A wave is described by `y=Asin(omegat-kx)`. The speeds of which of the following particles in the medium are increasing at time t=0?

Options

A. `a.x=0`
B. `x=(pi)/(4k)`
C. c.`x=(pi)/(2k)`
D. d.`x=(3pi)/(4k)`

Answer

To solve the problem of determining which particles in the medium have increasing speeds at time \( t = 0 \), we will analyze the wave function given by:

\[ y = A \sin(\omega t - kx) \]

### Step 1: Determine the velocity of the particles

The velocity \( v \) of the particles in the medium can be found by taking the partial derivative of \( y \) with respect to time \( t \):

\[ v = \frac{\partial y}{\partial t} = A \omega \cos(\omega t - kx) \]

At \( t = 0 \):

\[ v = A \omega \cos(-kx) = A \omega \cos(kx) \]

### Step 2: Determine the acceleration of the particles

The acceleration \( a \) can be found by taking the derivative of the velocity with respect to time:

\[ a = \frac{\partial v}{\partial t} = -A \omega^2 \sin(\omega t - kx) \]

At \( t = 0 \):

\[ a = -A \omega^2 \sin(-kx) = A \omega^2 \sin(kx) \]

### Step 3: Analyze conditions for increasing speed

A particle's speed is increasing when the velocity and acceleration have the same sign. Therefore, we need to analyze the signs of \( v \) and \( a \):

1. **For \( x = 0 \)**:
- \( v = A \omega \cos(0) = A \omega \) (positive)
- \( a = A \omega^2 \sin(0) = 0 \) (neither positive nor negative)
- **Conclusion**: Speed is constant.

2. **For \( x = \frac{\pi}{4k} \)**:
- \( v = A \omega \cos\left(\frac{\pi}{4}\right) = A \omega \cdot \frac{1}{\sqrt{2}} \) (positive)
- \( a = A \omega^2 \sin\left(\frac{\pi}{4}\right) = A \omega^2 \cdot \frac{1}{\sqrt{2}} \) (positive)
- **Conclusion**: Speed is increasing.

3. **For \( x = \frac{\pi}{2k} \)**:
- \( v = A \omega \cos\left(\frac{\pi}{2}\right) = 0 \) (zero)
- \( a = A \omega^2 \sin\left(\frac{\pi}{2}\right) = A \omega^2 \) (positive)
- **Conclusion**: Speed is increasing (from zero).

4. **For \( x = \frac{3\pi}{4k} \)**:
- \( v = A \omega \cos\left(\frac{3\pi}{4}\right) = -A \omega \cdot \frac{1}{\sqrt{2}} \) (negative)
- \( a = A \omega^2 \sin\left(\frac{3\pi}{4}\right) = A \omega^2 \cdot \frac{1}{\sqrt{2}} \) (positive)
- **Conclusion**: Speed is decreasing.

### Final Conclusion

The particles with increasing speeds at \( t = 0 \) are located at:
- \( x = \frac{\pi}{4k} \) (both velocity and acceleration are positive)
- \( x = \frac{\pi}{2k} \) (velocity is zero, but acceleration is positive)

Thus, the correct options are **B and C**.