The `p^(th)` overtone of an organ pipe open at both ends has a frequency `n_(1)` . What one end of the pipe is closed , its `q^(th)` overtone has a frequency `n_(2)`. What is the value of `n_(1)/n_(2)` ?

Options

A. `(2q + 1)/(2(p + 1))`
B. `(p+1)/(2(2q + 1)) `
C. `(2 ( p + 1))/(2q + 1)`
D. `(2(2q + 1))/(p + 1) `

Answer

C

Explanation

L is the length of the pipe and v is the speed of sound in air . When the pipe is open at both ends, the fundamental frequency ` N = v/(2L)`
and the frequency of the `p^(th)` overtone `n_(1) = (p + 1) N`
` :. N_(1) = (p + 1) v/(2L) `
` :. v/(2L) = n_(1)/(p + 1) ` ...(1)
Similarly when the pipe is closed at one end, the fundamental frequency `N' = v/(4L) `
as only odd harmonics are heard,the frequency of the `q^(th)` overtone `n_(2) = (2q + 1) v/(4L)`
` :. n_(2) = ((2q + 1))/2 (v/(2L))`
` :. " using " (1) , n_(2) = ((2q + 1)/2) (n_(1)/(p + 1))`
` :. n_(1)/n_(2) = (2(p+1))/((2q + 1))`