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Olenka [21]
2 years ago
10

The displacement x with respect to time t of a particle moving in simple harmonic motion is given by x = 5cos(16pi*t). where x i

s in mm and t is in seconds. If the particle starts at x = 5mm and t = 0s. At what time t does it first pass through its equilibrium position?
A) 1/32 s
B) 1/16 s
C) 1/5 s
D) 4 s
E) 8 s
Physics
2 answers:
Kitty [74]2 years ago
5 0

Answer:

A that's is the answer

letter A is the answer

Strike441 [17]2 years ago
4 0
T
x=0.05sin6t
(a) the amplitude of the oscillations
A
=
0.05
m
A=0.05m
the period of oscillations
T
=
2
π
ω
=
2
π
6
=
2.1
s
T=
ω
2π
​
=
6
2π
​
=2.1s
the maximum acceleration
a
max
⁡
=
A
ω
2
=
0.05
∗
6
2
=
1.8
m
/
s
2
a
max
​
=Aω
2
=0.05∗6
2
=1.8m/s
2

(b)
x
¨
+
ω
2
x
=
0
x
¨
+ω
2
x=0
ω
=
k
/
m
ω=
k/m
​
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Answer:

<em>v = 381 m/s</em>

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<u>Linear Speed</u>

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Where θ is the angular displacement and t is the time. Solving for t:

\displaystyle t=\frac{\theta}{\omega}

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I'll assume that the child is on Earth at sea level at the equator, so that his distance from the geocenter is 6378000 meters.

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An approximate answer is found from an equation from constant acceleration kinematics:

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Now, the above method is an approximation that makes the technically incorrect assumption that the acceleration of gravity is a constant throughout the entire fall. You get away with it because the drop is very short. In another situation, it might not be. So it would be nice to develop a more accurate method that does not assume constant gravitational acceleration. For that, we begin with the Vis Viva equation:

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Here,

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h = 44.0954 meters

So,

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Now we can calculate that

v = 29.396988 m/s

This is the more nearly correct answer because it takes into account the variability of the gravitational acceleration during the fall.

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