Question

A wave travels on a light string. The equation of the wave is $Y=A\sin (kx-\omega t+30+180)$. 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. $Y=0.8A\sin (kx-\omega t+30+180)$
• B. $Y=0.8A\sin (kx+\omega t+30+180)$
• C. $Y=0.8A\sin (kx+\omega t-30)$
• D. $Y=0.8A\sin (kx+\omega t+30)$

Answer 1

The correct option is D $Y=0.8A\sin (kx+\omega t+30)$

$Y=A\sin (kx-wt+30+180)$

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 equ of the reflected wave is

$y=0.8Asin(kx+wt+{30}^0+2\pi )$

$\Rightarrow [y=0.8Asin(kx+wt+{30}^0)]$

Hence Option (D) is correct.