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?

You're on a roll — The displacement of a pendulum is given by the equation y(t) = A ( t + ). We are given that the phase constant = (2)/(3). So the equation becomes y(t) = A ( t + (2)/(3)). Step 1: Evaluate the displacement at t=0. Substitute t=0 into the equation: y(0) = A ((0) + (2)/(3)) y(0) = A ((2)/(3)) We know that ((2)/(3)) = (120^) = sqrt(3)2. So, y(0) = A 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. Step 2: Evaluate the slope (velocity) at t=0. The velocity is the derivative of displacement with respect to time: v(t) = (dy)/(dt) = (d)/(dt) [A ( t + (2)/(3))] v(t) = A ( t + (2)/(3)) Substitute t=0: v(0) = A ((0) + (2)/(3)) v(0) = A ((2)/(3)) We know that ((2)/(3)) = (120^) = -(1)/(2). So, v(0) = A (-(1)/(2)) = -(A )/(2) Since A and are positive, v(0) must be negative. This means the graph should be decreasing at t=0. Step 3: Compare with the given options. 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 (B). Drop the next question.