A wave travels on a light string. The equation of the waves is `y= A sin (kx - omegat + 30^@)`. It is reflected from a heavy string tied to an end of the light string at x = 0. If 64% of the incident energy is reflected, then the equation of the reflected wave is

Options

A. `y = 0.8 A sin (kx - omegat + 30^@ + 180^@)`
B. `y = 0.8 A sin (kx + omegat + 30^@ + 180^@)`
C. `y= 0.8 A sin (kx - omegat - 30^@)`
D. `y= 0.8 A sin (kx - omegat + 30^@)`

Explanation

To solve the problem, we need to determine the equation of the reflected wave when a wave traveling on a light string is reflected at the boundary of a heavy string. The given wave equation is:

\[ y = A \sin(kx - \omega t + 30^\circ) \]

### Step 1: Understand the Reflection of Waves
When a wave travels from a lighter medium to a heavier medium, part of the wave is reflected back. The amplitude of the reflected wave can be determined based on the percentage of energy reflected. The energy reflected is given as 64%.

### Step 2: Relate Energy Reflection to Amplitude
The intensity of a wave is proportional to the square of its amplitude. If \( I \) is the intensity, then:

\[ I \propto A^2 \]

If 64% of the energy is reflected, then the intensity of the reflected wave \( I_r \) is:

\[ I_r = 0.64 I_i \]

Where \( I_i \) is the intensity of the incident wave. Therefore, we can express this in terms of amplitudes:

\[ \frac{A_r^2}{A_i^2} = 0.64 \]

### Step 3: Calculate the Amplitude of the Reflected Wave
Let \( A_i \) be the amplitude of the incident wave and \( A_r \) be the amplitude of the reflected wave. Then:

\[ A_r^2 = 0.64 A_i^2 \]

Taking the square root gives:

\[ A_r = \sqrt{0.64} A_i = 0.8 A_i \]

### Step 4: Determine the Phase Change
When a wave reflects off a boundary from a lighter to a heavier medium, it undergoes a phase shift of \( 180^\circ \) (or \( \pi \) radians). This means we need to adjust the phase of the reflected wave.

### Step 5: Write the Equation of the Reflected Wave
The reflected wave travels in the opposite direction, so we replace \( (kx - \omega t) \) with \( (kx + \omega t) \) and include the phase shift:

The equation of the reflected wave becomes:

\[ y_r = A_r \sin(kx + \omega t + 30^\circ + 180^\circ) \]

### Step 6: Simplify the Equation
Since \( 180^\circ \) can be added to the phase:

\[ y_r = 0.8 A_i \sin(kx + \omega t + 210^\circ) \]

### Final Answer
Thus, the equation of the reflected wave is:

\[ y_r = 0.8 A \sin(kx + \omega t + 210^\circ) \]