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 the light string at $x=0$. If $64$% of the incident energy is reflected 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 $Y=0.8A\sin (kx+\omega t+{30}^{\circ }+{180}^{\circ })$

$y=A\sin (Kx-wt+{30}^0)$

Intensity of the reflected wave $=64$%$=0.64$

Since $I\alpha A^2$

Amplitude of the reflected wave $=\sqrt{I}=\sqrt{0.64}$

$=0.8$ of the original

i.e. $80$%

The reflected wave always have a phase difference of $\pi$

Therefore equation of the reflected wave is

$[y=0.8Asin(Kx+wt+{30}^0+{180}^0)]$

Hence Option (B) is correct.