Two wires are fixed on a sonometer with their tensions are in the ratio 8:1, the length are in the ratio 36,35, the diameters in the ratio 4:1 and densities in the ratio 1:2. If the note of higher pitch has a frequency of 360Hz, then the frequency of other string will be

Explanation

For the two wires, we have
`n_(1) = (I)/(2l_(1)) sqrt(((T_(2))/(pi r_(2)^(2)p_(2)))) sqrt(((T_(1))/(pir_(1)^(2)p_(1))))`
`n_(2) = (I)/(2l_(1)) sqrt(((T_(2))/(pi r_(2)^(2)p_(2))))` `(l_(2))/(l_(1)) sqrt[[(T_(1))/(T_(2)) ((r_(2))/(r_(1)))^(2) (p_(2))/(p_(1))]]`
According to the given problem
`(l_(2))/(l_(1)) = (36)/(35) , (T_(1))/(T_(2)) = (8)/(1) , (r_(1))/(r_(2))`
Substituting these values in equation - (6.242)
`(n_(1))/(n_(2)) = (35)/(36) sqrt([((8)/(1))((1)/(16))((2)/(1))]] = (35)/(36)`
It is clear from equation that `n_(2)` has higher pitch so
`n_(2) = 360`
`(n_(1))/(360)` = `(35)/(36)`
`n_(1)` = `350 Hz`
Thus beat frequency is given as