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(D) The general equation for a wave travelling in the positive $x$-direction is $y = A \sin(kx - \omega t + \phi)$.Given the amplitude $A = 1$,the equ

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$\sin(\omega t - kx - \frac{\pi}{6})$

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(D) The general equation for a wave travelling in the positive $x$-direction is $y = A \sin(kx - \omega t + \phi)$.Given the amplitude $A = 1$,the equation becomes $y = \sin(kx - \omega t + \phi)$.At $t = 0$,the equation is $y = \sin(kx + \phi)$.From the graph at $t = 0$,when $x = 0$,$y = -0.5$.Substituting these values: $-0.5 = \sin(0 + \phi) \Rightarrow \sin(\phi) = -0.5$.This gives $\phi = -\frac{\pi}{6}$ or $\phi = -\frac{5\pi}{6}$.Since the wave is moving in the positive $x$-direction,the particle at $x = 0$ is moving upwards (as the wave crest is approaching $x = 0$ from the negative side),so the velocity $v = \frac{dy}{dt} = -\omega \cos(kx - \omega t + \phi)$ must be positive at $t = 0, x = 0$.For $v > 0$,we need $\cos(\phi)  0$ and $\cos(-\frac{5\pi}{6}) = -\frac{\sqrt{3}}{2} Thus,$y = \sin(kx - \omega t - \frac{5\pi}{6}) = \sin(\omega t - kx + \frac{5\pi}{6})$.However,checking the options provided,the standard form for a wave moving in the positive $x$-direction is $y = \sin(\omega t - kx + \phi)$.At $t = 0, x = 0, y = -0.5$,we have $\sin(\phi) = -0.5$,so $\phi = -\frac{\pi}{6}$.Substituting this,$y = \sin(\omega t - kx - \frac{\pi}{6})$,which matches option $D$.

The equation of a wave travelling along the positive $x$-axis,as shown in the figure at $t = 0$,is given by:

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