Question

The number of possible natural oscillations of an air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are (velocity of sound = 340 ms^-1):

(a) 4

(b) 5

(c) 7

(d) 6

Answer 1

Correct option is : 6

For a closed pipe at one end,

\(f_n = n(\frac{v}{4l})\) here n is an odd number

\( = n(\frac{340}{4\times85\times10^{-2}}) = n (100)\)

Here, n is an odd number, so for the given condition, n can go upto n = 11

i.e., n = 1, 3, 5, 7, 9, 11

So, number of possible natural oscillations could be 6, which are below 1250 Hz.