The displacement $$y(t) = A\sin(\omega t + \phi)$$ of a pendulum for $$\phi = \frac{2\pi}{3}$$ is correctly represented by

$$y(t) = A\sin(\omega t + \phi)$$, $$\phi = \frac{2\pi}{3}$$

$$y(0) = A\sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}A > 0$$

Differentiating to find initial velocity:

$$v(t) = \frac{dy}{dt} = A\omega\cos(\omega t + \phi)$$

$$v(0) = A\omega\cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}A\omega < 0$$

Finding the time intercept where $$y(t) = 0$$:

$$\omega t + \frac{2\pi}{3} = \pi \implies \omega t = \frac{\pi}{3} \implies t = \frac{\pi}{3\omega} > 0$$

Condition matches only in Option (A): $$y(0) > 0$$, slope is negative at $$t=0$$, and $$y=0$$ at a small positive time.