Question

$Y=A\sin (\omega t+{\varphi }_0)$is the time – displacement equation of an SHM, At t = 0, the displacement of the particle is $Y=\frac{A}{2}$and it is moving along negative x-direction. Then, the initial phase angle ${\varphi }_0$will be.

• A. $\frac{\mathrm{\pi }}{6}$
• B. $\frac{\mathrm{\pi }}{3}$
• C. $\frac{2\mathrm{\pi }}{3}$
• D. $\frac{5\mathrm{\pi }}{6}$

Answer 1

The correct option is A

$\frac{\mathrm{\pi }}{6}$

Step 1: Given data,

Displacement- time equation $Y=A\sin (\omega t+{\varphi }_0)\dots \left(i\right)$,

where $A$is the amplitude,

$\omega$ is the angular frequency,

$t$ is the time taken

${\varphi }_0$ is the phase constant

Step 2: Calculating the initial phase angle,

1. If the displacement of the oscillatory object can be described as a function of sine or cosine of an angle varying with time, the motion is said to be harmonic.
2. SHM is a type of oscillatory motion in which the restoring backward force is directly proportional to the displacement from the mean position and in the opposite direction of the displacement.

At $t=0,Y=\frac{A}{2}$, substituting these values in the equation $\left(i\right)$ we get

$$\begin{gathered} Y=\frac{A}{2}=A\sin (0+{\varphi }_0)=A\sin {\varphi }_0 \\ \Rightarrow \sin {\varphi }_0=\frac{1}{2} \\ \Rightarrow {\varphi }_0=\frac{\mathrm{\pi }}{6} \end{gathered}$$

Hence, option (A) is correct.