The number of possible natural oscillations of air column in a pipe closed at one end of length $85 \,\,cm$ whose frequencies lie below $1250\,\, Hz$ are (Velocity of sound $= 340 \,\,m s^{-1}$)
AIPMT 2014,JEE MAIN 2014,AIIMS 2018, Medium
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Fundamental frequency of the closed organ pipe is
$v=\frac{v}{4 L}$
Here, $v=340 \mathrm{m} \mathrm{s}^{-1}, L=85 \mathrm{cm}=0.85 \mathrm{m}$
$\therefore \quad v=\frac{340 \mathrm{ms}^{-1}}{4 \times 0.85 \mathrm{m}}=100 \mathrm{Hz}$
The natural frequencies of the closed organ pipe will be
$v_{n} =(2 n-1) v=v, 3 v, 5 v, 7 v, 9 v, 11 v, 13 v, \ldots$
$=100 \mathrm{Hz}, 300 \mathrm{Hz}, 500 \mathrm{Hz}, 700 \mathrm{Hz}, 900 \mathrm{Hz}$
$1100 \mathrm{Hz}, 1300 \mathrm{Hz}, \ldots$ and so on
Thus, the natural frequencies lies below the $1250 \mathrm{Hz}$ is $6.$
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