The position vector of a particle moving in `x-y` plane is given by
`vec(r) = (A sinomegat)hat(i) + (A cosomegat)hat(j)` then motion of the particle is :

Options

A. `SHM`
B. on a circle
C. on a straight line
D. with constant acceleration

Answer

To analyze the motion of the particle given by the position vector \(\vec{r} = A \sin(\omega t) \hat{i} + A \cos(\omega t) \hat{j}\), we can follow these steps:

### Step 1: Identify the components of the position vector
The position vector can be broken down into its components:
- \(x(t) = A \sin(\omega t)\)
- \(y(t) = A \cos(\omega t)\)

### Step 2: Relate the components using trigonometric identities
We can use the Pythagorean identity, which states that \(\sin^2(\theta) + \cos^2(\theta) = 1\). By manipulating the equations for \(x\) and \(y\):
\[ \frac{x}{A} = \sin(\omega t) \quad \text{and} \quad \frac{y}{A} = \cos(\omega t) \]
Squaring both equations gives:
\[ \left(\frac{x}{A}\right)^2 + \left(\frac{y}{A}\right)^2 = \sin^2(\omega t) + \cos^2(\omega t) = 1 \]

### Step 3: Formulate the equation of motion
From the above manipulation, we can derive the equation:
\[ \frac{x^2}{A^2} + \frac{y^2}{A^2} = 1 \]
Multiplying through by \(A^2\) leads to:
\[ x^2 + y^2 = A^2 \]

### Step 4: Interpret the equation
The equation \(x^2 + y^2 = A^2\) represents a circle of radius \(A\) centered at the origin in the \(x-y\) plane. This indicates that the particle moves along a circular path.

### Conclusion
Thus, the motion of the particle is circular motion.