A wire, kept between two bridges 25 cm apart, is stretched through a linear expansion of 0.04 cm . Find the fundamental frequency of vibration of the wire. Given, the density and the Young's modulus of the material of the wire are 10 `g*cm^(-3)` and `9xx 10^(11) dyn * cm^(-2)` , respectively.

Explanation

Longitudinal strain ` = ("linear expansion")/("original length") = (x)/(l)`
Longitudinal stress ` = ("tension in the wire")/("area of cross-section") = (T)/(alpha)`
So, Young.s modulus ,
`Y = ("longitudinal stress")/("longitudinal strain") = ((T)/(a))/((x)/(l))=(Tl)/(ax)`
or , ` T = (Yalpha x)/(l)`
Mass per unit length of the wire, m = density ` xx ` area of cross section = `rho alpha`
`:.` Fundamental frequency.
`n = (1)/(2l) sqrt((T)/(m))=(1)/(2l)sqrt((Yalphax)/(l)*(1)/(rhoalpha))=(1)/(2l),sqrt((Yx)/(lrho))`
`=(1)/(2 xx 25) xx sqrt(((9xx 10^(11))xx 0.04)/(25 xx 10))=240` Hz.