Final answer:
The question is about finding the constants A and B in the vertical displacement equation of an object experiencing damped harmonic motion. Initial conditions of position and velocity are applied to the displacement equation and its derivative to provide two simultaneous equations that can then be solved for A and B.
Step-by-step explanation:
The student's question deals with an object in a viscous fluid attached to a spring experiencing damped harmonic motion. Given the vertical displacement equation y = Ae−0.216t + Be−0.429t, and the initial conditions of the system being 3.62 cm above the equilibrium and descending at 9.85 cm/s, we are asked to find the constants A and B in the displacement equation.
To solve for A and B, we will use the initial conditions. At t = 0, y(0) = A + B should equal the initial displacement of 3.62 cm. To find the velocities, we derive the displacement equation to get v(t) = -0.216Ae−0.216t - 0.429Be−0.429t. With v(0) = -9.85 cm/s, we get another equation -0.216A - 0.429B = -9.85. Solving these two equations simultaneously will provide the values of A and B.
To find the values of A and B in the equation y = Ae-0.216t + Be-0.429t, we can use the initial conditions given in the problem.
Given that y = 3.62 cm above the equilibrium position and the object is descending at a velocity of -9.85 cm/s, we can find the values of A and B by solving a system of equations.
Using the initial position, we have: 3.62 = A + B
Using the initial velocity, we have: -9.85 = -0.216A - 0.429B
Solving these equations, we find that A = 2.65 and B = 0.97.