For a body executing SHM with amplitude A, time period T, maximum velocity `v_(max)` and phace constant zero, which of the following statements are correct for `0letle(T)/(4)` (y is displacement from mean position)?

Options

A. At `y = (A//2), v gt (v_(max)//2)`
B. for `v = (v_(max)//2), y gt (A//2)`
C. For `t = (T//8), (A//2)`
D. For `y = (A//2), t lt (T//8)`

Answer

To solve the question regarding a body executing Simple Harmonic Motion (SHM) with given parameters, we need to analyze the behavior of the displacement \( y \) as a function of time. The body has an amplitude \( A \), time period \( T \), maximum velocity \( v_{\text{max}} \), and a phase constant of zero.

### Step-by-Step Solution:

1. **Understanding SHM Equation**:
The displacement \( y \) of a body in SHM can be expressed as:
\[ y(t) = A \sin\left(\frac{2\pi}{T} t\right) \]
where \( A \) is the amplitude and \( T \) is the time period.

2. **Maximum Velocity**:
The maximum velocity \( v_{\text{max}} \) in SHM is given by:
\[ v_{\text{max}} = A \cdot \frac{2\pi}{T} \]

3. **Displacement at Specific Time**:
We need to evaluate the displacement \( y \) at \( t = \frac{T}{4} \):
\[ y\left(\frac{T}{4}\right) = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{4}\right) = A \sin\left(\frac{\pi}{2}\right) = A \]
Thus, at \( t = \frac{T}{4} \), the displacement is equal to the amplitude \( A \).

4. **Displacement at \( t = 0 \)**:
At \( t = 0 \), the displacement is:
\[ y(0) = A \sin(0) = 0 \]

5. **Displacement at \( t = \frac{T}{2} \)**:
At \( t = \frac{T}{2} \):
\[ y\left(\frac{T}{2}\right) = A \sin\left(\frac{2\pi}{T} \cdot \frac{T}{2}\right) = A \sin(\pi) = 0 \]

6. **Summary of Displacement Values**:
- At \( t = 0 \), \( y = 0 \)
- At \( t = \frac{T}{4} \), \( y = A \)
- At \( t = \frac{T}{2} \), \( y = 0 \)

### Conclusion:
From the analysis, we can conclude that:
- The displacement \( y \) reaches its maximum value \( A \) at \( t = \frac{T}{4} \).
- The displacement is zero at \( t = 0 \) and \( t = \frac{T}{2} \).