A particle of mass 100 g is executing S.H.M. with amplitude of 10 cm. When the particle passes through the mean position at t = 0. Its kinetic energy is 8 mJ. The equation of simple harmonic motion, if initial phase is zero is
A. x = 0.1 sin 4t
B. x = 0.1 cos 4t
C. x = 0.1 sin 2t
D. x = 0.1 cos 2t

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Correct Answer - A
`K.E.=(1//2)momega^(2)A^(2)`
`10^(-3)xx8=(1)/(2)xx0.1xx0.01omega^(2)`
`therefore omega^(2)=16`
`omega=4`
`therefore x=0.1sinomegat`

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