Here, at $t = 0, x = 1$ cm and $v = \omega\,\text{cm}\,\text{s}^{-1}; \omega = \pi\,\text{s}^{-1}$Given : $x = A \cos(\omega t + \phi)$ ... (i)Since at $t = 0, x = 1$, we get$1 = A \cos(\pi \times 0 + \phi) = A \cos \phi$The instantaneous particle velocity is given by$v = \frac{dx}{dt} = \frac{d}{dt}[A \cos(\omega t + \phi)]$ or $v = -A \omega \sin(\omega t + \phi)$Since at $t = 0, v = \omega\,\text{cm}\,\text{s}^{-1}$, we get$\omega = -A \omega \sin(\pi \times 0 + \phi)$or $A \sin \phi = -1$ ...(ii)Squaring and adding the equations (i) and (ii), we get$A^{2} \cos^{2} \phi + A^{2} \sin^{2} \phi = 1^{2} + (-1)^{2}$$A^{2} (\cos^{2} \phi + \sin^{2} \phi) = 2$ or $A^{2}(1) = 2$or $A = \sqrt{2}\,\text{cm}$Dividing the equation (ii) by (i), we get$\frac{A \sin \phi}{A \cos \phi} = \frac{-1}{1}$ or tan ϕ $= -1$ or $\phi = \frac{7\pi}{4}$When sine function is used to describe the simple harmonic motion.Here, $x = B \sin(\omega t + \alpha)$Since at $t = 0, x = 1$; we get$1 = B \sin(\pi \times 0 + \alpha)$or $B \sin \alpha = 1$ ...(iii)The instantaneous particle velocity is given by$v = \frac{dx}{dt} = \frac{d}{dt}[B \sin(\omega t + \alpha)]$$v = B \omega \cos(\omega t + \alpha)$Since at $t = 0, v = \omega\,\text{cm}\,\text{s}^{-1}$, we get$\omega = B \omega \cos(\pi \times 0 + \alpha)$or $B \cos \alpha = 1$ ...(iv)Squaring and adding the equations (iii) and (iv), we get$B^{2} \sin^{2} \alpha + B^{2} \cos^{2} \alpha = 1^{2} + 1^{2}$or $B^{2} (\sin^{2} \alpha + \cos^{2} \alpha) = 2$or $B^{2}(1) = 2$or $B = \sqrt{2}\,\text{cm}$Dividing the equation (iii) by (iv), we get$\frac{B \sin \alpha}{B \cos \alpha} = \frac{1}{1}$ or $\tan \alpha = 1$ or $\alpha = \frac{\pi}{4}$