Discussion Forum

A pipe of length 85 cm is closed from one end. Find the number of possible  natural oscillations  of air column in the pipe whose frequencies lie below 1250 Hz . The velocity of sound in air is 340 m/s.

Answer:

Explanation:

For closed organ pipe 

     = $\frac{(2n+1)v}{4l}$         [n=0,1,2,...]

$\frac{(2n+1)v}{4l}<1250$

  $(2n+1)<1250\times\frac{4\times0.85}{340}$

  $(2n+1)<1.25\Rightarrow 2n<11.50$

  n<5.25

 n=0,1,2,3,...5

 we have 6 possibilities