A light elastic string AB has natural length l and modulus of elasticity 2mg. A second light elastic string CD has natural length l and modulus of elasticity 4mg. The end B of the first string is connected to the end C of the other, and the end A is attached to a fixed point. A particle of mass m is suspended from end D and is at rest in equilibrium. Find the length AD.
My Working:
l of AB = l of CD = l
1/(combined spring constant)=1/k(AB)+1/k(CD)
k=lambda/l
Let lambda equal the combined modulus of elasticity
1/k=l/lambda=l/4mg+l/2mg
l cancels
1/lambda=1/4mg+1/2mg
1/lambda=1/4mg+2/4mg=3/4mg
lambda=4mg/3
Resolving vertically:
Let T be the combined tensions in the spring combination
T-mg=0; T=mg
mg=(4mg/3)*x/2l
mg=4mgx/6l
6lmg=4mgx
mg cancels
6l=4x
x=1.5l
Therefore the total length is 2l+1.5l=3.5l
Markscheme says 11l/4. How come?