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Let it be displaced upwards by dy due to tension
Then change in length dl = (dx^{2}+dy^{2})  dx
= dx[1+(dy/dx)^{2}]^{1/2}  dx
Expanding binomially and neglecting higher order terms we get
dl = (1/2)(dy/dx)^{2}.dx
Hence the work done due to tension = T.dl
= (T/2)(y/x)^{2}.dx
since y/x = Akcos(wtkx) = (T/2)(Ak)^{2}cos^{2}(wtkx).dx
Integrating it over time period.
Best Wishes
ok, now i understand , this is the same cse as that of constrained motion , isn't it ? that's why  when the element is lifted by a distance dy then strings from both side will come into the picture and hence the acceleration of center of mass will be halved ????
am i thinking correct ??
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