In a resonance tube, the first resonance with tuning fork occurs at 16 cm and second at 49 cm. If the velocity of sound is 330 m/s, the frequency of tuning fork is

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Aicharya · Accepted Answer

For closed pipe, $l_1 = \frac{v}{4f}$, $l_2 = \frac{3v}{4f}$ Now, $v = 2f(l_2 - l_1)$ $\Rightarrow f = \frac{v}{2(l_2 - l_1)} = \frac{330}{2(0.49 - 0.16)} = 500	ext{ Hz}$

In a resonance tube, the first resonance with tuning fork occurs at 16 cm and second at 49 cm. If the…

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