Discussion Forum

A pipe closed at one end has a length of 83 cm. The number of possible natural oscillations of air column whose frequencies lie below 1000 Hz are (take, the velocity of sound in air =332 m/s)

Answer:

Explanation:

Fundamental freequency of one  end closed pipe,

 $v=\frac{nv}{4L}$

 Here, n=1

 or   $v_{1}=\frac{332}{4\times 83\times 10^{-2}}$

 or v₁  = 100

As  odd harmonics alone are produced in closed organ  pipe, therefore  possible frequencies are

    3v₁ =300 Hz. 5 v₂ =500 Hz

  7 v₂ = 700 hz , 9 v₂ =900 Hz

 Hence, the number of the possible natural oscillation of air column is 5