If `lambda_(1), lambda_(2), lambda_(3)` are the wavelengths of the waves giving resonance in the fundamental, first and second overtone modes respecively in a open organ pipe, then the ratio of the wavelengths `lambda_(1) : lambda_(2) : lambda_(3)`, is :

Options

A. `1 : 3 : 5`
B. `1 : 2 : 3`
C. `5 : 3 : 1`
D. `1 : 1/3 : 1/5`

Answer

d

Explanation

For closed organ pipe,
`v=(2n+1)v/(4l)`
`therefore v/v=lambda = 4/((2n+1))`
n= 0 for fundamental mode
`lambda_(1)=4l`
Similarly n = 1 for lst overtone `lambda_(2)=(4l)/3`
n = 2 for second overtone,
Thus, `lambda_(1) : lambda_(2): lambda_(3)=1:1/3:1/5`