow λ = 2 (l_{2} - l_{1})$

The speed of sound is given by $v = nλ$

Substituting the value of $λ$ , we get:

$v = 2 n (l_{2} - l_{1})$

Answer: $2 n (l_{2} - l_{1})$

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