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

At positlon, t = 0,

The given function is $x(t)=A \cos (\omega t+\phi)…..(1)$

$\begin{array}{l}
1=A \cos (\omega \times 0+\phi)=A \cos \phi \\
A \cos \phi=1
\end{array}$

Differentiating equation (1) with respect to t, we get,

Velocity, $\mathrm{V}=\mathrm{dx} / \mathrm{dt}$

$\begin{array}{l}
v=-\mathrm{A} \omega \sin (\omega t+\phi) \\
\mathrm{t}=0 \text { and } \mathrm{v}=\omega \\
1=-\mathrm{A} \sin (\omega \times 0+\phi)=-\mathrm{A} \sin \phi \\
\mathrm{A} \sin \phi=-1
\end{array}$

On Squaring and adding equations (2) and (4), we get,

$\begin{array}{l}
A^{2}\left(\sin ^{2} \phi+\cos ^{2} \phi\right)=1+1 \\
A=\sqrt{2} \mathrm{~cm}
\end{array}$

Dividing equation (4) by equation (2),
$(A \sin \phi / A \cos \phi)=-1 / 1$
$\tan \phi=-1$

$\begin{array}{l}
\Rightarrow \phi=3 \pi / 4,7 \pi / 4 \\
\text { If sine function is used } \\
\mathrm{x}=\mathrm{B} \sin (\omega \mathrm{t}+\mathrm{a}) \\
\text { At } \mathrm{t}=0, \mathrm{x}=1 \text { and } \mathrm{v}=\omega \text { we get } \\
1=\mathrm{Bsin}(\omega \times 0+\alpha]=1+1 \\
\mathrm{~B} \text { sina }=1-(5) \\
\text { Velocity, } \mathrm{v}=\mathrm{dx} / \mathrm{dt}=\mathrm{B} \omega \cos (\omega t+\alpha) \\
\text { taking } \mathrm{v}=\omega \\
1=\mathrm{Bcos}(\omega(0)+\mathrm{a})=\mathrm{B} \cos \alpha-(6)
\end{array}$

Squaring and adding equations(5) and (6), we get:

$\begin{array}{l}
B^{2}\left[\sin ^{2} \alpha+\cos ^{2} \alpha\right]=1+1 \\
B^{2}=2 \\
B=\sqrt{2} \mathrm{~cm}
\end{array}$

Dividing equation (5) by equation (6), we get: $B \sin \alpha / B \cos \alpha=1 / 1$
$\tan \alpha=1$

As a result, $\alpha=\pi / 4,5 \pi / 4, \ldots \ldots$