A simple harmonic progressive wave in a gas has a particle displacement of y=a at time `t=T//4` at the orgin of the wave and a particle velocity of y =v at the same instant but at a distance `x=lambda//4` from the orgin where T and `lambda` are the periodic time and wavelength of the wave respectively. then for this wave.

Options

A. the amplitude `A` of the wave is `A=2a`
B. the amplitude `A` of the wave is `A=a`
C. the equation of the wave can be represented by
`y =asin(v)/(a)[(t)-(x)/(V)]`
D. The equation of the wave can be represented by
`y= 2a cos(v)/(a)[t-(x)/(V)]`

Explanation

To solve the problem, we need to analyze the given information about the simple harmonic progressive wave in a gas.

### Step-by-Step Solution:

1. **Understanding the Wave Characteristics**:
- We know that a simple harmonic wave can be described by the equation:
\[ y(x, t) = A \sin(kx - \omega t) \]
where \( A \) is the amplitude, \( k \) is the wave number, and \( \omega \) is the angular frequency.

2. **Given Conditions**:
- At time \( t = \frac{T}{4} \) (where \( T \) is the period of the wave), the particle displacement at the origin (i.e., \( x = 0 \)) is \( y(0, \frac{T}{4}) = a \).
- The particle velocity at \( x = \frac{\lambda}{4} \) at the same time is \( v = \frac{dy}{dt} \).

3. **Finding the Displacement at the Origin**:
- At \( x = 0 \):
\[ y(0, \frac{T}{4}) = A \sin\left(0 - \omega \frac{T}{4}\right) = A \sin\left(-\frac{\pi}{2}\right) = -A \]
- Given that this displacement is equal to \( a \), we have:
\[ -A = a \implies A = -a \]
- Since amplitude is always taken as a positive quantity, we can say:
\[ A = a \]

4. **Finding the Velocity at \( x = \frac{\lambda}{4} \)**:
- The wave number \( k \) is given by:
\[ k = \frac{2\pi}{\lambda} \]
- The angular frequency \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} \]
- The velocity of the particle is given by:
\[ v(x, t) = \frac{dy}{dt} = A \omega \cos(kx - \omega t) \]
- At \( x = \frac{\lambda}{4} \) and \( t = \frac{T}{4} \):
\[ v\left(\frac{\lambda}{4}, \frac{T}{4}\right) = A \omega \cos\left(k \frac{\lambda}{4} - \omega \frac{T}{4}\right) \]
- Substituting \( k = \frac{2\pi}{\lambda} \) and \( \omega = \frac{2\pi}{T} \):
\[ v\left(\frac{\lambda}{4}, \frac{T}{4}\right) = A \cdot \frac{2\pi}{T} \cos\left(\frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} - \frac{2\pi}{T} \cdot \frac{T}{4}\right) \]
- This simplifies to:
\[ v\left(\frac{\lambda}{4}, \frac{T}{4}\right) = A \cdot \frac{2\pi}{T} \cos\left(\frac{\pi}{2} - \frac{\pi}{2}\right) = A \cdot \frac{2\pi}{T} \cos(0) = A \cdot \frac{2\pi}{T} \]

5. **Conclusion**:
- The amplitude \( A \) of the wave is equal to \( a \). Therefore, the correct option is:
\[ \text{Amplitude } A = a \]

### Final Answer:
The amplitude \( A \) of the wave is \( a \).