Quasi frequency = 4√6
Quasi period = π√6/12
t ≈ 0.4045
<u>Explanation:</u>
Given:
Mass, m = 20g
τ = 400 dyn.s/cm
k = 3920
u(0) = 2
u'(0) = 0
General differential equation:
mu" + τu' + ku = 0
Replacing the variables with the known value:
20u" + 400u' + 3920u = 0
Divide each side by 20
u" + 20u' + 196u = 0
Determining the characteristic equation by replacing y" with r², y' with r and y with 1 in the differential equation.
r² + 20r + 196 = 0
Determining the roots:
![r = \frac{-20 +- \sqrt{(20)^2 - 4(1)(196)} }{2(1)}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7B-20%20%2B-%20%5Csqrt%7B%2820%29%5E2%20-%204%281%29%28196%29%7D%20%7D%7B2%281%29%7D)
r = -10 ± 4√6i
The general solution for two complex roots are:
y = c₁ eᵃt cosbt + c₂ eᵃt sinbt
with a the real part of the roots and b be the imaginary part of the roots.
Since, a = -10 and b = 4√6
u(t) = c₁e⁻¹⁰^t cos 4√6t + c₂e⁻¹⁰^t sin 4√6t
u(0) = 2
u'(0) = 0
(b)
Quasi frequency:
μ = ![\frac{\sqrt{4km - y^2} }{2m}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B4km%20-%20y%5E2%7D%20%7D%7B2m%7D)
![= \frac{\sqrt{4(3929)(20) - (400)^2} }{2(20)} \\\\= 4\sqrt{6}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Csqrt%7B4%283929%29%2820%29%20-%20%28400%29%5E2%7D%20%7D%7B2%2820%29%7D%20%5C%5C%5C%5C%3D%204%5Csqrt%7B6%7D)
(c)
Quasi period:
T = 2π / μ
![T = \frac{2\pi }{4\sqrt{6} } \\\\T = \frac{\pi\sqrt{6} }{12}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%20%7D%7B4%5Csqrt%7B6%7D%20%7D%20%5C%5C%5C%5CT%20%3D%20%5Cfrac%7B%5Cpi%5Csqrt%7B6%7D%20%20%7D%7B12%7D)
(d)
|u(t)| < 0.05 cm
u(t) = |2e⁻¹⁰^t cos 4√6t + 5√6/6 e⁻¹⁰^t sin 4√6t < 0.05
solving for t:
τ = t ≈ 0.4045
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Answer:
Approximately 21 km.
Explanation:
Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:
- No intersection: There's nothing that blocks the camera's view of the top of the building.
- Two intersections: The planet blocks the camera's view of the top of the building.
- One intersection: The point at which the top of the building appears or disappears.
There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.
The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle
which corresponds to this minor arc.
This angle comes can be split into two parts:
.
Also,
.
The radius of this circle is:
.
The lengths of segment DC, AC, BC can all be found:
In the two right triangles
and
, the value of
and
can be found using the inverse cosine function:
![\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cangle%20%5Cmathrm%7BB%5Chat%7BC%7DA%7D%20%3D%20%5Ccos%5E%7B-1%7D%7B%5Crm%20%5Cfrac%7BAC%7D%7BBC%7D%7D%20)
![\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cangle%20%5Cmathrm%7BD%5Chat%7BC%7DA%7D%20%3D%20%5Ccos%5E%7B-1%7D%7B%5Crm%20%5Cfrac%7BAC%7D%7BDC%7D%7D%20)
.
The length of the minor arc will be:
.
Answer:
1. no
2. it would be different because of gravity.
3. no
4. they would be different because of the unit of measurment used. Mexico and any other country would use the metric system.