Question

The displacement of the medium in a sound wave is given by the equation y = A cos (ax + bt) where A, a and b are constants. The wave is reflected by an obstacle situated at x = 0. The intensity of the reflected wave is 0.64 times that of the incident wave. The equation for the reflected wave is :

• A. 0.64 A cos (ax-bt)
• B. 0.8A cos (-ax+bt)
• C. 0.6A cos (ax-bt)
• D. 0.36A cos (bt-ax)

Answer 1

The correct option is B 0.8A cos (-ax+bt)

The given equation of the incident sound wave ,

$Y=A\cos (ax+bt)=A\cos (bt+ax)$ ,

this wave is moving along the negative direction of X-axis ,

If this wave is reflected by an obstacle (rigid surface) , then equation of the reflected wave ,

$Y_r=A_r\cos (bt-ax+\pi )=-A_r\cos (bt-ax)$ , (moving along the positive direction of X-axis) ,

now we have , intensity , $I\propto A^2$ ,

therefore , $I_r/I=A_r^2/A^2$ ,

given $I_r=0.64I$ ,

hence , $A_r/A=\sqrt{I_r/I}=\sqrt{0.64I/I}=0.8$ ,

or $A_r=0.8A$ ,

therefore equation of reflected wave ,

$Y=-0.8A\cos (bt-ax)$