Passage V) In SHM displacement, velocity and acceleration all oscillate simple harmonically with same angular frequency `omega`. Phase difference between any two is `pi/2` except that between displacement and acceleration which is `pi`.
Displacement -time equation of a particle is given as `x=Asin(omegat+pi/6)-A cos(omegat+pi/6)`, then:

Options

A. the motion of the particle is not simple harmonic
B. at t=0, acceleration of particles is negative
C. at t-0, velocity of particle is negative
D. None of the above.

Answer

To solve the problem step by step, we will analyze the given displacement-time equation of a particle in simple harmonic motion (SHM) and determine the characteristics of the motion.

### Step 1: Analyze the Displacement Equation
The displacement-time equation is given as:
\[ x = A \sin(\omega t + \frac{\pi}{6}) - A \cos(\omega t + \frac{\pi}{6}) \]

### Step 2: Simplify the Displacement Equation
We can use the trigonometric identity to combine the sine and cosine terms. Recall that:
\[ A \sin(\theta) - A \cos(\theta) = A \left( \sin(\theta) - \cos(\theta) \right) \]
We can rewrite the equation:
\[ x = A \left( \sin(\omega t + \frac{\pi}{6}) - \cos(\omega t + \frac{\pi}{6}) \right) \]

### Step 3: Find the Velocity Equation
To find the velocity, we differentiate the displacement with respect to time:
\[ v = \frac{dx}{dt} = A \omega \left( \cos(\omega t + \frac{\pi}{6}) + \sin(\omega t + \frac{\pi}{6}) \right) \]

### Step 4: Evaluate Velocity at \( t = 0 \)
Substituting \( t = 0 \) into the velocity equation:
\[ v(0) = A \omega \left( \cos(\frac{\pi}{6}) + \sin(\frac{\pi}{6}) \right) \]
Using the values \( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \) and \( \sin(\frac{\pi}{6}) = \frac{1}{2} \):
\[ v(0) = A \omega \left( \frac{\sqrt{3}}{2} + \frac{1}{2} \right) = A \omega \left( \frac{\sqrt{3} + 1}{2} \right) \]

### Step 5: Find the Acceleration Equation
The acceleration is the derivative of the velocity:
\[ a = \frac{dv}{dt} = -A \omega^2 \left( \sin(\omega t + \frac{\pi}{6}) - \cos(\omega t + \frac{\pi}{6}) \right) \]

### Step 6: Evaluate Acceleration at \( t = 0 \)
Substituting \( t = 0 \) into the acceleration equation:
\[ a(0) = -A \omega^2 \left( \sin(\frac{\pi}{6}) - \cos(\frac{\pi}{6}) \right) \]
Using the values \( \sin(\frac{\pi}{6}) = \frac{1}{2} \) and \( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \):
\[ a(0) = -A \omega^2 \left( \frac{1}{2} - \frac{\sqrt{3}}{2} \right) = -A \omega^2 \left( \frac{1 - \sqrt{3}}{2} \right) \]

### Step 7: Determine the Signs of Velocity and Acceleration
From the calculations:
- The velocity at \( t = 0 \) is positive.
- The acceleration at \( t = 0 \) is negative (since \( 1 - \sqrt{3} < 0 \)).

### Conclusion
Based on the analysis:
1. The displacement is negative.
2. The velocity is positive.
3. The acceleration is negative.

Thus, the correct option is **Option 4**.