A particle is executing simple harmonic motion in a conservative force field. The total energy of simple harmonic motion is given by `E=ax^(2)+bv^(2)`where ‘x’ is the displacement from mean position x = 0 and v is the velocity of the particle at x then choose the INCORRECT statements.{Potential energy at mean position is assumed to be zero}

Options

A. amplitude of S.H.M. is `sqrt(E/a)`
B. Maximum velocity of the particle during S.H.M. is `sqrt(E/b)`
C. Time period of motion is `2pi sqrt(b/a)`
D. displacement of the particle is proportional to the velocity of the particle.

Explanation

To solve the problem, we need to analyze the total energy of a particle executing simple harmonic motion (SHM) in a conservative force field, given by the equation \( E = ax^2 + bv^2 \). We will identify the incorrect statements based on the properties of SHM.

### Step-by-Step Solution:

1. **Understanding Total Energy in SHM**:
The total energy \( E \) in simple harmonic motion is the sum of kinetic energy (KE) and potential energy (PE). The standard forms are:
\[
KE = \frac{1}{2} mv^2 \quad \text{and} \quad PE = \frac{1}{2} kx^2
...