Total energy of a particle executing oscillating motionis 3 joule and given by
`E=x^(2)+2x +v^(2)-2v`
Where x is the displacement from origin at x=0 and v is velocity of particle at x. Then choose the correct statements)

Options

A. Amplitude of osciallation is 1m
B. Maximum velocity of particles during oscillation is 3m/sec.
C. Amplitude of oscillation is 2m
D. Maximum velocity of particle during oscillation is `(sqrt(5)+1)` m/sec.

Answer

To solve the problem, we need to analyze the given expression for the total energy of a particle executing oscillatory motion:

\[ E = x^2 + 2x + v^2 - 2v \]

We know that the total energy \( E \) is given as 3 Joules.

### Step 1: Set Up the Equation
We start by substituting the total energy into the equation:

\[ 3 = x^2 + 2x + v^2 - 2v \]

### Step 2: Analyze at Extreme Position
At the extreme position of oscillation, the velocity \( v \) is zero. Thus, we can simplify the equation:

\[ 3 = x^2 + 2x + 0^2 - 2(0) \]
\[ 3 = x^2 + 2x \]

### Step 3: Rearranging the Equation
Rearranging gives us:

\[ x^2 + 2x - 3 = 0 \]

### Step 4: Factor the Quadratic Equation
Next, we can factor this quadratic equation:

\[ (x + 3)(x - 1) = 0 \]

### Step 5: Solve for Displacement
Setting each factor to zero gives us the possible values for \( x \):

1. \( x + 3 = 0 \) → \( x = -3 \)
2. \( x - 1 = 0 \) → \( x = 1 \)

Since \( x \) represents displacement, the maximum displacement (amplitude) is:

\[ A = 1 \, \text{meter} \]

### Step 6: Maximum Velocity Calculation
To find the maximum velocity, we analyze the total energy at the mean position (where \( x = 0 \)). Substituting \( x = 0 \) into the energy equation:

\[ 3 = 0^2 + 2(0) + v^2 - 2v \]
\[ 3 = v^2 - 2v \]

### Step 7: Rearranging for Velocity
Rearranging gives us:

\[ v^2 - 2v - 3 = 0 \]

### Step 8: Factor the Velocity Equation
Factoring this quadratic equation:

\[ (v - 3)(v + 1) = 0 \]

### Step 9: Solve for Velocity
Setting each factor to zero gives us the possible values for \( v \):

1. \( v - 3 = 0 \) → \( v = 3 \, \text{m/s} \)
2. \( v + 1 = 0 \) → \( v = -1 \, \text{m/s} \)

Since we are interested in maximum velocity, we take:

\[ V_{max} = 3 \, \text{m/s} \]

### Conclusion
From the analysis, we conclude:

- The amplitude of oscillation is \( 1 \, \text{meter} \).
- The maximum velocity is \( 3 \, \text{m/s} \).

Thus, the correct statements are options 1 and 2.