Question

A wave travels on a light string. The equation of the wave is $y=A\sin (kx-\omega t+{30}^{\circ }$). It is reflected from a heavy string tied to the end of the light string at $x=0$. If $64\%$ of the incident energy is reflected, then the equation of the reflected wave is

• A. $y=0.8A\sin (kx-\omega t+{30}^{\circ }+{180}^{\circ })$
• B. $y=0.8A\sin (kx+\omega t+{30}^{\circ }+{180}^{\circ })$
• C. $y=0.8A\sin (kx-\omega t+{30}^{\circ })$
• D. $y=0.8A\sin (kx+\omega t+{30}^{\circ })$

Answer 1

The correct option is B $y=0.8A\sin (kx+\omega t+{30}^{\circ }+{180}^{\circ })$

We know that when a wave is reflected from a denser medium, there is phase change of ${180}^{\circ }$ and the reflected wave travels in opposite direction.
Also, for the same string and same wave, intensity of wave is proportional to its amplitude square.
Energy of the wave, $E\propto I\propto A^2$
$\Rightarrow \frac{I_r}{I_i}=\frac{A_r^2}{A_i^2}$
$\Rightarrow \frac{0.64I}{I}=\frac{A_r^2}{A^2}$
$\Rightarrow A_r=0.8A$
Hence, equation of the reflected wave is $y=0.8A\sin (kx+\omega t+{30}^{\circ }+{180}^{\circ })$