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pallavi pradeep bhardwaj Grade: 12
        

A particle of mass 10 gm is describing shm along a straight line with the period of 2sec and amplitude of 10 cm its kinetic energy when it is at 5 cm from its equillibrium position

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

Dear Pallavi


let equation of SHM is x=Xsinwt


wher w=2∏/T         and T= 2sec


X=10 cm


w=√k/m


where k is spring constant


k=w2m


  =(2∏/T)2m


   =∏2m


K.E at x=5 cm  +P.E at x=5  =P.E at x=10cm(extreme position)


  K.E at x=5 cm = 1/2 k 102  -  1/2 k52


                      =1/2 k(100 -25)


                      =1/2 ∏2m *75


      now solve





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Badiuddin

7 years ago
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