A particle performs harmonic oscillations along the `x` axis about the equilibrium position `x=0`. The oscillation frequency is `omega=4.00s^(-1)`. At a certain moment of time the particle has a coordinate`x_(0)=25.0 cm` and its velocity is equal to `v_(x0)=100cm//s`.
Find the coordinate `x` and the velocity `v_(x)` of the particle `t=2.40s` after that moment.

Explanation

Let the general equationof S.H.M. be
`x=a cos (omegat+alpha) ………..(1)`
So, `v_(x)=-a sin (omegat+alpha) ……(2)`
Let us assume that at `t=0, x=x_(0)` and `v_(x)=v_(x_(0)). `
Therefore `tan alpha=-(v_(x_(0)))/(omega x_(0))` and `a=sqrt(x_(0)^(2)+((v_(x_(0)))/(omega))^(2))=35.35 cm`
Under our assumption Equns `(1)` and `(2)` give the sought `x` and `v_(x)` if
`t=t=2.40s, a =sqrt(x_(0)^(2)+(v_(x_(0))//omega)^(2))` and `alpha=tan^(-1)(-(v_(x))/(omegax_(0)))=-(pi)/(4)`
Putting all the given numberical values, we get `:`
`x=-29cm `and `v_(x)=-81 cm//s`