The energy of a particle executing 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 correct statement(s)

Options

A. amplitude of SHM is `sqrt((2E)/A)`
B. maximum velocity of the particle during S.H.M. is `sqrt(EB)`
C. Time period of motion is`2pisqrt(B/A)`
D. displacement of the particle is proportional to the velocity of the particle.

Answer

To solve the problem, we need to analyze the given energy expression for a particle executing simple harmonic motion (SHM) and derive relevant parameters such as the time period and amplitude.

### Step-by-Step Solution:

1. **Understanding the Energy Expression**:
The energy of a particle in SHM is given by:
\[ E = Ax^2 + Bv^2 \]
where \(x\) is the displacement from the mean position and \(v\) is the velocity of the particle.

2. **Standard Energy Expression in SHM**:
The total mechanical energy \(E\) in SHM can also be expressed as:
\[ E = \frac{1}{2} k x^2 + \frac{1}{2} m v^2 \]
where \(k\) is the spring constant and \(m\) is the mass of the particle.

3. **Comparing the Two Expressions**:
By comparing the two expressions for energy, we can equate coefficients:
- From \(Ax^2\), we have:
\[ A = \frac{1}{2} k \quad \Rightarrow \quad k = 2A \]
- From \(Bv^2\), we have:
\[ B = \frac{1}{2} m \quad \Rightarrow \quad m = 2B \]

4. **Finding the Time Period**:
The time period \(T\) of SHM is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Substituting the values of \(m\) and \(k\):
\[ T = 2\pi \sqrt{\frac{2B}{2A}} = 2\pi \sqrt{\frac{B}{A}} \]

5. **Finding the Amplitude**:
At the extreme position of SHM, the velocity \(v = 0\) and all the energy is potential:
\[ E = A A^2 + B(0)^2 \quad \Rightarrow \quad E = A A^2 = A^3 \]
Rearranging gives:
\[ A = \sqrt{\frac{E}{A}} \]

6. **Maximum Velocity**:
The maximum velocity \(v_{max}\) in SHM occurs at the mean position and can be expressed as:
\[ v_{max} = A\omega = A\sqrt{\frac{k}{m}} = A\sqrt{\frac{2A}{2B}} = A\sqrt{\frac{A}{B}} \]

7. **Displacement and Velocity Relationship**:
In SHM, the displacement \(x\) is not directly proportional to the velocity \(v\). The velocity is maximum at the mean position and zero at the extremes, indicating that the relationship is not linear.

### Conclusion:
Based on the analysis, we can conclude that:
- The time period \(T = 2\pi \sqrt{\frac{B}{A}}\).
- The amplitude \(A\) can be expressed in terms of energy \(E\).
- The statement regarding the displacement being proportional to the velocity is incorrect.