Let the displacement of the particle at any time t be represented by $x = A \cos (\omega t + \phi_0)$...(i) where A = amplitude, $\phi_0 =$ initial phaseIf $v$ be the velocity of the particle at time t, Then At $t = 0, x = x_0$ and $v = v_0$By putting t = 0, from (i) and (ii), we get$x_0 = A \cos \phi_0$, $v_0 = -A \cos \phi_0$ $\Rightarrow v_0 = -\omega \sqrt{(A \sin \phi_0)^2}$$= -\omega \sqrt{A^2 (1 - \cos^2 \phi_0)}$ $= -\omega \sqrt{A^2 - A^2 \cos^2 \phi_0}$$= -\omega \sqrt{A^2 - x_0^2} \ldots (iii)$Equation (iii) shows that initial velocity is negative. Squaring on both sides of equation (iii), we get$v_0^2 = \omega (A^2 - x_0^2)$ ; $A^2 - x_0^2 = \frac{v_0^2}{\omega^2}$$\Rightarrow A^2 = x_0^2 + \frac{v_0^2}{\omega^2};$$A = \sqrt{x_0^2 + \frac{v_0^2}{\omega^2}}$