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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
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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