The displacement of a particle executing simple harmonic motion is $y=\mathrm{A} \sin (2 \mathrm{t}+\phi) m$…
The displacement of a particle executing simple harmonic motion is $y=\mathrm{A} \sin (2 \mathrm{t}+\phi) m$, where $t$ is time in second and $\phi$ is phase angle. At time $t=0$, the displacement and velocity of the particle are 2 m and $4 \mathrm{~ms}^{-1}$. The phase angle, $\phi=$
$60^{\circ}$
$30^{\circ}$
$45^{\circ}$
$90^{\circ}$
Solution
$\begin{aligned} & \text {For } S H M, y=A \sin (2 t+\phi) \\ \therefore & v=2 A \cos (2 t+\phi) \\ \text {At } & t=0, y=2 m, v=4 ms^{-1}\end{aligned}$
$\therefore 2=a \sin (2 \times 0+\phi) \Rightarrow \sin \phi=\frac{2}{A}...(i)$
Also, $4=2 \mathrm{~A} \cos (0+\phi) \Rightarrow \cos \phi=\frac{2}{\mathrm{~A}}...(ii)$
From eqs (i) and (ii), we get
$\sin \phi=\cos \phi \Rightarrow \phi=45^{\circ}$