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. 0.2
B. 0.4
C. 0.6
D. 0.3

Answer

A

Explanation

Equation of reflected wave is `y=0.8Asin(ax-bt+(3pi)/(2))`, due to reflection at denser medium a pahse difference of `pi` occurs.
Equation to resulting wave is `Y=Asin(ax+bt+pi//2)+0.8Asin(ax-bt+(3pi)/(2))`
`implies y=Acos(ax+bt)-0.8A(ax-bt)=0.8A[cos(ax+bt)-cos(ax-bt)]+0.2Acos(ax+bt)`
`=-1.6Asinaxsinbt+0.2Acos(ax+bt)implies C=0.2`