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=8xx10^(-3)J, phi_(0)=45^(@)=(pi)/(4)rad. `
Kinetice energy is SHM is
`K=(1)/(2)momega^(2)(a^(2)-y^(2))`
when, `y=0,K=8xx10^(-3)J`
`:. 8xx10^(-3)=(1)/(2)xx0.1xxomega^(2)(0.1^(2)-0^(2))`
or `omega =4rad//s`
Equation of SHM is `y=asin[omegat+phi__(0)]`
or `y=0.1sin[4t+pi//4]`