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(B) The energy of a wave is proportional to the square of its amplitude $(E \propto A^2)$.Given that $64\%$ of the incident energy is reflected,the ra

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$Y = 0.8 A \sin (kx + \omega t + 30^o + 180^o)$

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(B) The energy of a wave is proportional to the square of its amplitude $(E \propto A^2)$.Given that $64\%$ of the incident energy is reflected,the ratio of the reflected amplitude $(A_r)$ to the incident amplitude $(A_i)$ is given by $\frac{A_r}{A_i} = \sqrt{0.64} = 0.8$.Thus,$A_r = 0.8 A$.Since the wave is reflected from a heavier string (denser medium) at $x = 0$,there is a phase change of $180^o$ (or $\pi$ radians).The incident wave travels in the positive $x$-direction $(kx - \omega t)$,so the reflected wave must travel in the negative $x$-direction $(kx + \omega t)$.The equation for the reflected wave is $Y_r = A_r \sin (kx + \omega t + \phi + 180^o)$.Substituting the values: $Y_r = 0.8 A \sin (kx + \omega t + 30^o + 180^o)$.

$A$ wave travels on a light string. The equation of the wave is $Y = A \sin (kx - \omega t + 30^o)$. 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:

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