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**jiboom** a particle of weight W is attached by 2 light inextensible strings each of length a to 2 fixed points,A and B, distant a apart in a horizontal line. write down tension in either string.

one of the strings is now replaced by an elastic string of the same natural length and it is found that in the new position of equilibrium this string has stretched to a length 5a/4. prove modulus of elasticity of this string is 7W/(rt(39)

first part is done.

for second part:

i have triangle with sides a,a,5a/4

let fixed points be A and B and weight at C so AB=BC=a AC=5a/4

using cos rule i find cos CAB =5/8 so

sin CAB=rt(39)/8

define PE=0 at the line AB so here pe=epe=ke=0

at equilibrium

h/(5a/4)=sin CAB

so h=5a(rt{39})/32}

hence PE=-(W5a(rt(39))/32

natural length of string =a

at equilibrium length =5a/4 so extension= a/4

giving

epe= $\displaystyle \frac{\lambda x^2}{2a}$=$\displaystyle \frac{\lambda a} {32}$

then PE+EPE=0 by conservation of energy

gives

$\displaystyle \lambda $=5rt(39)w

now this is miles from the answer so where have i gone wrong? I can get the answer by resolving,but this question was in energy chapter so i want to do it by energy method. IF this is not possible can someone explain why this problem is not doable via enrgy?