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.
3. Position of second antinode is

Options

A. `x = pi/(3a)`
B. `x = (3pi)/a`
C. `x = (3pi)/(2a)`
D. `x = (2pi)/(3a)`

Explanation

To solve the problem step by step, we will analyze the given wave equations and find the position of the second antinode.

### Step 1: Identify the standing wave component
The given wave equation is:
\[ y = -1.6 \sin(ax) \sin(bt) + cA \cos(bt + ax) \]
The standing wave component is represented by:
\[ y_s = -1.6 \sin(ax) \sin(bt) \]
This is the part of the wave that will help us find the nodes and antinodes.

### Step 2: Determine the condition for antinodes
Antinodes occur where the amplitude of the standing wave is maximum. The amplitude of the standing wave is given by the coefficient of the sine term, which is 1.6. Therefore, we set:
\[ -1.6 \sin(ax) = 1.6 \]
This simplifies to:
\[ \sin(ax) = 1 \]

### Step 3: Solve for \( ax \)
The sine function equals 1 at specific angles:
\[ ax = \frac{\pi}{2} + 2n\pi \]
where \( n \) is any integer (0, 1, 2, ...).

### Step 4: Find positions of antinodes
To find the positions of the antinodes, we can express \( x \) in terms of \( n \):
\[ x = \frac{\frac{\pi}{2} + 2n\pi}{a} \]
For the first antinode (when \( n = 0 \)):
\[ x_1 = \frac{\frac{\pi}{2}}{a} = \frac{\pi}{2a} \]
For the second antinode (when \( n = 1 \)):
\[ x_2 = \frac{\frac{\pi}{2} + 2\pi}{a} = \frac{\frac{\pi}{2} + \frac{4\pi}{2}}{a} = \frac{\frac{5\pi}{2}}{a} = \frac{5\pi}{2a} \]

### Step 5: Conclusion
Thus, the position of the second antinode is:
\[ x = \frac{5\pi}{2a} \]

### Final Answer
The position of the second antinode is:
\[ \frac{5\pi}{2a} \]

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