For a particle moving in the x-y plane, the x and y coordinates are changing as x=a sin `omega t ` and `y = a ( 1-cos omega t ) `, where 'a' and `omega ` constants. Then, what can be inferred for the trajectory of the particle ?

Options

A. The trajectory is parabolic.
B. The trajectory is elliptical
C. The trajectory is such that distance moved by the particle is directly proportional to time.
D. The trajectory of the particle is equidistant from the origin at every instant of time.

Answer

To solve the problem, we need to analyze the given equations for the x and y coordinates of the particle:

1. **Given Equations**:
\[ x = a \sin(\omega t) \]
\[ y = a(1 - \cos(\omega t)) \]

2. **Expressing Trigonometric Functions**:
We can use the Pythagorean identity for sine and cosine. We know that:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]

3. **Rearranging the y Equation**:
From the equation for y, we can express \(\cos(\omega t)\) in terms of y:
\[ y = a(1 - \cos(\omega t)) \implies \cos(\omega t) = 1 - \frac{y}{a} \]

4. **Substituting into the Sine Identity**:
Now, we can substitute \(\cos(\omega t)\) into the sine identity:
\[ \sin^2(\omega t) = 1 - \cos^2(\omega t) \]
\[ \sin^2(\omega t) = 1 - \left(1 - \frac{y}{a}\right)^2 \]

5. **Expanding the Cosine Term**:
Expanding the squared term:
\[ \sin^2(\omega t) = 1 - \left(1 - 2\frac{y}{a} + \frac{y^2}{a^2}\right) \]
\[ \sin^2(\omega t) = 2\frac{y}{a} - \frac{y^2}{a^2} \]

6. **Substituting for x**:
Now substitute \(x = a \sin(\omega t)\):
\[ \sin^2(\omega t) = \frac{x^2}{a^2} \]

7. **Combining the Two Equations**:
Now we can set the two expressions for \(\sin^2(\omega t)\) equal to each other:
\[ \frac{x^2}{a^2} = 2\frac{y}{a} - \frac{y^2}{a^2} \]

8. **Multiplying through by \(a^2\)**:
To eliminate the fractions, multiply through by \(a^2\):
\[ x^2 = 2ay - y^2 \]

9. **Rearranging the Equation**:
Rearranging gives us:
\[ x^2 + y^2 - 2ay = 0 \]

10. **Completing the Square**:
Completing the square for the y terms:
\[ x^2 + (y - a)^2 = a^2 \]

11. **Identifying the Shape**:
This is the equation of a circle with:
- Center at \((0, a)\)
- Radius \(a\)

### Conclusion:
The trajectory of the particle is a circle centered at \((0, a)\) with radius \(a\).