The displacement of a particle is given by `vecr=A(vecicosomegat+vecjsinomegat)`. The motion of the particle is

Options

A. simple harmonic
B. on a straight line
C. on a circle
D. with constant acceleration

Answer

To determine the motion of a particle given its displacement as \( \vec{R} = A (\vec{i} \cos \omega t + \vec{j} \sin \omega t) \), we can analyze the components of the displacement vector.

### Step-by-Step Solution:

1. **Identify the Components**:
The displacement vector can be broken down into its components:
\[ \vec{R} = A \cos \omega t \, \vec{i} + A \sin \omega t \, \vec{j} \]
Here, \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \vec{i} \) and \( \vec{j} \) are the unit vectors in the x and y directions, respectively.

2. **Recognize the Form of the Equation**:
The equation resembles the parametric equations of a circle:
\[ x = A \cos \omega t \quad \text{and} \quad y = A \sin \omega t \]

3. **Use the Pythagorean Identity**:
To confirm that this represents circular motion, we can use the Pythagorean identity:
\[ x^2 + y^2 = (A \cos \omega t)^2 + (A \sin \omega t)^2 = A^2 (\cos^2 \omega t + \sin^2 \omega t) = A^2 \]
This shows that the path traced by the particle is a circle of radius \( A \).

4. **Determine the Type of Motion**:
Since both components \( A \cos \omega t \) and \( A \sin \omega t \) have the same amplitude \( A \), the motion is circular. If the amplitudes were different, the motion would be elliptical.

5. **Conclusion**:
Therefore, the motion of the particle is circular.

### Final Answer:
The motion of the particle is circular.

---