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(B) From the given figure,we can observe the phase relationship between the waves.Wave $A$ is represented by $y = a \sin(\omega t)$.Wave $B$ starts at

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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) From the given figure,we can observe the phase relationship between the waves.Wave $A$ is represented by $y = a \sin(\omega t)$.Wave $B$ starts at $t=0$ with a negative slope and reaches its mean position at $t=0$,which corresponds to a phase shift of $+\pi/2$ relative to $A$ (since it is a negative sine wave,$y_B = a \sin(\omega t + \pi/2) = a \cos(\omega t)$).Wave $C$ reaches its mean position at $t = \pi/2\omega$,which is $\pi/2$ later than wave $A$. Thus,wave $C$ lags behind wave $A$ by a phase angle of $\pi/2$ $(y_C = a \sin(\omega t - \pi/2))$.Therefore,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 given below shows four progressive waves $A, B, C$ and $D$ with their phases expressed with respect to the wave $A$. It can be calculated from the figure that

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