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 determine the motion of the particle described 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
The position vector can be broken down into its components:
- The x-component is given by \(x = A \sin(\omega t)\).
- The y-component is given by \(y = A \cos(\omega t)\).

### Step 2: Express Sine and Cosine in Terms of x and y
From the equations for x and y, we can express sine and cosine:
- From \(x = A \sin(\omega t)\), we can write \(\sin(\omega t) = \frac{x}{A}\).
- From \(y = A \cos(\omega t)\), we can write \(\cos(\omega t) = \frac{y}{A}\).

### Step 3: Use the Pythagorean Identity
We know from trigonometry that:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
Applying this identity, we substitute \(\sin(\omega t)\) and \(\cos(\omega t)\):
\[ \left(\frac{x}{A}\right)^2 + \left(\frac{y}{A}\right)^2 = 1 \]

### Step 4: Rearranging the Equation
Multiplying through by \(A^2\) gives:
\[ x^2 + y^2 = A^2 \]
This is the equation of a circle with radius \(A\) centered at the origin \((0, 0)\).

### Step 5: Conclusion
Since the position vector describes a circular path in the x-y plane, we conclude that the motion of the particle is circular motion.

### Final Answer
The motion of the particle is on a circle.

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