Incident wave `y= A sin (ax + bt+ pi/2)` is reflected by an obstacle at x = 0 which reduces intensity of reflected wave by 36%. Due to superposition, the resulting wave consists of a standing wave and a travelling wave given by
`y= -1.6 sin ax sin bt + cA cos (bt + ax)`
where A, a, b and c are positive constants.
2. Value of c is

Options

A. 0.2
B. 0.4
C. 0.6
D. 0.3

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Incident Wave
The incident wave is given as:
\[ y = A \sin(ax + bt + \frac{\pi}{2}) \]
This can be rewritten using the sine addition formula:
\[ y = A \cos(ax + bt) \]
since \(\sin(x + \frac{\pi}{2}) = \cos(x)\).

### Step 2: Determine the Intensity Reduction
The problem states that the intensity of the reflected wave is reduced by 36%. This means that the intensity of the reflected wave is 64% of the incident wave's intensity.

Let \(I_I\) be the intensity of the incident wave and \(I_R\) be the intensity of the reflected wave:
\[ I_R = 0.64 I_I \]
The intensity is proportional to the square of the amplitude, so:
\[ I_I \propto A^2 \]
\[ I_R \propto A_R^2 \]
Thus:
\[ A_R^2 = 0.64 A^2 \]
Taking the square root gives:
\[ A_R = A \sqrt{0.64} = A \cdot 0.8 \]

### Step 3: Determine the Amplitude of the Transmitted Wave
The intensity of the transmitted wave is 36% of the incident wave's intensity:
\[ I_T = 0.36 I_I \]
Similarly, we have:
\[ A_T^2 = 0.36 A^2 \]
Taking the square root gives:
\[ A_T = A \sqrt{0.36} = A \cdot 0.6 \]

### Step 4: Analyze the Resulting Wave
The resulting wave is given as:
\[ y = -1.6 \sin(ax) \sin(bt) + cA \cos(bt + ax) \]
The first term represents a standing wave, and the second term represents a traveling wave.

### Step 5: Relate the Amplitudes
From the standing wave term, the amplitude of the standing wave is:
\[ 2A_R = 2 \cdot 0.8A = 1.6A \]
This matches the coefficient of the standing wave term in the resulting wave equation.

### Step 6: Determine the Value of \(c\)
The traveling wave term has an amplitude of \(cA\). From our earlier calculation, we found:
\[ A_T = 0.6A \]
Thus, we can equate:
\[ cA = 0.6A \]
Dividing both sides by \(A\) (assuming \(A \neq 0\)):
\[ c = 0.6 \]

### Final Answer
The value of \(c\) is:
\[ \boxed{0.6} \]

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