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)

Question

A pipe open at both ends has a fundamental frequency $f$ in air. The pipe is now dipped vertically in water drum to half of its length. The fundamental frequency of the air column is now equal to:

Answer 1

Step 1: Pattern of resonances:
Closed tube: $l = \frac\lambda4, \frac{3\lambda}4, \frac{5\lambda}4, \dots$. Neighbouring lengths differ by $\frac\lambda2$.

Step 2: Use the wave equation:
$v = n\lambda$ with $\lambda = 2(l_2 - l_1)$.

Step 3: Check with numbers:
If $n = 512$ Hz and $l_2 - l_1 = 0.33$ m, then $v = 2\times512\times0.33 = 338$ m/s, a sensible speed of sound.

Final Answer:
v equals 2n(l2 - l1), option (A). \[ \boxed{2n(l_2-l_1)} \]

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)

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The Correct Option is A

Solution and Explanation

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