A wave travels on a light string. The equatio | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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:

Similar Questions

The wave pattern on a stretched string is shown in the figure. Interpret what kind of wave this is and find its wavelength.

$A$ rod of length $1.2\,m$ is clamped at the midpoint and its fundamental frequency is $2\,MHz$. What is the speed of the wave inside the rod?

The equation of a stationary wave is $y = 0.8 \cos \left( \frac{\pi x}{20} \right) \sin (200 \pi t)$,where $x$ is in $cm$ and $t$ is in $sec$. The separation between consecutive nodes will be .... $cm$.

Explain the reflection of a wave at a rigid support.

The equation of a stationary wave is $ y = 2 \sin \left( \frac{\pi x}{15} \right) \cos (48 \pi t) $. The distance between a node and its next antinode is (in $units$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module