Question

The displacement of the medium in a sound wave is given by the equation; $y_1=Acos(ax+bt)$ where A, a & b are positive 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. In the resultant wave formed after reflection, find the maximum & minimum values of the particle speed in the medium.

Answer 1

Let the intensity of first wave is $I$ , therefore intensity of second wave will be $0.64I$ ,

now we have , $I\propto A^2$ (where A is amplitude ) ,

hence , $I_2/I_1=A_2^2/A_1^2$ ,

given , $I_1=I,I_2=0.64I,A_1=A$ ,

so $A_2^2=0.64A^{2}I/I$ ,

or $A_2=0.8A$ ,

now equation of reflected wave (moving in +ive -direction , after reflection)

$y_2=0.8A\cos (bt-ax)$ ,

therefore particle velocity is ,

$u=d{y}_2/dt=d\left(0.8\cos \right(bt-ax\left)\right)/dt$ ,

or $u=-0.8bA\sin (bt-ax)$ ,

now maximum particle velocity (when $\sin (bt-ax)=1$ , is maximum)

$u_{max}=0.8bA$ ,

minimum particle velocity (when $\sin (bt-ax)=0$ , is minimum)

$u_{min}=0$