Question

Find the times (to the nearest hundredth of a second) that the weight is halfway to its maximum negative position over the interval 0<_t<_0.5. 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

Answer:

0.20 and 0.36

Step-by-step explanation:

y(t) = 2 sin (4π t) + 5 cos (4π t)

We wish to convert this to:

y = A sin(ωt + φ)

We know that ω = 4π.  We also know the following:

5 = A sin φ

2 = A cos φ

Divide the first equation by the second equation:

5/2 = tan φ

φ = tan⁻¹(5/2)

Now, square the two equations and add them together.

5² + 2² = (A sin φ)² + (A cos φ)²

29 = A²

A = √29

The equation of the wave is therefore:

y = √29 sin(4π t + tan⁻¹(5/2))

The maximum negative position is -√29.  And half of that is -½√29.

-½√29 = √29 sin(4π t + tan⁻¹(5/2))

-½ = sin(4π t + tan⁻¹(5/2))

7π/6 + 2kπ or 11π/6 + 2kπ = 4π t + tan⁻¹(5/2)

7 + 12k or 11 + 12k = 24t + 6 tan⁻¹(5/2) / π

t = (7 + 12k − 6 tan⁻¹(5/2) / π) / 24 or (11 + 12k − 6 tan⁻¹(5/2) / π) / 24

Trying different integer values of k, we find there are two possible values for t between 0 and 0.5, both when k = 0.

t = (7 − 6 tan⁻¹(5/2) / π) / 24 or (11 − 6 tan⁻¹(5/2) / π) / 24

t ≈ 0.20 or 0.36