Question

The displacement of the medium in a sound wave is given by the equation $y_1=A\cos (ax+bt)$ 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 if the incident wave.
(a) What are the wavelength and frequency of the incident wave ?
(b) Write the equation for the reflected wave.
(c) In the resultant wave formed after reflection, find the minimum value 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 the amplitude of wave) ,

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

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

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

or $A_2=0.8A$ ,

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

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

therefore particle velocity is ,

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

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

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

$u_{max}=0.8Ab$ ,

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

$u_{min}=0$