Question

HELLLLLLPPPPPP MEEEE PLEASEEEEE!!!!!! Find the times (to the nearest hundredth of a second) that the weight is halfway to its maximum negative position over the interval . Solve algebraically, and show your work and final answer in the response box. Hint: Use the amplitude to determine what y(t) must be when the weight is halfway to its maximum negative position. Graph the equation and explain how it confirms your solution(s).

Answer 1

0.20, 0.36 seconds

Explanation:

We have already seen that the equation for y(t) can be written as ...

y(t) = √29·sin(4πt +arctan(5/2))

The sine function will have a value of -1/2 for the angles 7π/6 and 11π/6. Then the weight will be halfway from its equilibrium position to the maximum negative position when ...

4πt +arctan(5/2) = 7π/6 or 11π/6

t = (7π/6 -arctan(5/2))/(4π) ≈ 0.196946 . . . seconds

and

t = (11π/6 -arctan(5/2))/(4π) ≈ 0.363613 . . . seconds

The weight will be halfway from equilibrium to the maximum negative position at approximately 0.20 seconds and 0.36 seconds and every half-second thereafter.