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Question

Physics Question on Oscillations

A closed organ pipe has length I. The air in it is vibrating in 3rd overtone with a maximum amplitude of A. Find the amplitude at a distance of l14\frac{l}{14} from closed end of the pipe

A

A

B

zero

C

A2\frac{A}{\sqrt2}

D

32A\frac{\sqrt3}{2}A

Answer

A2\frac{A}{\sqrt2}

Explanation

Solution

n=7ν4l;λ=4l7Y=Acoskxsinωtn = \frac{7\nu}{4l} ; \lambda = \frac{4l}{7} \, Y = A \, cos \, k \, x \, sin \, \omega t amplitude = AcosKxA \, cos \, Kx