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).

This equation for y(t) can be expressed as follows:

was its value of the given equation for angles .

The weight will be midway between its equilibrium point and its maximum negative position if...

and

The weight would be halfway from equilibrium towards the maximum negative position around–approximately , and that every half-second subsequently.

Therefore the final answer is "0.20 second and 0.36 second".

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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

c

step-by-step explanation: