A particle isolated simultaneously by mutually perpendicular simple harmonic motion x = a cos `omegat` and y = a sin `omegat`. The trajectory of motion of the particle will be:

Options

A. An ellipse
B. A parabola
C. A circle
D. A straight line

Explanation

To solve the problem of finding the trajectory of a particle undergoing mutually perpendicular simple harmonic motions given by the equations \( x = a \cos(\omega t) \) and \( y = a \sin(\omega t) \), we can follow these steps:

### Step-by-Step Solution

1. **Write down the equations of motion**:
- The equations given are:
\[ x = a \cos(\omega t) \]
\[ y = a \sin(\omega t) \]

2. **Express \(\cos(\omega t)\) and \(\sin(\omega t)\) in terms of \(x\) and \(y\)**:
- From the first equation, we can express \(\cos(\omega t)\):
\[ \cos(\omega t) = \frac{x}{a} \]
- From the second equation, we can express \(\sin(\omega t)\):
\[ \sin(\omega t) = \frac{y}{a} \]

3. **Use the Pythagorean identity**:
- We know from trigonometry that:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
- Applying this identity to our expressions:
\[ \left(\frac{y}{a}\right)^2 + \left(\frac{x}{a}\right)^2 = 1 \]

4. **Rearranging the equation**:
- Squaring both sides gives:
\[ \frac{y^2}{a^2} + \frac{x^2}{a^2} = 1 \]
- Multiplying through by \(a^2\) results in:
\[ x^2 + y^2 = a^2 \]

5. **Identify the shape of the trajectory**:
- The equation \(x^2 + y^2 = a^2\) represents a circle with radius \(a\) centered at the origin (0, 0).

### Final Answer
The trajectory of the particle is a **circle** with center at (0, 0).

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