The displacement-time equation of a particle executing SHM is `A-Asin(omegat+phi)`. At tme t=0 position of the particle is x=`A/2` and it is moving along negative x-direction. Then the angle `phi` can be

Options

A. `pi/6`
B. `(pi)/3`
C. `(2pi)/3`
D. `(5pi)/6`

Answer

To find the angle \( \phi \) in the displacement-time equation of a particle executing Simple Harmonic Motion (SHM), we start with the given equation:

\[ x = A - A \sin(\omega t + \phi) \]

At time \( t = 0 \), the position of the particle is given as \( x = \frac{A}{2} \). Therefore, we can substitute \( t = 0 \) into the equation:

\[ \frac{A}{2} = A - A \sin(\phi) \]

Now, we can simplify this equation:

1. Rearranging the equation gives:
\[ A \sin(\phi) = A - \frac{A}{2} \]

2. Simplifying the right side:
\[ A \sin(\phi) = \frac{A}{2} \]

3. Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \sin(\phi) = \frac{1}{2} \]

The sine function equals \( \frac{1}{2} \) at two angles in the range of \( 0 \) to \( 360 \) degrees: \( 30^\circ \) and \( 150^\circ \).

Next, we need to determine which of these angles is valid given the condition that the particle is moving in the negative x-direction.

4. Since the particle is at \( x = \frac{A}{2} \) and moving towards the mean position (which is at \( x = 0 \)), it must be in the second quadrant (where the sine is positive and the cosine is negative). Therefore, the angle \( \phi \) must be:
\[ \phi = 180^\circ - 30^\circ = 150^\circ \]

Thus, the angle \( \phi \) can be expressed in radians as:
\[ \phi = \frac{5\pi}{6} \]

So the final answer is:
\[ \phi = 150^\circ \text{ or } \frac{5\pi}{6} \text{ radians} \]