A particle is moving along the x-axis and force acting on it is given by `F=F_0sinomegaxN`, where `omega` is a constant. The work done by the force from `x=0` to `x=2` will be

Options

A. `(F_0)/(omega(1-cosomega))`
B. `(F_0)/(2omega(1-cos2omega))`
C. `(F_0)/(omega(1-cos2omega))`
D. `(2F_0sin^2omega)/omega`

Answer

To solve the problem of finding the work done by the force \( F = F_0 \sin(\omega x) \) from \( x = 0 \) to \( x = 2 \), we will follow these steps:

### Step 1: Understand the Work Done Formula
The work done \( W \) by a force is given by the integral of the force over the distance moved:
\[ W = \int_{x_1}^{x_2} F \, dx \]
In our case, \( x_1 = 0 \) and \( x_2 = 2 \).

### Step 2: Set Up the Integral
Substituting the expression for the force into the work done formula:
\[ W = \int_{0}^{2} F_0 \sin(\omega x) \, dx \]

### Step 3: Factor Out Constants
Since \( F_0 \) is a constant, we can factor it out of the integral:
\[ W = F_0 \int_{0}^{2} \sin(\omega x) \, dx \]

### Step 4: Integrate the Sine Function
The integral of \( \sin(\omega x) \) is:
\[ \int \sin(\omega x) \, dx = -\frac{1}{\omega} \cos(\omega x) + C \]
Thus, we can evaluate the definite integral:
\[ W = F_0 \left[-\frac{1}{\omega} \cos(\omega x)\right]_{0}^{2} \]

### Step 5: Apply the Limits
Now we substitute the limits into the integral:
\[ W = F_0 \left[-\frac{1}{\omega} \cos(2\omega) + \frac{1}{\omega} \cos(0)\right] \]
Since \( \cos(0) = 1 \), this simplifies to:
\[ W = F_0 \left[-\frac{1}{\omega} \cos(2\omega) + \frac{1}{\omega}\right] \]
\[ W = \frac{F_0}{\omega} \left[1 - \cos(2\omega)\right] \]

### Step 6: Use Trigonometric Identity
We can use the trigonometric identity:
\[ 1 - \cos(2\theta) = 2 \sin^2(\theta) \]
So, substituting \( \theta = \omega \):
\[ W = \frac{F_0}{\omega} \cdot 2 \sin^2(\omega) \]
Thus, the final expression for the work done is:
\[ W = \frac{2 F_0 \sin^2(\omega)}{\omega} \]

### Final Answer
The work done by the force from \( x = 0 \) to \( x = 2 \) is:
\[ W = \frac{2 F_0 \sin^2(\omega)}{\omega} \]

---