The motion of a particle executing simple harmonic motion is described by the displacement function,$x(t) = A \cos (\omega t + \phi)$. If the initial $(t = 0)$ position of the particle is $1 \; cm$ and its initial velocity is $\omega \; cm/s$,what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi \; s^{-1}$. If instead of the cosine function,we choose the sine function to describe the $SHM$: $x = B \sin (\omega t + \alpha)$,what are the amplitude and initial phase of the particle with the above initial conditions?

(A) Initially,at $t = 0$:
Displacement,$x = 1 \; cm$
Initial velocity,$v = \omega \; cm/s$
Angular frequency,$\omega = \pi \; rad/s$
For $x(t) = A \cos (\omega t + \phi)$:
$1 = A \cos (\omega \times 0 + \phi) = A \cos \phi \implies A \cos \phi = 1 \dots (i)$
Velocity $v = \frac{dx}{dt} = -A \omega \sin (\omega t + \phi)$
$\omega = -A \omega \sin (\phi) \implies A \sin \phi = -1 \dots (ii)$
Squaring and adding $(i)$ and $(ii)$:
$A^2 (\cos^2 \phi + \sin^2 \phi) = 1^2 + (-1)^2 = 2$
$A = \sqrt{2} \; cm$
Dividing $(ii)$ by $(i)$:
$\tan \phi = -1 \implies \phi = \frac{3\pi}{4} \; rad$
For $x = B \sin (\omega t + \alpha)$:
$1 = B \sin (\alpha) \implies B \sin \alpha = 1 \dots (iii)$
Velocity $v = \frac{dx}{dt} = B \omega \cos (\omega t + \alpha)$
$\omega = B \omega \cos (\alpha) \implies B \cos \alpha = 1 \dots (iv)$
Squaring and adding $(iii)$ and $(iv)$:
$B^2 (\sin^2 \alpha + \cos^2 \alpha) = 1^2 + 1^2 = 2$
$B = \sqrt{2} \; cm$
Dividing $(iii)$ by $(iv)$:
$\tan \alpha = 1 \implies \alpha = \frac{\pi}{4} \; rad$