The magnitude of displacement of a particle moving in a circle of radius a with constant angular speed `omega` varries with time t is

Options

A. `2 a sin omega t`
B. `2a " sin " (omegat)/(2)`
C. `2a cos omega t`
D. `2a " cos " (omega t)/(2)`

Answer

To find the magnitude of displacement of a particle moving in a circle of radius \( a \) with constant angular speed \( \omega \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Motion**:
- The particle moves in a circular path with radius \( a \) and constant angular speed \( \omega \).
- At time \( t = 0 \), the particle is at point \( P \) on the circumference of the circle.

2. **Determine the Angular Displacement**:
- The angular displacement \( \theta \) after time \( t \) can be expressed as:
\[ \theta = \omega t \]

3. **Identify Points**:
- Let \( O \) be the center of the circle, \( P \) be the initial position of the particle, and \( R \) be the final position after time \( t \).
- The displacement we need to find is the straight-line distance \( PQ \) between points \( P \) and \( R \).

4. **Use Geometry**:
- Draw a perpendicular from point \( R \) to line \( OP \), meeting at point \( Q \).
- The lengths \( OP \) and \( OR \) are both equal to the radius \( a \).

5. **Calculate Lengths**:
- The vertical component \( QR \) can be expressed as:
\[ QR = a \sin(\theta) = a \sin(\omega t) \]
- The horizontal component \( RP \) can be expressed as:
\[ RP = a - a \cos(\theta) = a(1 - \cos(\omega t)) \]

6. **Apply the Pythagorean Theorem**:
- The displacement \( PQ \) can be calculated using the Pythagorean theorem:
\[ PQ = \sqrt{QR^2 + RP^2} \]
- Substitute the expressions for \( QR \) and \( RP \):
\[ PQ = \sqrt{(a \sin(\omega t))^2 + (a(1 - \cos(\omega t)))^2} \]

7. **Simplify the Expression**:
- Factor out \( a^2 \):
\[ PQ = a \sqrt{\sin^2(\omega t) + (1 - \cos(\omega t))^2} \]
- Expand the second term:
\[ (1 - \cos(\omega t))^2 = 1 - 2\cos(\omega t) + \cos^2(\omega t) \]
- Combine the terms:
\[ PQ = a \sqrt{\sin^2(\omega t) + 1 - 2\cos(\omega t) + \cos^2(\omega t)} \]
- Use the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ PQ = a \sqrt{2 - 2\cos(\omega t)} = a \sqrt{2(1 - \cos(\omega t))} \]

8. **Use the Half-Angle Identity**:
- Recall that \( 1 - \cos(\theta) = 2\sin^2(\frac{\theta}{2}) \):
\[ PQ = a \sqrt{2 \cdot 2 \sin^2\left(\frac{\omega t}{2}\right)} = 2a \sin\left(\frac{\omega t}{2}\right) \]

9. **Final Result**:
- Therefore, the magnitude of displacement of the particle is:
\[ PQ = 2a \sin\left(\frac{\omega t}{2}\right) \]