A particle performs harmonic oscillations along the  x– axis about the equilibrium position $x=0$ .The oscillation frequency is $\omega =4.00 s^{-1}$.At a certain moment of time the particle has a coordinate $x_0=25.0 cm$ and its velocity is equal to$v_{x_0}=100 cm/s.$Then at $t=2.40 s$ after that moment ,

Suppose

 $$\begin{gathered} x=a \cos \left(\omega t+\alpha \right) \\ \therefore v_x=\frac{dx}{dt} \\ =-\omega a\sin \left(\omega t+\alpha \right) \end{gathered}$$

Given at

 $$\begin{gathered} t=0, \\ {x}_0=25 cm \\ and v_{x0}= 100 cm/s \\ \therefore x_0=a\cos \alpha ...\left(i\right) \\ and v_{x0}=-\omega a\sin \alpha \\ \therefore \frac{v_{x0}}{\omega }=-a\sin \alpha ...\left(ii\right) \end{gathered}$$

From equations, $\left(i\right)$ and $\left(ii\right)$,we have

 $$\begin{gathered} a=\sqrt{{x_0}^2+{\left(\frac{v_{x0}}{\omega }\right)}^2} \\ and \tan \alpha =-\left(\frac{v_{x0}}{{\omega }_{x0}}\right) \end{gathered}$$

After substituting the values, we get $x=-29 cm and v_x=-81 cm$