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(B) The equation of wave $A$ is given by $y_A = A \sin(\omega t)$.From the graph,wave $B$ starts at its maximum positive value at $t=0$,so its equatio

Vedclass · Accepted Answer

The wave $C$ lags behind by a phase angle of $\pi /2$ and the wave $B$ is ahead by a phase angle of $\pi /2$.

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(B) The equation of wave $A$ is given by $y_A = A \sin(\omega t)$.From the graph,wave $B$ starts at its maximum positive value at $t=0$,so its equation is $y_B = A \sin(\omega t + \pi/2)$. This means wave $B$ is ahead of wave $A$ by a phase angle of $\pi/2$.Wave $C$ starts at its minimum value at $t=0$,so its equation is $y_C = A \sin(\omega t - \pi/2)$. This means wave $C$ lags behind wave $A$ by a phase angle of $\pi/2$.Wave $D$ starts at $t=0$ with a negative slope,representing a phase shift of $\pi$ relative to $A$,so $y_D = A \sin(\omega t - \pi)$.Thus,wave $C$ lags behind by a phase angle of $\pi/2$ and wave $B$ is ahead by a phase angle of $\pi/2$.

The figure shows four progressive waves $A, B, C$ and $D$ with their phases expressed with respect to the wave $A$. It can be concluded from the figure that

Similar Questions

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