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 transmitted wave is :

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

Answer 1

The correct option is C 0.6 A 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 transmitted wave ,

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

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

therefore , $I_t/I=A_t^2/A^2$ ,

given $I_r=0.64I$ , (where $I=$ intensity of incident wave , and $I_r=$ intensity of reflected wave)

as intensity of reflected wave is $I_r$ , therefore intensity of transmitted wave , $I_t=I-I_r=I-0.64I=0.36I$

hence , $A_t/A=\sqrt{I_t/I}=\sqrt{0.36I/I}=0.6$ ,

or $A_t=0.6A$ ,

therefore equation of reflected wave ,

$Y=0.6A\cos (bt+ax)$