The displacement of a particle is given at time t, by:
`x=A sin (-2 omega t ) + B sin^(2) omega t` Then,

Options

A. the motion of the particle is SHM with an amplitude of `sqrt(A^(2)+(B^(2))/(4))`
B. the motion of the particle is not SHM, but oscillatory with a time period of `T=pi//omega`
C. the motion of the particle is oscillatory with a time period of `T=pi//2omega`
D. none

Answer

To solve the problem, we need to analyze the given displacement equation of a particle:

**Given:**
\[ x = A \sin(-2\omega t) + B \sin^2(\omega t) \]

### Step 1: Simplify the equation
We start by simplifying the first term. Recall that:
\[ \sin(-\theta) = -\sin(\theta) \]
Thus, we can rewrite the equation as:
\[ x = -A \sin(2\omega t) + B \sin^2(\omega t) \]

### Step 2: Use the identity for \(\sin^2(\omega t)\)
Next, we can use the trigonometric identity:
\[ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \]
Applying this identity to \(\sin^2(\omega t)\):
\[ \sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2} \]
Substituting this into our equation gives:
\[ x = -A \sin(2\omega t) + B \left(\frac{1 - \cos(2\omega t)}{2}\right) \]
\[ x = -A \sin(2\omega t) + \frac{B}{2} - \frac{B}{2} \cos(2\omega t) \]

### Step 3: Rearranging the equation
Now, we can rearrange the equation:
\[ x = \frac{B}{2} - A \sin(2\omega t) - \frac{B}{2} \cos(2\omega t) \]

### Step 4: Identify the form of the equation
The equation can be expressed in a form that resembles simple harmonic motion (SHM). The terms involving \(\sin(2\omega t)\) and \(\cos(2\omega t)\) can be combined. We can express this as:
\[ x - \frac{B}{2} = -A \sin(2\omega t) - \frac{B}{2} \cos(2\omega t) \]

### Step 5: Recognize the SHM characteristics
The expression on the left indicates that the particle undergoes simple harmonic motion about the mean position \(\frac{B}{2}\). The coefficients of \(\sin\) and \(\cos\) indicate that the motion is oscillatory.

### Step 6: Calculate the amplitude
The amplitude of the resultant SHM can be calculated using the formula for the resultant amplitude when two perpendicular components are combined:
\[ R = \sqrt{A^2 + \left(\frac{B}{2}\right)^2} \]
Thus, the amplitude \(R\) of the motion is:
\[ R = \sqrt{A^2 + \left(\frac{B}{2}\right)^2} \]

### Conclusion
The motion of the particle is simple harmonic motion (SHM) with an amplitude of:
\[ R = \sqrt{A^2 + \left(\frac{B}{2}\right)^2} \]
and the mean position is \(\frac{B}{2}\).