Question

Y = A sin($\omega$t+$\phi_0$) is the time-displacement equation of a 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 $\phi_0$ will be:

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

Hint:

When determining the initial phase in SHM, the displacement information gives two possible angles. The velocity information (direction of motion) is crucial to select the correct angle from the two possibilities. A positive velocity implies the cosine of the phase is positive (quadrants I or IV), while a negative velocity implies the cosine is negative (quadrants II or III).

Answer 1

The Correct Option is B

The displacement equation is given by $Y(t) = A \sin(\omega t + \phi_0)$.
At $t=0$, the displacement is $Y(0) = A/2$.
Substituting into the equation: $A/2 = A \sin(0 + \phi_0) \implies \sin(\phi_0) = 1/2$.
This condition is satisfied for $\phi_0 = \frac{\pi}{6}$ and $\phi_0 = \frac{5\pi}{6}$ in the range $[0, 2\pi]$.
The velocity of the particle is given by the derivative of the displacement:
$v(t) = \frac{dY}{dt} = A\omega \cos(\omega t + \phi_0)$.
At $t=0$, the particle is moving in the negative direction, which means $v(0)<0$.
$v(0) = A\omega \cos(\phi_0)<0$.
Since A and $\omega$ are positive, this requires $\cos(\phi_0)<0$.
Now we check the two possible values for $\phi_0$:
If $\phi_0 = \frac{\pi}{6}$, then $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$, which is positive. This is not the correct phase.
If $\phi_0 = \frac{5\pi}{6}$, then $\cos(\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2}$, which is negative. This matches the condition.
Therefore, the initial phase angle is $\phi_0 = \frac{5\pi}{6}$.