A uniform wire of linear density 0.004 per kg-m, when stretched between two rigid supports, with a tension `3.6 xx 10^(2)` N, resonates with a frequency of 420 hz. The next harmonic frequency with which the wire resonates is 490 Hz. The length of the wire in metre is

Options

A. 1.41
B. 2.41
C. 2.14
D. 3.14

Answer

C

Explanation

First frequency of vibration is
` v_(1) = 2/(2l) .sqrt(T/mu)`
New harmonic is
` v_(2) = (n+1)/(2l) .sqrt(T/mu)`
Form Eqs (i) and (ii)
` (v_(1))(2l) xx sqrt(mu/T) = (v_(2)) (2l) sqrt(T/mu)-1`
` Rightarrow (420) (2l) xx sqrt((0.004)/(3.6 xx 10^(2)))`
` = (490)(2l)sqrt((0.004)/(3.6 xx 10^(2))) -1 `
`Rightarrow (2l)xx 2/10 [420/60 -490/60]=-1`
` Rightarrow (4l)/10 [ 7 - 49/6] =-1`
` Rightarrow l = (10(-1))/(4(-7)) = 10/28 or l=2.8 m`