1.

A particle ofmass 0.1 kg is executing S.H.M. with amplitude 0.1 m. At the mean position its K.E. is 8xx10^(-3) J. If the intial phase is 45^(@). What is the equation of its motion ?

Answer»

`y=0.1 sin(4t)`
`y=0.1 sin PIT`
`y=0.1sin[4t+(pi)/(4)]`
`y=0.1sin[4t-(pi)/(4)]`.

Solution :Here
`(1)/(2)m omega^(2)A^(2)=8xx10^(-3)" J"`
or `(1)/(2)xx(0.1)^(2)xx0.1xx omega^(2)=8xx10^(-3)`
`omega^(2)=16` or `omega=4` rad `s^(-1)`
and `phi=(+pi)/(4)`or `y=0.1sin(4t+(pi)/(4))`.
Correct CHOICE is (b).


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