Young moduli of two rods of equal length and equal cross section and `Y_1` and `Y_2`. These rods are joined end to end forming a composite rod-system. Prove that the equivalent Young's modulus of the composite system of rods `=(2Y_1Y_2)/(Y_1+Y_2)`

Explanation

Let the applied force F, elongations of the two rods be `l_1`and `l_2` respectively [Fig.1.22]
`thereforel_1=(FL)/(Y_1A),l_2=(FL)/(Y_2A)`
A = area of cross section of each rod, L= original length of each rod
Now, equivalent Young.s modulus of the composite rod system.
`Y=(F/A)/((l_1+l_2)/(2L))=(2FL)/(A(l_1+l_2))`
=`(2FL)/(A[(FL)/(Y_1A)+(FL)/(Y_2A)])`
`=2/(1/Y_1+1/Y_2)=(2Y_1Y_2)/(Y_1+Y_2)`