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. `1:2`
B. `1:3`
C. `1:4`
D. `1:6`

Answer

D

Explanation

The frequency of vibration of a string is `upsilon=(p)/(lD)=sqrt((T)/(pirho))` where D is the diameter of string, l is the length of string, T is the tension in the string, `rho` is density of material of the string and p is the number of loops.
For first overtone (`2^(nd)` harmonic) p = 2 and for second overtone (`3^(rd)` harmonic) p = 3
`therefore upsilon_(A)=(2)/(l_(A)D_(A))sqrt((T)/(pirho))andupsilon_(B)=(3)/(l_(B)D_(B))sqrt((T)/(pirho))`
As `upsilon_(A)=upsilon_(B)` (Given)
`therefore (2)/(l_(A)D_(A))sqrt((T)/(pirho))=(3)/(l_(B)D_(B))sqrt((T)/(pirho))`
or `(2)/(l_(A)D_(A))=(3)/(l_(B)(D_(A))/(2))or(l_(A))/(l_(B))=(2)/(6)=(1)/(3)(...D_(A)=2D_(B))`