So give the complete and correct question if you please.
Let it be displaced upwards by dy due to tension
Then change in length dl = (dx2+dy2) - 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
since y/x = -Akcos(wt-kx) = (T/2)(Ak)2cos2(wt-kx).dx
Integrating it over time period.
So tell me the steps how u came to this answer then only I can help u.
This happens only when a is small.
here dy/dx = -Akcos(wt-kx) is not small everwhere.
Hope u got it.
Actually u have taken work to be T Sina * dY
but tension is lifting the centre of mass by dy/2.
Your thinking is good and i praise it.
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 ??
when u straighten it up vertically then its lower remain as it is but the upper part is displaced by the lenght of the rod in the vertical direction.
And the centre of mass is at height of length/2.