For the transverse wave equation `y = A sin(pix + pit)`, choose the correct option at `t =0`

Options

A. points at `x = 0 and x = 1` are at mean positions
B. points at `x = 0.5 and x = 1.5` have maximum accelerations
C. points at `x = 0.5 and x = 1.5` are at rest
D. the given wave is travelling in negative x-direction

Explanation

To solve the problem, we need to analyze the given transverse wave equation and evaluate the options provided at \( t = 0 \).

### Given Wave Equation:
\[ y = A \sin(\pi x + \pi t) \]

### Step 1: Evaluate the wave equation at \( t = 0 \)
At \( t = 0 \), the equation simplifies to:
\[ y = A \sin(\pi x) \]

### Step 2: Analyze the wave at specific points
Now let's evaluate the wave at the specified points in the options:

#### Option A: Points at \( x = 0 \) and \( x = 1 \)
- For \( x = 0 \):
\[ y(0) = A \sin(\pi \cdot 0) = A \sin(0) = 0 \]

- For \( x = 1 \):
\[ y(1) = A \sin(\pi \cdot 1) = A \sin(\pi) = 0 \]

Both points are at the mean position (equilibrium position). Thus, **Option A is correct**.

#### Option B: Points at \( x = 0.5 \) and \( x = 1.5 \) have maximum acceleration
To find acceleration, we need to find the velocity first:
\[ v = \frac{dy}{dt} = A \pi \cos(\pi x + \pi t) \]
At \( t = 0 \):
\[ v = A \pi \cos(\pi x) \]

Now, we check the points:
- For \( x = 0.5 \):
\[ v(0.5) = A \pi \cos(\pi \cdot 0.5) = A \pi \cos\left(\frac{\pi}{2}\right) = 0 \]

- For \( x = 1.5 \):
\[ v(1.5) = A \pi \cos(\pi \cdot 1.5) = A \pi \cos\left(\frac{3\pi}{2}\right) = 0 \]

Both points have zero velocity, which indicates they are at rest, but we need to check for maximum acceleration. The acceleration is given by:
\[ a = \frac{d^2y}{dt^2} = -A \pi^2 \sin(\pi x + \pi t) \]
At \( t = 0 \):
\[ a = -A \pi^2 \sin(\pi x) \]

- For \( x = 0.5 \):
\[ a(0.5) = -A \pi^2 \sin(\pi \cdot 0.5) = -A \pi^2 \cdot 1 = -A \pi^2 \] (maximum)

- For \( x = 1.5 \):
\[ a(1.5) = -A \pi^2 \sin(\pi \cdot 1.5) = -A \pi^2 \cdot (-1) = A \pi^2 \] (maximum)

Thus, **Option B is correct**.

#### Option C: Points at \( x = 0.5 \) and \( x = 1.5 \) are at rest
As calculated, both points have zero velocity at \( t = 0 \):
- \( v(0.5) = 0 \)
- \( v(1.5) = 0 \)

Thus, both points are indeed at rest. Therefore, **Option C is correct**.

#### Option D: The given wave is traveling in the negative x direction
The wave equation is of the form:
\[ y = A \sin(kx + \omega t) \]
Where \( k = \pi \) and \( \omega = \pi \). The wave travels in the negative x-direction because of the positive sign in the argument of the sine function. Thus, **Option D is correct**.

### Conclusion:
All options A, B, C, and D are correct.