A point particle if mass `0.1 kg` is executing SHM of amplitude `0.1 m`. When the particle passes through the mean position, its kinetic energy is `8 xx 10^(-3)J`. Write down the equation of motion of this particle when the initial phase of oscillation is `45^(@)`.

Explanation

Here, `m=0.1kg,A=0.1m,`
`(K.E.)_(max)=8xx10^(-3)J, phi_(0)=45^(@)=(pi)/(4)rad`
KE at the mean position =Max. KE
`=(1)/(2)momega^(2)A^(2)`
`8xx10^(-3)=(1)/(2)xx0.1xxomega^(2)xx(0.1)^(2)`
or `omega^(2)=16 or omega=4rads^(-1)`
The equation of motion for the particle is
`y=Asin(omegat)+phi_(0)=0.1sin(4t+pi//4)`