Question
The motion of a particle executing simple harmonic motion is described by the displacement function, $x ( t )= A$ $\cos \left(\omega_t+\omega\right)$.
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 $SHM : x = B \sin (\omega t+a)$, what are the amplitude and initial phase of the particle with the above initial conditions.

Answer

Displacement of the particles is given by -
$
y=a \sin w t
$
Here $a$ is the amplitude of oscillation and $2 \pi / \omega$ is the period of oscillation and $\omega$ is the angular frequency of the wave.
Initially, at $t =0$ :
Displacement, $x =1 cm$
Initial velocity, $v =\omega cm / sec$.
Angular frequency, $\omega=\pi rad / s ^{-1}$
It is given that:
$
\begin{aligned}
& x(x)=A \cos (\omega t+\phi) \\
& 1=A \cos (\omega \times 0+\phi)=A \cos \phi \\
& A \cos \phi=1 \ldots \text { (i) }
\end{aligned}
$
Velocity, $v=\frac{d x}{d t}$
$
\begin{aligned}
& \omega=-A \omega \sin (\omega t+\phi) \\
& 1=A \sin (\omega \times 0+\phi)=A \sin \phi \\
& A \sin \phi=-1 \ldots \text {...(ii) }
\end{aligned}
$
Squaring and adding equations (i) and (ii), we get:
$
\begin{aligned}
& A^2\left(\sin ^2 \phi+\cos ^2 \phi\right)=1+1 \\
& A^2=2 \\
& \therefore A=\sqrt{2} cm
\end{aligned}
$
Dividing equation (ii) by equation (i), we get:
$
\begin{aligned}
& \tan \phi=-1 \\
& \therefore \phi=\frac{3 \pi}{4}, \frac{7 \pi}{4}, \ldots \ldots
\end{aligned}
$
SHM is given as:
$
x=B \sin (\omega t+a)
$
Putting the given values in this equation, we get:
$
1=B \sin [\omega \times 0+a]
$
B $\sin a =1 \ldots$ (iii)
Velocity, $v=\omega B \cos (\omega t+a)$
Substituting the given values, we get:
$
\pi=\pi B \sin a
$
B $\sin a =1$...(iv)
Squaring and adding equations (iii) and (iv), we get:
$
\begin{aligned}
& B^2\left[\sin ^2 a+\cos ^2 a\right]=1+1 \\
& B^2=2 \\
& \therefore B=\sqrt{2} cm
\end{aligned}
$
Dividing equation (iii) by equation (iv), we get:
$
\begin{aligned}
& \frac{B \sin a}{B \cos a}=\frac{1}{1} \\
& \tan a=1=\tan \frac{\pi}{4} \\
& a=\frac{\pi}{4}, \frac{5 \pi}{4}, \ldots \ldots
\end{aligned}
$

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