The velocity v and displacement x of a particle executing simple harmonic motion are related as
`v (dv)/(dx)= -omega^2 x`.
`At x=0, v=v_0.` Find the velocity v when the displacement becomes x.

Options

A. `sqrt(v_(0)^(2)+omega^(2)x^(2))`
B. `sqrt(v_(0)^(2)-omega^(2)x^(2))`
C. `v =root(3)(v_(0)^(3)+omega^(2)x^(3))`
D. `v=v_(0)-(omega^(3)x^(3)e^(x^(3)))^(1//3)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Start with the given equation
The relationship between velocity \( v \) and displacement \( x \) for a particle in simple harmonic motion is given by:
\[ v \frac{dv}{dx} = -\omega^2 x \]

### Step 2: Rearrange the equation
We can rearrange the equation to separate variables:
\[ v \, dv = -\omega^2 x \, dx \]

### Step 3: Integrate both sides
Now, we will integrate both sides. The left side integrates with respect to \( v \) and the right side with respect to \( x \):
\[ \int v \, dv = \int -\omega^2 x \, dx \]

### Step 4: Perform the integration
The integrals yield:
\[ \frac{v^2}{2} = -\frac{\omega^2 x^2}{2} + C \]
where \( C \) is the constant of integration.

### Step 5: Determine the constant of integration
We know that at \( x = 0 \), \( v = v_0 \). Plugging these values into the equation gives:
\[ \frac{v_0^2}{2} = C \]
Thus, the equation becomes:
\[ \frac{v^2}{2} = -\frac{\omega^2 x^2}{2} + \frac{v_0^2}{2} \]

### Step 6: Simplify the equation
Multiplying through by 2 to eliminate the fractions:
\[ v^2 = v_0^2 - \omega^2 x^2 \]

### Step 7: Solve for \( v \)
Taking the square root of both sides, we find:
\[ v = \sqrt{v_0^2 - \omega^2 x^2} \]

### Final Result
The velocity \( v \) when the displacement is \( x \) is given by:
\[ v = \sqrt{v_0^2 - \omega^2 x^2} \]

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