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.
1. Amplitude of reflected wave is

Options

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

Answer

C

Explanation

`sinax=+-1 implies ax=(pi)/(2),(3pi)/(2),(5pi)/(2), . . . implies x=(3pi)/(2a)` for 2nd antinodes.