Find the ratio of the fundamental tone frequencies of two identical strings after one of them was stretched by `eta_(1)=2.0 %` and the other, by `eta_(2)=4.0 %`. The tension is assumed to be proportional to the elongation.

Explanation

We have `n=(1)/(n l ) sqrt((T)/(m))=(1)/( 2l)sqrt((Tl)/( M))` Where `M=` total mass of the wire. When the wire is streched,total mass of the wire remains consta. For the first wire the new length `=l+ eta_(1)l` and for the record wire, the length is `l=eta_(2)l`. Also `T_(1)=alpha(eta_(1)l)` where `alpha` is a constant and `T_(2)=alpha( eta_(2)l)`. Subtituting in the above formula.
`v_(1)=(1)/( 23(l+eta_(1)l))sqrt(((alphaeta_(1)l)(l + eta_(1)l))/( M))`
`v_(1)=(1)/( 23(l+eta_(2)l))sqrt(((alphaeta_(2)l)(l + eta_(2)l))/( M))`
`:. (v_(2))/(v_(1))=(1+eta_(1))/( 1+eta_(2))sqrt((eta_(2))/( eta_(1)).(1+eta_(2)l)/(1+eta_(1)))`
`(v_(2))/( v_(1))=sqrt((eta_(2)(1+eta_(1)))/(eta_(2)(1+eta_(2))))=sqrt((0.04(1+0.02))/(0.02(1+0.04)))=1.4`