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1 year 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]1 year ago

5 0

Answer:

A that's is the answer

letter A is the answer

Strike441 [17]1 year 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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