Question
The number of possible natural oscillation of air column in a pipe closed at one end of length $$85\,cm$$ whose frequencies lie below $$1250\,Hz$$ are: (velocity of sound $$ = 340\,m{s^{ - 1}}$$ )
A.
4
B.
5
C.
7
D.
6
Answer :
6
Solution :
In case of closed organ pipe frequency,
$$\eqalign{
& {f_n} = \left( {2n + 1} \right)\frac{v}{{4l}} \cr
& {\text{for}} \cr
& n = 0,{f_0} = 100\,Hz \cr
& n = 1,{f_1} = 300\,Hz \cr
& n = 2,{f_2} = 500\,Hz \cr
& n = 3,{f_3} = 700\,Hz \cr
& n = 4,{f_4} = 900\,Hz \cr
& n = 5,{f_5} = 1100\,Hz \cr
& n = 6,{f_6} = 1300\,Hz \cr} $$
Hence possible natural oscillation whose frequencies $$ < 1250\,Hz = 6\left( {n = 0,1,2,3,4,5} \right)$$