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 analyze the trajectory of a particle moving in the x-y plane with the given equations \( x = a \sin(\omega t) \) and \( y = a(1 - \cos(\omega t)) \), we can follow these steps:

### Step 1: Identify the equations for x and y
We have:
- \( x = a \sin(\omega t) \)
- \( y = a(1 - \cos(\omega t)) \)

### Step 2: Express \( \sin(\omega t) \) and \( \cos(\omega t) \) in terms of x and y
From the equation for \( x \):
\[ \sin(\omega t) = \frac{x}{a} \]
From the equation for \( y \):
\[ y = a(1 - \cos(\omega t)) \implies \cos(\omega t) = 1 - \frac{y}{a} \]

### Step 3: Use the Pythagorean identity
We know from trigonometry that:
\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]
Substituting our expressions for \( \sin(\omega t) \) and \( \cos(\omega t) \):
\[ \left(\frac{x}{a}\right)^2 + \left(1 - \frac{y}{a}\right)^2 = 1 \]

### Step 4: Simplify the equation
Expanding the equation:
\[ \frac{x^2}{a^2} + \left(1 - \frac{y}{a}\right)^2 = 1 \]
\[ \frac{x^2}{a^2} + \left(1 - \frac{2y}{a} + \frac{y^2}{a^2}\right) = 1 \]
Multiplying through by \( a^2 \) to eliminate the denominators:
\[ x^2 + (a^2 - 2ay + y^2) = a^2 \]
Rearranging gives:
\[ x^2 + y^2 - 2ay = 0 \]

### Step 5: Identify the trajectory
Rearranging the equation:
\[ x^2 + (y - a)^2 = a^2 \]
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 \( a \).