Question

A pipe of length 85cm closed from one end. Find then number of possible natural oscillation of air column in pipe

Answer 1

Answer:

6

Step-by-step explanation:

f = v/λ=3404×0.85=100 Hz. ∴ Possible frequency = 100 Hz, 300 Hz, 500 Hz, 700Hz, 900 Hz, 1100 Hz below 1250 Hz. thus number of possible natural oscillations whose frequency is below 1250 is 6.