The position of a particle moving along x- axis is given by `x=x_0 cos^2(omegat)` . Its when it is at mean position is

Options

A. `2x_0omega`
B. `x_0omega^2`
C. `(x_0omega)/2`
D. `x_0omega`

Answer

To solve the problem, we need to find the velocity of a particle moving along the x-axis, given its position as \( x = x_0 \cos^2(\omega t) \), and determine the velocity when the particle is at its mean position.

### Step-by-Step Solution:

1. **Identify the Position Function**:
The position of the particle is given by:
\[ x = x_0 \cos^2(\omega t) \]

2. **Differentiate the Position Function**:
To find the velocity, we need to differentiate the position function with respect to time \( t \):
\[ v = \frac{dx}{dt} \]
Using the chain rule, we differentiate \( \cos^2(\omega t) \):
\[ v = x_0 \cdot \frac{d}{dt}(\cos^2(\omega t)) = x_0 \cdot 2\cos(\omega t) \cdot \frac{d}{dt}(\cos(\omega t)) \]
The derivative of \( \cos(\omega t) \) is:
\[ \frac{d}{dt}(\cos(\omega t)) = -\omega \sin(\omega t) \]
Therefore, substituting this back, we have:
\[ v = x_0 \cdot 2\cos(\omega t)(-\omega \sin(\omega t)) = -2x_0 \omega \cos(\omega t) \sin(\omega t) \]

3. **Use the Trigonometric Identity**:
We can use the identity \( 2\sin(a)\cos(a) = \sin(2a) \) to simplify the expression for velocity:
\[ v = -x_0 \omega \sin(2\omega t) \]

4. **Determine the Mean Position**:
The mean position corresponds to the maximum velocity. The maximum value of \( \sin(2\omega t) \) is 1. Therefore, the maximum velocity occurs when:
\[ \sin(2\omega t) = -1 \quad \text{(to make the negative sign positive)} \]
Thus, the maximum velocity \( v_{\text{max}} \) is:
\[ v_{\text{max}} = x_0 \omega \]

5. **Conclusion**:
The velocity of the particle when it is at its mean position is:
\[ v = x_0 \omega \]

### Final Answer:
The velocity of the particle when it is at the mean position is \( x_0 \omega \).