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 an end of light string at x = 0. If $36\%$ of the incident energy is reflected, then the equation of the reflected wave is:

• A. $y=0.6Asin(kx+\omega t-{30}^{\circ })$
• B. $y=0.6Asin(kx-\omega t+{30}^{\circ }+{180}^{\circ })$
• C. $y=0.6Asin(kx+\omega t+{30}^{\circ }+{180}^{\circ })$
• D. $y=0.6Asin(kx+\omega t+{30}^{\circ })$

Answer 1

The correct option is A $y=0.6Asin(kx+\omega t-{30}^{\circ })$

As E $\propto A^2$
$\therefore A^{\prime }=0.6A$
$y=0.6Asin\left(k\right(-x)-\omega t+{30}^{\circ }+{180}^{\circ })$
$y=0.6Asin(kx+\omega t-{30}^{\circ })$