The number of possible natural oscillations of air column in a pipe closed at one end of length 85 c m whose frequencies lies below 1250 H z are: (velocity of sound = 340 m s - 1 )NEET 2014

Question

TestHub · Accepted Answer

Fundamental frequency of a closed organ pipe is $f_{1} = \frac{V}{4 l} = \frac{340}{4 \times 0.85} = 100 H z$
The natural frequencies of the organ pipe will be $f = 100 H z, 300 H z, 500 H z, 700 H z, 900 H z, 1100 H z$ which are below $1250 H z$

Physics - Sound Wave Question with Solution | TestHub

Options:

Doubts & Discussion