The plane wave represented by an eqution of the form y = f(x-vt) implies the propagation along the positive x-axis without chang of shape with constant velocity v. Then

Options

A. `(dely)/(delt)=-v((dely)/(delx))`
B. `(dely)/(delt)=-v((del^(2)y)/(delx^(2)))`
C. `(del^(2)y)/(delt^(2))=-v^(2)((del^(2)y)/(delx^(2)))`
D. `(del^(2)y)/(delt^(2))=v^(2)((del^(2)y)/(delx^(2)))`

Answer

To solve the problem regarding the plane wave represented by the equation \( y = f(x - vt) \), we will analyze the wave properties and derive the relationships between the derivatives of the wave function.

### Step-by-Step Solution:

1. **Understanding the Wave Equation**:
The equation \( y = f(x - vt) \) represents a wave traveling in the positive x-direction with a constant velocity \( v \). The function \( f \) describes the shape of the wave, which does not change as it propagates.

2. **Consider a Specific Wave Function**:
Let's take a specific form of the wave function, such as:
\[ y = A \sin(\omega t - kx) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( k \) is the wave number.

3. **Calculate the First Derivative with Respect to Time**:
We need to find \( \frac{dy}{dt} \):
\[ \frac{dy}{dt} = A \omega \cos(\omega t - kx) \]

4. **Calculate the First Derivative with Respect to Position**:
Next, we calculate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = -A k \cos(\omega t - kx) \]

5. **Relate the Derivatives**:
Now, we can relate \( \frac{dy}{dt} \) to \( \frac{dy}{dx} \):
\[ \frac{dy}{dt} = -\omega k \frac{dy}{dx} \]
We know that \( v = \frac{\omega}{k} \), so we can rewrite this as:
\[ \frac{dy}{dt} = -v \frac{dy}{dx} \]

6. **Calculate the Second Derivative with Respect to Time**:
We find \( \frac{d^2y}{dt^2} \):
\[ \frac{d^2y}{dt^2} = -A \omega^2 \sin(\omega t - kx) \]
This can be expressed in terms of \( y \):
\[ \frac{d^2y}{dt^2} = -\omega^2 y \]

7. **Calculate the Second Derivative with Respect to Position**:
Now, we calculate \( \frac{d^2y}{dx^2} \):
\[ \frac{d^2y}{dx^2} = -A k^2 \sin(\omega t - kx) \]
This can also be expressed in terms of \( y \):
\[ \frac{d^2y}{dx^2} = -k^2 y \]

8. **Relate the Second Derivatives**:
We can relate the second derivative with respect to time to the second derivative with respect to position:
\[ \frac{d^2y}{dt^2} = -\frac{\omega^2}{k^2} \frac{d^2y}{dx^2} \]
Since \( v^2 = \frac{\omega^2}{k^2} \), we have:
\[ \frac{d^2y}{dt^2} = -v^2 \frac{d^2y}{dx^2} \]

### Conclusion:
From the analysis, we can conclude that:
- Option A is correct: \( \frac{dy}{dt} = -v \frac{dy}{dx} \)
- Option C is correct: \( \frac{d^2y}{dt^2} = -v^2 \frac{d^2y}{dx^2} \)