A particle moves in `xy` plane accordin to the law `x=a sin w t` and `y=a (1- cos w t )` where `a` and `w` are costant . The particle traces.

Options

A. a parabola
B. a straight line equally inclined to `x` and `y` axes
C. a circle
D. a distane proportional to time.

Answer

To solve the problem, we need to analyze the motion of a particle in the xy-plane given by the equations:

1. \( x = a \sin(\omega t) \)
2. \( y = a(1 - \cos(\omega t)) \)

### Step 1: Rewrite the equations
We can rewrite the second equation for clarity:

\[ y = a - a \cos(\omega t) \]

### Step 2: Identify the type of motion
Both equations represent simple harmonic motion (SHM). The equation for \(x\) is a sine function, and the equation for \(y\) can be expressed as a cosine function.

### Step 3: Determine the phase difference
The sine and cosine functions have a phase difference of \(\frac{\pi}{2}\) radians. This indicates that the motion is circular because the two components are perpendicular to each other.

### Step 4: Analyze the motion
At \(t = 0\):
- For \(x\):
\[ x = a \sin(0) = 0 \]
- For \(y\):
\[ y = a(1 - \cos(0)) = a(1 - 1) = 0 \]

Thus, the particle starts at the origin (0, 0).

### Step 5: Determine the range of motion
As \(t\) varies, \(x\) oscillates between \(-a\) and \(a\) (since \(\sin(\omega t)\) varies between -1 and 1), and \(y\) oscillates between \(0\) and \(2a\) (since \(\cos(\omega t)\) varies between -1 and 1).

### Step 6: Identify the path traced by the particle
The equations \(x = a \sin(\omega t)\) and \(y = a - a \cos(\omega t)\) can be combined to eliminate \(t\). We can use the Pythagorean identity:

\[ \sin^2(\theta) + \cos^2(\theta) = 1 \]

By substituting \(\sin(\omega t)\) and \(\cos(\omega t)\) into the equations, we can derive the relationship between \(x\) and \(y\):

1. From \(x = a \sin(\omega t)\), we have:
\[ \sin(\omega t) = \frac{x}{a} \]

2. From \(y = a - a \cos(\omega t)\), we can express \(\cos(\omega t)\):
\[ \cos(\omega t) = 1 - \frac{y}{a} \]

3. Substitute into the Pythagorean identity:
\[ \left(\frac{x}{a}\right)^2 + \left(1 - \frac{y}{a}\right)^2 = 1 \]

4. Simplifying this gives us:
\[ \frac{x^2}{a^2} + \left(1 - \frac{y}{a}\right)^2 = 1 \]
\[ \frac{x^2}{a^2} + \left(\frac{y - a}{a}\right)^2 = 1 \]

This is the equation of a circle centered at \((0, a)\) with a radius of \(a\).

### Conclusion
The particle traces a circular path in the xy-plane.

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