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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

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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a)

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Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

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