If abody starts executing S.H.M. from mean position with amplitude 'A', maximum velocity `v_(0)` and time period 'T', then the correct statemens are (x is displacement from mean position)

Options

A. If`v= (v_(0))/(2)`, then `|x| gt (A)/(2)`
B. if `x=(A)/(2)` , then `V gt (v_(0))/(2)`
C. For `t=(T)/(8),x gt(A)/(2)`
D. For`x= (A)/(2) , t lt (T)/(8)`

Answer

To solve the problem, we need to analyze the statements regarding a body executing Simple Harmonic Motion (S.H.M.) starting from the mean position with given parameters: amplitude \( A \), maximum velocity \( v_0 \), and time period \( T \).

### Step-by-Step Solution:

1. **Understanding S.H.M. Parameters**:
- The displacement \( x \) from the mean position can be expressed as:
\[ x = A \sin(\omega t) \]
where \( \omega = \frac{2\pi}{T} \) is the angular frequency.

2. **Maximum Velocity**:
- The maximum velocity \( v_{\text{max}} \) in S.H.M. is given by:
\[ v_{\text{max}} = A \omega \]
- Given that \( v_{\text{max}} = v_0 \), we can write:
\[ v_0 = A \omega \]

3. **Velocity at Displacement \( x \)**:
- The velocity \( v \) at any displacement \( x \) is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]

4. **Analyzing the First Statement**:
- If \( v = \frac{v_0}{2} \):
\[ \frac{v_0}{2} = \omega \sqrt{A^2 - x^2} \]
Substituting \( v_0 = A \omega \):
\[ \frac{A \omega}{2} = \omega \sqrt{A^2 - x^2} \]
Dividing both sides by \( \omega \) (assuming \( \omega \neq 0 \)):
\[ \frac{A}{2} = \sqrt{A^2 - x^2} \]
Squaring both sides:
\[ \frac{A^2}{4} = A^2 - x^2 \]
Rearranging gives:
\[ x^2 = A^2 - \frac{A^2}{4} = \frac{3A^2}{4} \]
Thus:
\[ x = \frac{\sqrt{3}}{2} A \]
Since \( \frac{\sqrt{3}}{2} A \) is greater than \( \frac{A}{2} \), the first statement is **correct**.

5. **Analyzing the Second Statement**:
- If \( x = \frac{A}{2} \):
\[ v = \omega \sqrt{A^2 - \left(\frac{A}{2}\right)^2} = \omega \sqrt{A^2 - \frac{A^2}{4}} = \omega \sqrt{\frac{3A^2}{4}} = \frac{A \omega \sqrt{3}}{2} \]
Since \( v_0 = A \omega \), we have:
\[ v = \frac{v_0 \sqrt{3}}{2} \]
Since \( \sqrt{3} \approx 1.732 \), it follows that \( v > \frac{v_0}{2} \). Thus, the second statement is **correct**.

6. **Analyzing the Third Statement**:
- For \( t = \frac{T}{8} \):
\[ x = A \sin\left(\omega \frac{T}{8}\right) = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{8}\right) = A \sin\left(\frac{\pi}{4}\right) = A \cdot \frac{1}{\sqrt{2}} = \frac{A}{\sqrt{2}} \]
Since \( \frac{A}{\sqrt{2}} \approx 0.707 A \) which is greater than \( \frac{A}{2} \), the third statement is **correct**.

7. **Analyzing the Fourth Statement**:
- If \( x = \frac{A}{2} \):
\[ \frac{A}{2} = A \sin(\omega t) \implies \sin(\omega t) = \frac{1}{2} \implies \omega t = \frac{\pi}{6} \]
Thus:
\[ t = \frac{\pi}{6\omega} = \frac{\pi}{6 \cdot \frac{2\pi}{T}} = \frac{T}{12} \]
Since \( \frac{T}{12} < \frac{T}{8} \), the fourth statement is **correct**.

### Conclusion:
All statements are correct:
- Statement 1: Correct
- Statement 2: Correct
- Statement 3: Correct
- Statement 4: Correct