Question

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

Answer 1

The natural frequencies of the closed organ pipe will be

Vn = (2n -1)v = v , 3v ,5v ,7v ......so on.

Explanation:

Fundamental frequency of the closed organ pipe "v" = v / 4 L

Here v = 340 m/s

L = 85 cm = 0.85 m

v = 340 / 4 x 0.85

v  = 100 Hz

The natural frequencies of the closed organ pipe will be

Vn = (2n -1)v = v , 3v ,5v ,7v ......

                    = 100 Hz, 300 Hz, 500 Hz, 700 Hz ......

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