Respuesta :
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

This equation for y(t) can be expressed as follows:
[tex]\to \bold{y(t) = \sqrt{29} \cdot \sin(4\pi t +\ arc \tan(\frac{5}{2}))}[/tex]
[tex]-\frac{1}{2}[/tex] was its value of the given equation for angles [tex]\bold{\frac{7\pi}{6}} \ and \ \bold{\frac{11 \pi}{6}}[/tex].
The weight will be midway between its equilibrium point and its maximum negative position if...
[tex]\to \bold{4\pi t +\arc \tan(\frac{5}{2}) = \frac{7\pi }{6}\ or\ \frac{11\pi}{6}}\\\\ \to \bold{t = \frac{(\frac{7\pi}{6} -\ arc \tan(\frac{5}{2}))}{(4\pi)} \approx 0.196946 \ seconds}[/tex]
and
[tex]\to \bold{t = \frac{(\frac{11\pi}{6} - \ arc \tan(\frac{5}{2})) }{(4\pi)} \approx 0.363613 \ seconds}[/tex]
The weight would be halfway from equilibrium towards the maximum negative position around–approximately [tex]\bold{0.20\ seconds \ and \ 0.36\ seconds}[/tex], and that every half-second subsequently.
Therefore the final answer is "0.20 second and 0.36 second".
Learn more:
brainly.com/question/17107379
