For a pendulum with displacement y(t) = A (omega t + ), if the phase constant is = (2pi)/(3), what are the displacement and velocity at t=0?

Substitute $t=0$ into the equation:

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

We know that $\sin\left(\frac{2\pi}{3}\right) = \sin(120^\circ) = \frac{\sqrt{3}}{2}$.
So,

$$y(0) = A \frac{\sqrt{3}}{2}$$

Since $A$ is the amplitude (a positive value), $y(0)$ must be positive. This means the graph should start at a positive value on the y-axis.

The velocity is the derivative of displacement with respect to time:

$$v(t) = \frac{dy}{dt} = \frac{d}{dt} \left[A \sin\left(\omega t + \frac{2\pi}{3}\right)\right]$$ $$v(t) = A \omega \cos\left(\omega t + \frac{2\pi}{3}\right)$$

Substitute $t=0$:

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

We know that $\cos\left(\frac{2\pi}{3}\right) = \cos(120^\circ) = -\frac{1}{2}$.
So,

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

Since $A$ and $\omega$ are positive, $v(0)$ must be negative. This means the graph should be decreasing at $t=0$.

We need a graph that starts at a positive $y$ value and is decreasing at $t=0$.

• Graph (A) starts at $y=0$ and increases. Incorrect.
• Graph (B) starts at a positive $y$ value and decreases. This matches our findings.
• Graph (C) starts at $y=0$ and increases. Incorrect.
• Graph (D) starts at a positive $y$ value and increases. Incorrect.

Therefore, option (B) correctly represents the displacement.

The final answer is $\text{(B)}$.