An object of mass m is moving in unifrom circular motion in xy-plane. The circle has radius R and object is moving clockwise around the circle with speed v. The motion is projected onto the x-axis where it appears as simple harmonic motion accoding to `x(t) = Rcos (omegat + phi)`. The motion starts from pointof coordinates
`(0,R)`
In this projection `omega` is -

Options

A. `v//R`
B. `m^(2)//R`
C. `R//v`
D. None of these

Explanation

To solve the problem, we need to determine the angular frequency \( \omega \) of the object moving in uniform circular motion, which is projected onto the x-axis as simple harmonic motion (SHM).

### Step-by-step Solution:

1. **Understanding Circular Motion**:
- The object of mass \( m \) is moving in a circle of radius \( R \) with a constant speed \( v \).
- The coordinates of the object at any time \( t \) can be represented in terms of the angle \( \theta(t) \) it makes with the x-axis.

2. **Angular Frequency Definition**:
- The angular frequency \( \omega \) is defined as the rate of change of the angle with respect to time. It is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
where \( T \) is the time period of one complete revolution.

3. **Finding the Time Period**:
- The time period \( T \) can be calculated using the circumference of the circle and the speed:
\[ T = \frac{\text{Circumference}}{\text{Speed}} = \frac{2\pi R}{v} \]

4. **Substituting Time Period into Angular Frequency**:
- Now substituting \( T \) into the formula for \( \omega \):
\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{\frac{2\pi R}{v}} = \frac{v}{R} \]

5. **Conclusion**:
- Therefore, the angular frequency \( \omega \) of the motion projected onto the x-axis is:
\[ \omega = \frac{v}{R} \]

### Final Answer:
The angular frequency \( \omega \) is \( \frac{v}{R} \).