Two strings of the same length but different mass per unit length are used to suspend a rod whose density varies as `rho=rho_(0)[1+alphax]` as shown in the figure. The frequency of standing waves produced in the string is in the ratio `n : 1`, when both the strings are vibrating in their fundamental mode. If the mass per unit length of string `(l)` is `mu_(0)`, calculate the mass per unit length of the other string.

Explanation

`A_(0)e^(-lambdat)=N_(0)e^(-lambdat)`
& `y=L-x= (alphaL^(2)+3L)/(3[alphaL+2])`
`T_(1)L=Mgy` where `M=int_(0)^(L)arho_(0)[1+x]dx=(arho_(0)[2L+alphaL^(2)])/(2)`
`rArr T_(1)=(arho_(0)gL[3+alphaL])/(6)`, `T_(2)=(arho_(0)L[3+2alphaL])/(6)`
`V_(1)=sqrt((arho_(0)L[alphaL+3])/(6mu_(0)))`, `V_(2)=sqrt((arho_(0)gL[3+2alphaL])/(6mu_(2)))`
Dividing, `(V_(1))/(V_(2))=n=f_(1)//f_(2)`
`mu_(2)=(mu_(0)n^(2)[3+2alphaL])/(alphaL+3)`