A particle is performing S.H.M. Its total energy ie E When the displacement of the particle is half of its amplitude, its K.E. will be

Options

A. E
B. E/4
C. E/2
D. E/4

Answer

B

Explanation

`KE(1)/(2)momega^(2)(A^(2)-y^(2))`
`KE=(E)/(A^(2))(A^(2)-y^(2))"("becauseE=(1)/(2)momega^(2)A^(2)")"`
`EA=(E)/(A^(2))(A^(2)-(A^(2))/(4))`
`=(3E)/(4)`