A particle moves in x-y plane according to the equation `vecr=(Asinomegat+Bcosomegat)(hati+hatj)` :

Options

A. SHM and along a straight line
B. SHM and along an elliptical path
C. SHM and along a circular path
D. Not SHM, but periodic

Explanation

To solve the problem, we need to analyze the given equation of motion for the particle in the x-y plane:

\[ \vec{r} = (A \sin(\omega t) + B \cos(\omega t)) \hat{i} + (A \sin(\omega t) + B \cos(\omega t)) \hat{j} \]

### Step 1: Identify the components of the motion
The position vector can be separated into its x and y components:

- \( x = A \sin(\omega t) + B \cos(\omega t) \)
- \( y = A \sin(\omega t) + B \cos(\omega t) \)

### Step 2: Differentiate to find velocity
To check if the motion is simple harmonic, we need to find the velocity vector by differentiating the position vector with respect to time \( t \):

\[ \vec{v} = \frac{d\vec{r}}{dt} = \left( A \omega \cos(\omega t) - B \omega \sin(\omega t) \right) \hat{i} + \left( A \omega \cos(\omega t) - B \omega \sin(\omega t) \right) \hat{j} \]

### Step 3: Differentiate to find acceleration
Next, we differentiate the velocity vector to find the acceleration vector:

\[ \vec{a} = \frac{d\vec{v}}{dt} = \left( -A \omega^2 \sin(\omega t) - B \omega^2 \cos(\omega t) \right) \hat{i} + \left( -A \omega^2 \sin(\omega t) - B \omega^2 \cos(\omega t) \right) \hat{j} \]

### Step 4: Check the relationship between acceleration and position
We can express the acceleration in terms of the position vector \( \vec{r} \):

\[ \vec{a} = -\omega^2 \vec{r} \]

This shows that the acceleration is directly proportional to the negative of the position vector, which is a characteristic of simple harmonic motion (SHM).

### Step 5: Determine the path of motion
To determine the path of the particle, we need to analyze the phase difference between the x and y components. Since both components are derived from the same sine and cosine functions, we can conclude that the phase difference is zero.

### Step 6: Identify the type of path
- If the phase difference is \(0^\circ\) or \(180^\circ\), the motion is along a straight line.
- If the phase difference is \(90^\circ\), the motion is circular.
- If the phase difference is between \(0^\circ\) and \(90^\circ\), the motion is elliptical.

Since we have established that the phase difference is \(0^\circ\), the motion of the particle is along a straight line.

### Conclusion
The particle moves in a straight line, and the equation represents simple harmonic motion.