A travelling wave `y=Asin(kx-omegat+theta)` passes from a heavier string to a lighter string. The reflected wave has amplitude `0.5A`. The junction of the string is at `x=0` The equation of the reflected is :

Options

A. `y'=0.5A sin (kx+omegat+theta)`
B. `y'=-0.5A sin (kx+omegat+theta)`
C. `y'=-0.5A sin (omegat=kx-theta)`
D. `y'=-0.5A sin (kx+omegat-theta)`

Answer

To find the equation of the reflected wave when a traveling wave passes from a heavier string to a lighter string, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Incident Wave**: The given traveling wave is represented by the equation:
\[ y = A \sin(kx - \omega t + \theta) \]
Here, \(A\) is the amplitude, \(k\) is the wave number, \(\omega\) is the angular frequency, and \(\theta\) is the phase constant.

2. **Understand the Reflection**: When the wave travels from a heavier string to a lighter string, it reflects at the junction. The key points to note are:
- There is no phase change upon reflection when moving from a heavier to a lighter medium.
- The direction of the wave changes, but the amplitude of the reflected wave is given as \(0.5A\).

3. **Determine the Amplitude of the Reflected Wave**: The amplitude of the reflected wave is given as:
\[ A' = 0.5A \]

4. **Write the General Form of the Reflected Wave**: The reflected wave will have the same phase constant \(\theta\) but will change the direction of propagation. Hence, the reflected wave can be written as:
\[ y' = A' \sin(kx + \omega t + \theta) \]

5. **Substitute the Amplitude**: Now substituting the amplitude \(A' = 0.5A\) into the equation, we get:
\[ y' = 0.5A \sin(kx + \omega t + \theta) \]

6. **Final Equation of the Reflected Wave**: Therefore, the equation of the reflected wave is:
\[ y' = 0.5A \sin(kx + \omega t + \theta) \]