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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

pallavi pradeep bhardwaj , 15 Years ago
Grade 12
anser 1 Answers
Badiuddin askIITians.ismu Expert

Last Activity: 15 Years ago

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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