obennoskey (blo242) – hw11 – Demkov – (59910)
1
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44
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001
10.0 points
A particle oscillates up and down in simple
harmonic motion. Its height
y
as a function
of time
t
is shown in the diagram.
1
2
3
4
5
5
5
y
(cm)
t
(s)
At what time
t
in the period shown does
the particle achieve its maximum positive ac
celeration?
1.
t
= 1 s
correct
2.
t
= 2 s
3.
t
= 3 s
4.
t
= 4 s
5.
None of these; the acceleration is con
stant.
Explanation:
This oscillation is described by
y
(
t
) =

sin
parenleftbigg
π t
2
parenrightbigg
,
v
(
t
) =
d y
dt
=

π
2
cos
parenleftbigg
π t
2
parenrightbigg
a
(
t
) =
d
2
y
dt
2
=
parenleftBig
π
2
parenrightBig
2
sin
parenleftbigg
π t
2
parenrightbigg
.
The maximum acceleration will occur when
sin
parenleftbigg
π t
2
parenrightbigg
= 1, or at
t
= 1 s
.
From a noncalculus perspective, the veloc
ity is negative just before
t
= 1 s since the
particle is slowing down. At
t
= 1 s, the par
ticle is momentarily at rest and
v
= 0. Just
after
t
= 1 s , the velocity is positive since
the particle is speeding up.
Remember that
a
=
Δ
v
Δ
t
,
acceleration is a positive maximum
because the velocity is changing from a nega
tive to a positive value.
002
10.0 points
The equation of motion of a simple harmonic
oscillator is
d
2
x
dt
2
=

9
x ,
where
x
is displacement and
t
is time.
What is the period of oscillation?
1.
T
=
3
2
π
2.
T
=
9
2
π
3.
T
=
2
π
3
correct
4.
T
=
2
π
9
5.
T
= 6
π
Explanation:
d
2
x
dt
2
=

ω
2
x ,
where
ω
is the angular frequency, so the pe
riod of oscillation is
T
=
2
π
ω
=
2
π
√
9
=
2
π
3
.
003
10.0 points
A block rests on a flat plate that executes
vertical simple harmonic motion with a period
of 1
.
82 s.
What is the maximum amplitude of the
motion for which the block does not separate
from the plate? The acceleration of gravity is
9
.
8 m
/
s
2
.
Correct answer: 0
.
82226 m.
Explanation:
Let :
g
= 9
.
8 m
/
s
2
,
and
T
= 1
.
82 s
.
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obennoskey (blo242) – hw11 – Demkov – (59910)
2
At each instant, there are two forces acting
on the block: the force of gravity
F
g
=
m g
and the normal force
N
from the plate. The
block separates from the plate when
N
= 0.
By Newton 2
nd
law
summationdisplay
F
=

m g
+
N
=
m a ,
where up is positive. We see that
N
=
m
(
g
+
a
)
,
so for
N
to vanish, we must have
a
max
=

g ,
(1)
For simple harmonic motion
x
=
A
cos(
ω t
)
a
=
d
2
x
dt
2
=

A ω
2
cos(
ω t
)
,
and at the highest and lowest points, the ac
celeration is at a maximum. The acceleration
of the block is positive (up) at the lowest
point, and negative (down) at the highest
point. We want the negative situation, so
a
max
=

A
max
ω
2
.
Since
T
=
2
π
ω
A
max
ω
2
=
g
A
max
=
g
ω
2
,
and since
T
=
2
π
ω
,
A
max
=
g T
2
4
π
2
=
(9
.
8 m
/
s
2
) (1
.
82 s)
2
4
π
2
=
0
.
82226 m
.
004
10.0 points
An object can oscillate ONLY around
1.
any stable equilibrium point.
correct
2.
any point, provided the forces exerted on
it obey Hooke’s law.
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 Spring '08
 DEMOVK
 Energy, Force, Mass, Simple Harmonic Motion

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