11Since its symmetrical Each mass will move x/2 distance
Therefore the total increase in potential energy is kx2/2
But The potential energy is due to both the mass therefore the potential energy due to each mass is kx2/4
Therefore work done on the mass is -kx2/4
1but if we use following W= ∫[o to x/2] -kxdx we get W=-kx2/8................why??????
11See if by Mass 1 x/2 distance is pulled.
Then for mass 2 The integration will be from x/2 to x since it the spring is already pulled x/2 by mass 1. Due to this reason we integrate it from 0 to x and divide the work done equally among them
1no............din get as u din xplain it [1]...........wat u r saying i got ..........did u try calculating values.............did u get same values..........[3]..............then???????? actually u din xplain this u rewritten wat i wrote in first two lines[1] and in next u provided me an alternate to find answer.........that won't do for me [1]...........[1].............so xplain wat i asked!!!!!!
11See you cannot integrate from 0 to x/2 since Both the masses are moving simultaneously and therefore integrating from 0 to x/2 is wrong.
1YUP..........I AGREE SIR JI [1]......................JUST WANTED U TO CONFIRM..............TO BE 100% SURE!!!!!!
THANX
