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  4. A tube of certain diameter and of length 48 cm is open at both ends. Its fundamental frequency of resonance is found to be 320 Hz. If velocity of sound in air is 320 ms-1 the diameter of the tube is

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Q. A tube of certain diameter and of length 48 cm is open at both ends. Its fundamental frequency of resonance is found to be 320 Hz. If velocity of sound in air is $320 ms^{-1}$ the diameter of the tube is

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Solution:

$\frac{\lambda}{2} = 48 + 2 ( 0.3 d)$
where .d is diameter of the tube
but $\lambda = \frac{V}{n} = \frac{320}{320} = lm = 100 cm$
$\therefore \, \frac{100}{2} = 48 + (0.6) \, d \, \Rightarrow \, d = \frac{2}{0.6} = 3.33 cm$