The equation of a plane progressive wave is `y=0.04sin4pi[t-(x)/(20)]`. When it is reflected at a denser medium (medium with lesser wave velocity) at x=0, intensity of reflected wave is 81% of that of the incident wave. The equation of the relfected wave is:

Options

A. `y=0.04sin4pi[t-(x)/(20)]`
B. `y=0.04sin4pi[t+(x)/(20)]`
C. `y=+0.036sin4pi[t+(x)/(20)]`
D. `y=-0.036sin4pi[t+(x)/(20)]`

Explanation

To solve the problem step by step, we will analyze the given wave equation, determine the properties of the incident wave, and then derive the equation of the reflected wave.

### Step 1: Identify the parameters of the incident wave
The equation of the plane progressive wave is given as:
\[ y = 0.04 \sin\left(4\pi t - \frac{x}{20}\right) \]

From this equation, we can identify:
- Amplitude \( A_0 = 0.04 \)
- Angular frequency \( \omega = 4\pi \)
- Wave number \( k = \frac{4\pi}{20} = \frac{\pi}{5} \)
- The wave is traveling in the positive x-direction.

### Step 2: Calculate the intensity of the incident wave
The intensity \( I \) of a wave is proportional to the square of its amplitude:
\[ I \propto A^2 \]
Thus, the intensity of the incident wave \( I_0 \) can be expressed as:
\[ I_0 = k A_0^2 \]
Where \( k \) is a proportionality constant.

### Step 3: Determine the intensity of the reflected wave
According to the problem statement, the intensity of the reflected wave \( I_r \) is 81% of the intensity of the incident wave:
\[ I_r = 0.81 I_0 \]

### Step 4: Relate the amplitudes of the incident and reflected waves
Let \( A \) be the amplitude of the reflected wave. The intensity of the reflected wave can also be expressed as:
\[ I_r = k A^2 \]
Setting the two expressions for \( I_r \) equal gives:
\[ 0.81 I_0 = k A^2 \]

### Step 5: Substitute the expression for \( I_0 \)
Substituting \( I_0 = k A_0^2 \) into the equation gives:
\[ 0.81 k A_0^2 = k A^2 \]

### Step 6: Simplify and solve for the amplitude of the reflected wave
Dividing both sides by \( k \) (assuming \( k \neq 0 \)):
\[ 0.81 A_0^2 = A^2 \]
Taking the square root of both sides:
\[ A = A_0 \sqrt{0.81} = A_0 \cdot 0.9 \]

### Step 7: Calculate the new amplitude
Substituting \( A_0 = 0.04 \):
\[ A = 0.04 \cdot 0.9 = 0.036 \]

### Step 8: Determine the phase change upon reflection
When a wave reflects off a denser medium, there is a phase change of \( \pi \) (or 180 degrees). Therefore, the phase of the reflected wave will be:
\[ \text{Phase of reflected wave} = \pi \]

### Step 9: Write the equation of the reflected wave
The equation of the reflected wave will have the same angular frequency and wave number but will travel in the negative x-direction. Thus, the equation becomes:
\[ y = A \sin(\omega t + kx + \pi) \]
Substituting the values:
\[ y = 0.036 \sin(4\pi t + \frac{\pi}{5} x + \pi) \]

### Step 10: Simplify using the sine addition formula
Using the identity \( \sin(\theta + \pi) = -\sin(\theta) \):
\[ y = -0.036 \sin(4\pi t + \frac{\pi}{5} x) \]

### Final Answer
The equation of the reflected wave is:
\[ y = -0.036 \sin\left(4\pi t + \frac{x}{20}\right) \]