Question

A closed tube contains air at 27oc. Calculate the first two successive resonance lengths using a tuning fork whose frequency is 341Hz (neglect the end correction). The speed of sound in air at 0oc is 330m/s

Answer 1

Answer:

l₁ = 25.3 cm

l₂ = 76.1 cm

Explanation:

Speed of sound in air at temperature T is given by:

$v=\sqrt{\frac{(gamma)RT}{M} }$

Since speed of sound is 330 at 0⁰C,

$(330)^{2}=\frac{(gamma)R(273)}{M} \\\\398.9= \frac{(gamma)R}{M}----(1)$

∴ Speed of sound at 27⁰C is:

$v'=\sqrt{\frac{(gamma)R(300)}{M} }----(2) \\\\v'=\sqrt{(398.9)(300)} [From (1) and (2)]\\ \\v'=345.93$

Now using v' at 27⁰C, using resonance condition for an open tube

$f=\frac{(2n+1)v'}{4l} \\\\l=\frac{(2n+1)v'}{4f}$

l₁= $\frac{345.93}{4(341)}$$=0.253 m$

l₂=$\frac{3(345.93)}{4(341)} =0.761m$

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