A open pipe of length `l` is vibrating in 3rd overtone with maximum amplitude A. The amplitude at a distance of `(l)/(16) ` from any open end is

Options

A. A
B. 0
C. `(A)/(sqrt(2))`
D. `(sqrt(3A))/(2)`

Answer

C

Explanation

For open organ pipe wavelength of `n^(th)` overtone
`lambda_(n)= (2l)/(n+1) (n=3)`
`lambda = (2l)/(4)= (l)/(2)`
As pipe is open so antinode ( maximum amplitude ) will form at open ends . At l=0 amplitude = A
At `(l)/(16) or (lambda)/(8)` distance amplitude R = A cos `phi` where `phi` = phase angle .
`phi = (2pi)/(lambda) xx(lambda)/(8)= (pi)/(4)= 45^(@)`
R = A cos `45^(@)`
`R = (A)/(sqrt(2))`