Two very long string are tied together at the point x = 0 In region x lt 0, the wave speed is `v_(1)`,while in the region x gt0, the speed is `v_(2)`. A sinusoidal wave is incident on the knot from the left(x lt 0). Part of the wave is redlected and part is transmitted. For X lt 0 the the displacement of the wave is described by y(x,t) = A sin(`K_(1)x-wt`)+B sin (`k_(1)x+wt)`, while for x gt 0, y(x,t)=Csin (`K_(1))x-wt`, where`w//k_(1) =v_(1)"and "w//k_(2) = v_(2)`. Which of the following is /are correct .

Options

A. `C/A=(2v_(2))/(v_(1)+v_(2))`
B. `B/A=(v_(2)-v_(1))/(v_(1)+v_(2)`
C. `B^(2)+(v_(1))/(v_(2))C^(2)=A^(2)`
D. A^(2)+(v^(1))/(v_(2))C^(2)=B^(2)`

Answer

To solve the problem, we will analyze the wave behavior at the boundary between the two regions where the wave speeds differ. We will derive the reflection and transmission coefficients based on the given equations.

### Step-by-Step Solution:

1. **Identify the Wave Equations**:
- For \( x < 0 \) (incident and reflected wave):
\[ y(x,t) = A \sin(k_1 x - \omega t) + B \sin(k_1 x + \omega t) \]
- For \( x > 0 \) (transmitted wave):
\[ y(x,t) = C \sin(k_2 x - \omega t) \]

2. **Relate Angular Frequency and Wave Numbers**:
- Given that:
\[ \frac{\omega}{k_1} = v_1 \quad \text{and} \quad \frac{\omega}{k_2} = v_2 \]
- From this, we can express \( k_1 \) and \( k_2 \):
\[ k_1 = \frac{\omega}{v_1} \quad \text{and} \quad k_2 = \frac{\omega}{v_2} \]

3. **Calculate Reflection Coefficient (R)**:
- The reflection coefficient \( R \) is given by:
\[ R = \frac{B}{A} = \frac{k_1 - k_2}{k_1 + k_2} \]
- Substitute \( k_1 \) and \( k_2 \):
\[ R = \frac{\frac{\omega}{v_1} - \frac{\omega}{v_2}}{\frac{\omega}{v_1} + \frac{\omega}{v_2}} \]
- Factor out \( \omega \):
\[ R = \frac{\frac{1}{v_1} - \frac{1}{v_2}}{\frac{1}{v_1} + \frac{1}{v_2}} = \frac{v_2 - v_1}{v_2 + v_1} \]

4. **Calculate Transmission Coefficient (T)**:
- The transmission coefficient \( T \) is given by:
\[ T = \frac{C}{A} = \frac{2k_1}{k_1 + k_2} \]
- Substitute \( k_1 \) and \( k_2 \):
\[ T = \frac{2 \cdot \frac{\omega}{v_1}}{\frac{\omega}{v_1} + \frac{\omega}{v_2}} \]
- Factor out \( \omega \):
\[ T = \frac{2 \cdot \frac{1}{v_1}}{\frac{1}{v_1} + \frac{1}{v_2}} = \frac{2v_2}{v_2 + v_1} \]

5. **Final Results**:
- The reflection coefficient is:
\[ R = \frac{v_2 - v_1}{v_2 + v_1} \]
- The transmission coefficient is:
\[ T = \frac{2v_2}{v_2 + v_1} \]

### Summary of Results:
- Reflection Coefficient \( R = \frac{v_2 - v_1}{v_2 + v_1} \)
- Transmission Coefficient \( T = \frac{2v_2}{v_2 + v_1} \)