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

### Step 1: Understand the given wave equation
The wave traveling on the light string is given by:
\[ y = A \sin(kx - \omega t + 30^\circ) \]

### Step 2: Determine the reflection characteristics
When a wave is reflected from a heavier medium (like a heavy string), it undergoes a phase change of \(180^\circ\) (or \(\pi\) radians) and a change in amplitude based on the energy reflection coefficient.

### Step 3: Calculate the amplitude of the reflected wave
Given that 64% of the incident energy is reflected, we can find the ratio of the amplitudes. The energy of a wave is proportional to the square of the amplitude. Therefore, if \(E_i\) is the incident energy and \(E_r\) is the reflected energy:
\[ \frac{E_r}{E_i} = \frac{64}{100} = 0.64 \]
Since energy is proportional to the square of the amplitude:
\[ \frac{A_r^2}{A_i^2} = 0.64 \]
Taking the square root gives:
\[ \frac{A_r}{A_i} = \sqrt{0.64} = 0.8 \]
Thus, the amplitude of the reflected wave \(A_r\) is:
\[ A_r = 0.8 A \]

### Step 4: Write the equation for the reflected wave
The reflected wave will have the same wave number \(k\) and angular frequency \(\omega\), but the direction of propagation will change, and there will be a phase shift of \(180^\circ\). Therefore, the equation of the reflected wave can be written as:
\[ y_r = A_r \sin(kx + \omega t + 30^\circ + 180^\circ) \]
Substituting \(A_r = 0.8A\):
\[ y_r = 0.8A \sin(kx + \omega t + 210^\circ) \]

### Final Answer
Thus, the equation of the reflected wave is:
\[ y_r = 0.8A \sin(kx + \omega t + 210^\circ) \]

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