Animated Solution for Physics - Oscillations: $Y=A\sin (\omega t+{\varphi }_0)$ is the time-displacement equation of 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

Imagine a particle executing Simple Harmonic Motion (SHM). If I tell you that the particle is currently at a displacement of $Y=\frac{A}{2}$, do you know exactly what it is doing? The answer is no!

In any oscillatory motion, a particle crosses every non-extreme position twice during a single complete cycle. It crosses $Y=\frac{A}{2}$ once while moving away from the mean position towards the positive extreme, and it crosses it again while returning from the positive extreme towards the mean position.

This means that a single displacement value corresponds to two different phase angles. To pinpoint the exact state of the particle, we need a second piece of information: the direction of its velocity.

Let's translate this physical reality into mathematics. We are given the time-displacement equation:

We are told that at the very beginning of our observation, $t=0$, the displacement is $Y=\frac{A}{2}$. Let's substitute these initial conditions into our master equation:

The amplitude $A$ beautifully cancels out from both sides, leaving us with a simple trigonometric equation:

Now, we must ask ourselves: which angles in the first cycle $[0,2\pi ]$ have a sine value of $\frac{1}{2}$? There are exactly two possibilities: 1. ${\varphi }_0=\frac{\pi }{6}$ (or $30^{\circ }$), which lies in the first quadrant. 2. ${\varphi }_0=\frac{5\pi }{6}$ (or $150^{\circ }$), which lies in the second quadrant.

Both of these angles satisfy the displacement condition. But which one is the true initial phase?

This is where the second condition from the problem comes to our rescue: the particle is moving in the negative direction. This implies that its velocity must be less than zero ($v<0$).

To find the velocity, we differentiate our displacement equation with respect to time:

Now, let's test our two candidate angles at $t=0$:

Case 1: If ${\varphi }_0=\frac{\pi }{6}$ Since $\frac{\sqrt{3}}{2}$ is positive, the velocity is positive ($v>0$). This means the particle is moving in the positive direction. This contradicts our given condition.

Case 2: If ${\varphi }_0=\frac{5\pi }{6}$ Here, the cosine yields a negative value, making the velocity negative ($v<0$). The particle is moving in the negative direction, which perfectly matches the problem statement!

Therefore, the correct initial phase angle is ${\varphi }_0=\frac{5\pi }{6}$.

While the calculus approach is rigorous, you can solve this instantly using a Phasor Diagram.

Imagine a reference circle of radius $A$. The projection of a rotating vector (phasor) on the Y-axis represents our SHM. If you draw a horizontal line at $Y=\frac{A}{2}$, it cuts the circle at two points: one in the 1st quadrant ($30^{\circ }$) and one in the 2nd quadrant ($150^{\circ }$).

Because the phasor always rotates counter-clockwise, the point in the 1st quadrant is moving upwards (positive velocity), while the point in the 2nd quadrant is moving downwards (negative velocity). Without writing a single derivative, you can visually confirm that the 2nd quadrant angle, $\frac{5\pi }{6}$, is the only correct answer!