A wave travels on a light string. The equation of the wave is `Y = A sin (kx - omegat + 30^(@)+180^(@))`. It is relected from a heavy string tied to an end of the light string at `x = 0`. If `64%` of the incident energy is reflected 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^(@))`

Answer

To find the equation of the reflected wave when a wave traveling on a light string reflects off a heavy string, we can follow these steps:

### Step 1: Understand the given wave equation
The incident wave is given by the equation:
\[ Y = A \sin(kx - \omega t + 30^\circ + 180^\circ) \]

### Step 2: Simplify the wave equation
We can simplify the phase of the wave:
\[ 30^\circ + 180^\circ = 210^\circ \]
Thus, the equation becomes:
\[ Y = A \sin(kx - \omega t + 210^\circ) \]

### Step 3: Determine the percentage of energy reflected
We know that 64% of the incident energy is reflected. The intensity of a wave is proportional to the square of its amplitude. Therefore, if \( I \) is the intensity of the incident wave, the intensity of the reflected wave \( I_r \) can be expressed as:
\[ I_r = 0.64 I \]

### Step 4: Relate intensity to amplitude
Since intensity is proportional to the square of the amplitude, we have:
\[ \frac{I_r}{I} = \left(\frac{A_r}{A}\right)^2 \]
Substituting the values, we get:
\[ 0.64 = \left(\frac{A_r}{A}\right)^2 \]

### Step 5: Solve for the amplitude of the reflected wave
Taking the square root of both sides:
\[ \frac{A_r}{A} = \sqrt{0.64} \]
\[ A_r = A \cdot 0.8 \]

### Step 6: Determine the phase change upon reflection
When a wave reflects off a heavier medium, it undergoes a phase change of \( 180^\circ \) (or \( \pi \) radians). Therefore, the phase of the reflected wave will be:
\[ kx - \omega t + 210^\circ + 180^\circ = kx - \omega t + 390^\circ \]

### Step 7: Simplify the phase of the reflected wave
Since \( 390^\circ \) is equivalent to \( 30^\circ \) (because \( 390^\circ - 360^\circ = 30^\circ \)), we can rewrite the equation:
\[ Y_r = 0.8A \sin(kx - \omega t + 30^\circ) \]

### Final Equation of the Reflected Wave
Thus, the equation of the reflected wave is:
\[ Y_r = 0.8A \sin(kx - \omega t + 30^\circ) \]

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