Download Version 001 – shmgravityII – holland – (1570)

Version 001 – shmgravityII – holland – (1570)
This print-out should have 22 questions.
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before answering.
AP B 1993 MC 43
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.
5
y (cm)
t (s)
1
2
3
4
5
5
At what time t in the period shown does
the particle achieve its maximum positive acceleration?
1
particle is slowing down. At t = 1 s, the particle is momentarily at rest and v = 0. Just
after t = 1 s , the velocity is positive since
the particle is speeding up. Remember that
∆v
a=
, acceleration is a positive maximum
∆t
because the velocity is changing from a negative to a positive value.
AP B 1993 MC 9
002 10.0 points
When an object oscillating in simple harmonic
motion is at its maximum displacement from
the equilibrium position, which of the following is true of the values of its speed and the
magnitude of the restoring force?
Magnitude of
Speed
Restoring Force
1. Maximum
Zero
1. t = 4 s
2. Zero
Maximum correct
2. t = 1 s correct
3. Maximum
3. t = 3 s
4.
4. t = 2 s
5. Zero
5. None of these; the acceleration is constant.
Explanation:
This oscillation is described by
πt
y(t) = − sin
,
2
dy
π
πt
v(t) =
= − cos
dt
2
2
2
d y
a(t) = 2
dt
π 2
πt
.
sin
=
2
2
Themaximum
acceleration will occur when
πt
sin
= 1, or at t = 1 s .
2
From a non-calculus perspective, the velocity is negative just before t = 1 s since the
1
maximum
2
1
maximum
2
1
maximum
2
Zero
Explanation:
The maximum displacement occurs at the
turning points (where the velocity or speed
is zero). The magnitude of restoring force is
given by Hooke’s law
F = −k x ,
where k is the spring constant and x is the
displacement. Since x is a maximum, F is a
maximum.
From a different perspective, the displacement from the equilibrium position can be
written as
y = A sin θ ,
where θ is the phase of the oscillation. When
the object is at its maximum displacement
sin θ = 1
π
θ=
2
Version 001 – shmgravityII – holland – (1570)
so its speed is
π
v = ω A cos θ = ω A cos = 0
2
and the restoring force is
π
F = m A ω 2 sin θ = m A ω 2 = m A ω 2 ,
2
at its maximum value.
AP B 1998 MC 56
003 10.0 points
An object moves up and down the y-axis with
an acceleration given as a function of time t
by the expression a = A sin ω t, where A and
ω are constants.
What is the period of this motion?
1. T =
Which graph represents the kinetic energy
K of the object as a function of displacement
x?
K
Kmax
1.
−xmax
K
x
+xmax
Kmax
2.
2π
correct
ω
−xmax
K
x
+xmax
Kmax
2
2. T = ω A
3. T =
2
ω
2π
3.
4. T = 2 π ω
correct
−xmax
K
Kmax
5. T = ω
Explanation:
This motion is a simple harmonic motion.
1
ω = 2πf = 2π
T
2π
T =
.
ω
AP B 1998 MC 8
004 10.0 points
The graph below represents the potential energy U as a function of displacement x for
an object on the end of a spring (F = −k x)
oscillating in simple harmonic motion with
amplitude xmax .
U
4.
−xmax
K
x
+xmax
Kmax
5.
−xmax
K
Kmax
x
+xmax
6.
Umax
−xmax
x
−xmax
x
+xmax
+xmax
x
+xmax
Version 001 – shmgravityII – holland – (1570)
3
A) The block is hung from only one of the
two springs:
K
Kmax
7.
k
x
−xmax
+xmax
Explanation:
At the equilibrium point (x = 0), the velocity is maximum and the kinetic energy is Umax
due to conservation of energy. At the maximum displacement points +xmax and −xmax
the velocity is zero and the kinetic energy is
zero.
From a different perspective, in simple harmonic motion of an object on the end of a
spring, the total energy is conserved. At the
maximum displacement xmax , the kinetic energy is 0, so E = U (xmax ) = Umax .
m
B) The block is hung from the same two
springs, but the springs are connected in
series rather than parallel:
k
U + K = E = Umax
K(x) = Umax − U (x) .
k
m
Thus, K(x) looks like an upsidedown U (x).
AP M 1993 MC 24
005 10.0 points
Two identical massless springs are hung from
a horizontal support. A block of mass m is
suspended from the pair of springs, as shown.
C) An additional mass of m is attached to
the block:
k
k
k
k
2m
m
When the block is in equilibrium, each
spring is stretched an additional ∆x. Then
the block is set into oscillation with amplitude A; when it passes through its equilibrium
point it has a speed v.
In which of the following cases will the
block, when oscillating with amplitude A, also
have speed v when it passes through its equilibrium point? The acceleration of gravity is
g.
1. A and C only
2. A only
3. A, B, and C
4. B only
5. A and B only
Version 001 – shmgravityII – holland – (1570)
4
For springs in series:
k1
k2
6. C only
m
7. None of these correct
8. B and C only
Consider the forces from a spring’s point of
view. The oscillating mass exerts the same
force F (at some instant in time) on each
spring, so
Explanation:
Let
k1 = k ,
k2 = k .
and
Call the displacement of the mass x, and
choose the positive direction to be to the right.
For springs in parallel:
k1
k2
m
F = k 1 x1 ⇒ x1 =
F
k1
F = k 2 x2 ⇒ x2 =
F
.
k2
Now consider the effective spring constant
kseries , where x = x1 + x2 .
F
F
=
x
x1 + x2
F
k1 k2
k1 k2
=
·
=
,
F
F k1 k2
k2 + k1
+
k
k2
r1
r
2
k
k
=
.
=
2km
2m
kseries =
Hooke’s law is
F = −k x
and the frequency of oscillation is
ωseries
ω
f≡
.
2π
Using Eq. 2, the velocity is
The forces from the springs on the mass m
are to the left: F1 = −k1 x , F2 = −k2 x, and
F = F1 + F2 so that force equilibrium is
v=
dx
= ω A cos(ω t + δ) .
dt
At the equilibrium point
r
d2 x
−k1 x − k2 x = m a = m 2 .
dt
v = ωA =
This is a differential equation for x(t)
d2 x
k1 + k2
x = 0,
+
dt2
m
k
A.
m
Therefore the angular velocity ω as presented in the question should be the same in
cases A, B, and/or C.
The question presents the springs in parallel (Eq. 3), so
which has a sine solution of the form
x(t) = A sin(ω t + δ) ,
where the angular frequency ω is
r
r
k1 + k2
2k
=
.
ω=
m
m
ωparallel =
Note:
kparallel = k1 + k2 .
so
(1)
r
k1 + k2
=
m
r
2k
.
m
Case A: Only one spring is present:
r
r
k
2k
ωsingle =
6=
.
m
m
Version 001 – shmgravityII – holland – (1570)
Case B: Eq. 4, the springs are in series:
r
r
k
2k
ωseries =
6=
.
2m
m
Case C: Eq. 1, but the mass is doubled:
r
r
r
2k
k
k
ω2 m =
=
6=
.
2m
m
2m
Consequently, none of the choices is the
correct answer.
AP M 1993 MC 07 08
006 (part 1 of 2) 10.0 points
A block on a horizontal frictionless plane is
attached to a spring, as shown below. The
block oscillates along the x-axis with simple
harmonic motion of amplitude A.
k
vx0
m
−A
0
+A
Which statement about the block is correct?
1. At x = A, its displacement is at a maximum. correct
2. At x = A, its velocity is at a maximum.
3. At x = 0, its velocity is zero.
4. At x = 0, its acceleration is at a maximum.
5. At x = A, its acceleration is zero.
5
3. The potential energy of the spring is at a
minimum at x = A.
4. The kinetic energy of the block is always
equal to the potential energy of the spring.
5. The kinetic energy of the block is at a
maximum at x = A.
Explanation:
From conservation of energy, v = 0 at x =
±A, so the kinetic energy is zero and the
spring potential energy is at its maximum. At
x = 0, the spring potential energy is 0 and the
kinetic energy is at its maximum.
AP M 1993 MC 33
008 10.0 points
A simple pendulum consists of a brass sphere
of mass m = 1 kg suspended on a string of
length L ≈ 1 meter. The pendulum oscillates
with amplitude A = 1 cm and period T =
2 seconds.
Which of the following alterations will
change the period of the pendulum’s oscillations to T ′ = 1 second?
1. Using a longer string of length L′ ≈ 4 meters.
2. Using a longer string of length L′ ≈ 2 meters.
3. Reducing the oscillation amplitude to
A′ = 0.25 cm.
Explanation:
The block oscillates from maximum displacements at x = A and x = −A. At those
points the velocity is momentarily zero.
4. Using a lighter brass sphere of mass m′ ≈
0.25 kg.
007 (part 2 of 2) 10.0 points
Which statement about energy is correct?
5. Using a shorter string of length L′ ≈
0.25 meters. correct
1. The kinetic energy of the block is at a
minimum at x = 0.
6. Using a shorter string of length L′ ≈
0.5 meters.
2. The potential energy of the spring is at a
minimum at x = 0. correct
7. Increasing the oscillation amplitude to
A′ = 4 cm.
Version 001 – shmgravityII – holland – (1570)
′
8. Using a heavier brass sphere of mass m ≈
4 kg.
Explanation:
As long as the oscillation amplitude is small
compared to the string length, the pendulum’s period is given by a simple formula
s
L
T = 2π
.
g
Note that this period does not depend on
the mass of the pendulum’s bob (the brass
sphere) or the oscillation amplitude, but only
on the pendulum’s length L and the local
gravitational field g. And since we cannot
change g, the only way to change the period
is to change the
√ length L.
Since T ∝ L and we want to halve the
period,
T′
1s
1
=
= ,
T
2s
2
we need
√
2
1
L′
L′
1
1
√ =
⇒
=
=
2
L
2
4
L
1
and hence L′ = L ≈ 0.25 meters.
4
AP M 1998 MC 10
009 10.0 points
A pendulum with a period of 1 s on Earth,
where the acceleration due to gravity is g, is
taken to another planet, where its period is 2
s.
The acceleration due to gravity on the other
planet is most nearly
1. ag = g
2. ag = 2 g
3. ag = 4 g
g
2
g
5. ag = correct
4
Explanation:
4. ag =
6
For a pendulum, the relationship between
s
ℓ
the period and the acceleration is T = 2 π
.
g
1
T ∝ √ , so
g
r
g2
T1
=
T2
g1
r
1
g2
=
2
g1
1
g2
=
g1
4
1
g2 = g1 ,
4
which gives the acceleration on the other
g
planet as .
4
AP M 1998 MC 35
010 10.0 points
An ideal massless spring is fixed to the wall
at one end, as shown below. A block of mass
M attached to the other end of the spring
oscillates with amplitude A on a frictionless,
horizontal surface. The maximum speed of
the block is vm .
k
vm
m
−A
0
+A
What is the force constant k of the spring?
1. k =
2. k =
3. k =
4. k =
5. k =
2
m vm
2A
2
m vm
correct
A2
mg
A
m g vm
2A
2
m vm
2 A2
Explanation:
For the ideal harmonic oscillation of the
spring system, the kinetic energy maximum is
Version 001 – shmgravityII – holland – (1570)
equal to the potential energy maximum which
is also the total energy of the system, so we
obtain
1
1
2
k A2 = m vm
2
2
2
m vm
.
k=
A2
AP M 1998 MC 9
011 10.0 points
The equation of motion of a simple harmonic
oscillator is
d2 x
= −9 x ,
dt2
where x is displacement and t is time.
What is the period of oscillation?
3
2π
2π
2. T =
9
9
3. T =
2π
2π
4. T =
correct
3
1. T =
5. T = 6 π
Explanation:
d2 x
= −ω 2 x ,
dt2
where ω is the angular frequency, so the period of oscillation is
T =
2π
2π
2π
=√ =
.
ω
3
9
AP B 1993 MC 6
012 10.0 points
If Spacecraft X has twice the mass of Spacecraft Y , then what is true about X and Y ?
I) On Earth, X experiences twice the gravitational force that Y experiences;
II) On the Moon, X has twice the weight of
Y;
III) When both are in the same circular orbit,
X has twice the centripetal acceleration
of Y .
7
1. I, II, and III
2. III only
3. I only
4. I and II only correct
5. II and III only
Explanation:
I) gravitational force ∝ mass.
II) weight ∝ mass.
III) The centripetal acceleration is determined by
ac =
v2
,
r
so X and Y should have the same centripetal
acceleration when they are in the same circular orbit.
AP B 1993 MC 6 02
013 10.0 points
If Spacecraft X has twice the mass of Spacecraft Y , then true statements about X and Y
include which of the following?
I) On Earth, X experiences twice the gravitational force that Y experiences.
II) On the Moon, X has twice the weight of
Y.
III) When both are in the same circular orbit,
X has twice the centripetal acceleration
of Y .
1. II and III only
2. I only
3. I and II only correct
4. III only
5. I, II, and III
Explanation:
I) gravitational force ∝ mass
II) weight ∝ mass
III) The centripetal acceleration is deter-
Version 001 – shmgravityII – holland – (1570)
8
mined by
M
,
r2
where G is the gravitational constant, M is
the planet’s mass and r is the radius of the
spacecraft’s orbit. Thus X and Y should have
the same centripetal acceleration when they
are in the same circular orbit.
ac = G
AP B 1998 MC 39
014 10.0 points
An object has a weight W when it is on the
surface of a planet of radius R.
What will be the gravitational force on the
object after it has been moved to a distance
of 4 R from the center of the planet?
1. F = W
2. F =
1
W correct
16
3. F = 16 W
4. F = 4 W
1
W
4
Explanation:
On the surface of the planet,
5. F =
GM m
.
R2
When the object is moved to a distance 4 R
from the center of the planet, the gravitational
force on it will be
GM m
F =
(4 R)2
GM m
=
16 R2
1 GM m
=
16 R2
1
W .
=
16
W=
AP B 1993 MC 8
015 10.0 points
Two spheres have equal densities and are subject only to their mutual gravitational attraction.
Which quantity must have the same magnitude for both spheres?
1. kinetic energy
2. displacement from the center of mass
3. velocity
4. acceleration
5. gravitational force correct
Explanation:
Two spheres with the same density have
different masses due to their relative sizes.
Using Newton’s third law, F~1 = −F~2 .
All of the other quantities (acceleration, velocity, kinetic energy, and displacement from
the center of mass) have different magnitudes
because the two spheres have different masses.
AP B 1998 MC 39 01
016 10.0 points
An object has a weight W when it is on the
surface of a planet of radius R.
What will be the gravitational force on the
object after it has been moved to a distance
of 4 R from the center of the planet?
1
W
2
1
2. F =
W
64
1
3. F = W
4
1. F =
4. F = 16 W
5. F = 2 W
6. F = W
7. F =
1
W
8
Version 001 – shmgravityII – holland – (1570)
8. F = 4 W
1
W correct
16
Explanation:
On the surface of the planet,
9. F =
GM m
.
R2
When the object is moved to a distance 4 R
from the center of the planet, the gravitational
force on it will be
GM m
F =
(4 R)2
GM m
=
16 R2
1 GM m
=
16 R2
1
=
W .
16
W=
AP B 1998 MC 40
017 10.0 points
What is the kinetic energy of a satellite of
mass m that orbits the Earth of mass M in a
circular orbit of radius R?
1 GM m
correct
2
R
1 GM m
2. K =
2 R2
GM m
3. K =
R2
1 GM m
4. K =
4
R
1. K =
5. K = 0
Explanation:
The gravitational force on the satellite provides the centripetal force needed to keep it
in circular orbit:
GM m
v2
=
F
=
F
=
m
c
G
R2
R
G
M
m
, so
m v2 =
R
K=
1 GM m
1
m v2 =
.
2
2
R
9
AP M 1993 MC 22
018 10.0 points
A newly discovered planet has twice the mass
of the Earth, but the acceleration due to gravity on the new planet’s surface is exactly the
same as the acceleration due to gravity on the
Earth’s surface.
What is the radius Rp of the new planet in
terms of the radius R of Earth?
1. Rp = 4 R
2. Rp = 2 R
√
2
3. Rp =
R
2
1
4. Rp = R
2
√
5. Rp = 2 R correct
Explanation:
From Newton’s second law and the law of
universal gravitation, the gravitational force
near the surface is
Mm
r2
GM
g= 2 .
r
Fg = m g = G
Mp = 2 Me and gp = ge , so
G Mp
G Me
2 G Me
=
=
2
2
R
Rp
Rp2
2
1
= 2
2
R
Rp
√
Rp = 2 R .
AP M 1993 MC 32
019 10.0 points
A satellite is in an elliptical orbit around a
planet as shown, with r1 and r2 being its
closest and farthest distances, respectively,
from the center of the planet.
Version 001 – shmgravityII – holland – (1570)
1
r1
Planet
r2
~v2
Satellite 2
10
v
M
v
M
~v1
D
If the satellite has a speed v1 ≡ k~v1 k at
its closest distance, what is its speed at its
farthest distance?
r2
v1
1. k~v2 k =
r1
r1 + r2
2. k~v2 k =
v1
2
3. k~v2 k = (r2 − r1 ) v1
r22
v1
r12
r2
5. k~v2 k = 12 v1
r2
r2 − r1
6. k~v2 k =
v1
r1 + r2
r2 + r1
7. k~v2 k =
v1
r1 − r2
r1
8. k~v2 k =
v1 correct
r2
Explanation:
The angular momentum of the satellite S
relative to the planet P remains the same as
the satellite travels in the orbit, because the
torque on the satellite is zero:
4. k~v2 k =
m r1 v1 = m r2 v2
r1
v1 .
v2 =
r2
AP M 1998 MC 20
020 10.0 points
Two identical stars, a fixed distance D apart,
revolve in a circle about their mutual center
of mass, as shown below. Each star has mass
M and speed v.
Which of the following is a correct relationship among these quantities? G is the
universal gravitational constant.
2 G M2
D
GM
2. v 2 =
D
4 G M2
3. v 2 =
D
4GM
4. v 2 =
D
1. v 2 =
5. v 2 = M G D
2GM
D
GM
7. v 2 =
D2
GM
8. v 2 =
correct
2D
Explanation:
From circular orbital movement, the cen2 v2
v2
=
.
tripetal acceleration is a =
D
D
2
Using Newton’s second law of motion, the
acceleration is
6. v 2 =
a=
F
1 G M2
GM
=
·
=
2
M
M
D
D2
F
GM
2 v2
=a=
=
D
M
D2
GM
.
v2 =
2D
AP M 1998 MC 7 8
Version 001 – shmgravityII – holland – (1570)
021 (part 1 of 2) 10.0 points
A ball is tossed straight up from the surface
of a small, spherical asteroid with no atmosphere. The ball rises to a height equal to the
asteroid’s radius and then falls straight down
toward the surface of the asteroid.
What forces, if any, act on the ball while it
is on the way up?
1. Only an increasing gravitational force
that acts downward
2. No forces act on the ball.
3. Both a constant gravitational force that
acts downward and a decreasing force that
acts upward
4. Only a decreasing gravitational force that
acts downward correct
5. Only a constant gravitational force that
acts downward
Explanation:
There is no friction in the system, and the
ball doesn’t have any contact with other objects, so the only force acting on the ball is
the attractive gravitational force, which acts
downward.
~ = −G M m r̂, the force will deFrom F
r2
crease as the ball rises.
022 (part 2 of 2) 10.0 points
The acceleration of the ball at the top of its
path is
1. equal to one-half the acceleration at the
surface of the asteroid.
2. zero.
3. equal to one-fourth the acceleration at the
surface of the asteroid. correct
4. at its maximum value for the ball’s
flight.
5. equal to the acceleration at the surface of
the asteroid.
Explanation:
1
1
F = m a ∝ 2 , so a ∝ 2 and
r
r
1 1
1
1
=
∝ a.
a′ ∝
2
2
(2 r)
4r
4
11