Two uniform strings `A and B` made of steel are made to vibrate under the same tension. If the first overtone of `A` is equal to the second overtone of `B` and if the radius of `A` is twice that of `B`, the ratio of the lengths of the strings is

Options

A. `2:01`
B. `3:04`
C. `3:02`
D. `1:03`

Answer

d

Explanation

Frequency of 1st overtone of A
`n_(1) = 2/(2l_(1)) sqrt(T/m)= 2/(l_(1)D_(1))sqrt(T/(pip))`
Frequency of 2nd overtone of B
`n_(2) = 3/(2l_(2)) sqrt(T/m)= 3/(l_(2)D_(2))sqrt(T/(pip))`
As `n_(1) = n_(2)` (resonance condition )
`therefore 2/(l_(1)D_(1))sqrt(T/m)=3/(l_(2)D_(2)) sqrt(T/(pip)) rArr (l_(1)D_(1))/(l_(2)D_(2)) = 2/3`
or, `l_(1)/l_(2)= 2/3xx D_(2)/D_(1)=2/3D_(2)/(2D_(2))=or l_(1)/l_(2)=1 : 3`