Chapter 14 - Oscillations w./ QuickCheck Questions

Transcription

Chapter 14 - Oscillations
w./ QuickCheck Questions
© 2015 Pearson Education, Inc.
Anastasia Ierides
Department of Physics and Astronomy
University of New Mexico
November 17, 2015
Review of Last Time
Energy transformations and loss through
thermal energy - from one kind of energy to
another; heat transfer
Introduction to fluids; mass density & pressure
Hydrostatic equilibrium; measuring the pressure
Buoyant forces on objects in a fluid
Archmedes’ principle; Bernoulli Equation
Float or Sink?
The volume of fluid displaced by a floating
object of uniform density is Vf = ρo/ρf Vo.
QuickCheck Question 13.7
Which floating block is most
dense?
A.
B.
C.
D.
E.
Block a
Block b
Block c
Blocks a and b are tied.
Blocks b and c are tied.
Which floating block is most
dense?
A.
B.
C.
D.
E.
Block a
Block b
Block c
Blocks a and b are tied.
Blocks b and c are tied.
The fraction of the object below the surface of the liquid is the
object’s density as a fraction of the liquid’s density (if we can
ignore the density of the air). Since A has the greatest fraction
below the surface it is the most dense.
Boy-o-Buoyancy
These students are competing in a concrete canoe
contest
How can such heavy, dense objects stay afloat?
Boats
The hull of a boat is
really a hollow shell, so
the volume of water
displaced by the shell is
much larger than the
volume of the hull itself
The boat sinks until the weight of the displaced
water exactly matches the boat’s weight reaching
static equilibrium and floating
Balloons
The density of air is low
so the buoyant force is
generally negligible
Balloons cannot be filled
with regular air because
it would weigh the same
amount as the displaced
air and therefore have
no net upward force
For a balloon to
float, it must be
filled with a gas
that has a lower
density than that
of air
Blocks a, b, and c are all the same
size. Which experiences the largest
buoyant force?
Block a
Block b
Block c
All have the same buoyant force.
E. Blocks a and c have the same
buoyant force, but the buoyant
force on block b is different.
A.
B.
C.
D.
Blocks a, b, and c are all the same
size. Which experiences the largest
buoyant force?
Block a
Block b
Block c
All have the same buoyant force.
E. Blocks aprinciple
and c have
thethat
same
Archimedes’
states
the buoyant force on an object
buoyant
is equal
to theforce,
weightbut
of the
the buoyant
fluid displaced by the object. Each
on block
b isthe
different.
objectforce
displaces
exactly
same amount of fluid since each is
A.
B.
C.
D.
the same volume.
Blocks a, b, and c are all the same size. Which is the correct
order of the scale readings?
A.
B.
C.
D.
E.
a=b=c
c>a=b
c>a>b
b>c>a
a=c>b
Blocks a, b, and c are all the same size. Which is the correct
order of the scale readings?
A.
B.
C.
D.
E.
a=b=c
c>a=b
c>a>b
b>c>a
a=c>b
Remember: Same size
⇒ Same volume of liquid displaced
⇒ same buoyant force
but different masse have different weight — scale reads the weight
Fluids in Motion
For fluid dynamics we use a simplified model of an
ideal fluid. We assume
1. The fluid is incompressible. This is a very good
assumption for liquids, but it also holds
reasonably well for a moving gas, such as air.
For instance, even when a 100 mph wind slams
into a wall, its density changes by only about
1%.
Fluids in Motion
2. The flow is steady. That is, the fluid velocity at
each point in the fluid is constant; it does not
fluctuate or change with time. Flow under these
conditions is called laminar flow, and it is
distinguished from turbulent flow.
Fluids in Motion
3. The fluid is nonviscous. Water flows much more
easily than cold pancake syrup because the syrup
is a very viscous liquid. Viscosity is resistance to
flow, and assuming a fluid is nonviscous is
analogous to assuming the motion of a particle is
frictionless. Gases have very low viscosity, and
even many liquids are well approximated as
being nonviscous.
Fluids in Motion
Consider the smoke produced
when lighting incense
The rising smoke begins as
laminar flow, recognizable by
the smooth contours
At some point, the smoke
undergoes a transition to
turbulent flow
Fluids in Motion
A laminar-to-turbulent
transition is not uncommon in
fluid flow
Our model of fluids can only
be applied to laminar flow
The Equation of Continuity
When an
incompressible fluid enters a tube, an
equal volume of the
fluid must leave the
tube
The velocity of the molecules will change with
different cross-section areas of the tube
ΔV1 = A1 Δx1 = A1 v1 Δt = ΔV2 = A2 Δx2 = A2 v2 Δt
The velocity of the molecules
will change with different
cross-section areas of the tube
ΔV1 = A1 Δx1 = A1 v1 Δt = ΔV2 = A2 Δx2 = A2 v2 Δt
Dividing both sides of the previous equation by
∆t gives the equation of continuity:
The volume of an incompressible fluid entering
one part of a tube or pipe must be matched by an
equal volume leaving downstream
A consequence of the equation of continuity is
that flow is faster in narrower parts of a tube,
slower in wider parts
The volume of an incompressible fluid entering
one part of a tube or pipe must be matched by an
equal volume leaving downstream
A consequence of the equation of continuity is
that flow is faster in narrower parts of a tube,
slower in wider parts
The rate at which fluid
flows through a tube
(volume per second) is
called the volume flow
rate Q = vA
The SI units of Q are
m3/s.
Another way to express
the meaning of the
equation of continuity is
to say that the volume
flow rate is constant at
all points in the tube
Water flows from left to right through this pipe. What can you
say about the speed of the water at points 1 and 2?
A. v1 > v2
B. v1 = v2
C. v1< v2
Water flows from left to right through this pipe. What can you
say about the speed of the water at points 1 and 2?
A. v1 > v2
B. v1 = v2
C. v1< v2
Representing Fluid Flow: Streamlines
and Fluid Elements
A streamline is the path
or trajectory followed
by a “particle of fluid”.
Representing Fluid Flow: Streamlines
and Fluid Elements
A fluid element is a
small volume of a fluid,
a volume containing
many particles of fluid
A fluid element has a
shape that can change
and a volume that
remains constant
Fluid Dynamics
A fluid element
changes velocity as it
moves from the wider
part of a tube to the
narrower part
This acceleration of
the fluid element must
be caused by a force
The fluid element is
pushed from both
ends by the
surrounding fluid, that
is, by pressure forces
Fluid Dynamics
To accelerate the fluid
element, the pressure
must be higher in the
wider part of the tube
A pressure gradient is
a region where there is
a change in pressure
from one point in the
fluid to another
Fluid Dynamics
An ideal fluid
accelerates whenever
there is a pressure
gradient
Fluid Dynamics
The pressure is
higher at a point
along a stream line
where the fluid is
moving slower,
lower where the
fluid is moving faster
This property of fluids was discovered by Daniel
Bernoulli and is called the Bernoulli effect
Fluid Dynamics
The speed of a fluid can be measured by a
Venturi tube
Gas flows from left to right through this pipe, whose interior is
hidden. At which point does the pipe have the smallest inner
diameter?
A.
B.
C.
D.
E.
Point a
Point b
Point c
The diameter doesn’t change.
Not enough information to tell.
Gas flows from left to right through this pipe, whose interior is
hidden. At which point does the pipe have the smallest inner
diameter?
A.
B.
C.
D.
E.
Point a
Point b
Point c
The diameter doesn’t change.
Not enough information to tell.
The higher the velocity of fluid, the
lower the pressure at a given point in
the pipe, and the smaller the diameter.
Fluids in Motion
Moving fluids
can exert large
forces — the
air passing
this massive
airplane’s
wings can lift
it into the air
Applications of the Bernoulli Effect
Lift is the upward force
on the wing of an
airplane that makes
flight possible
The wing is designed
such that above the
wing the air speed
increases and the
pressure is low
Below the wing, the air
is slower and the
pressure is high
The high pressure
below the wing pushes
more strongly than the
low pressure from
above, causing lift
In a hurricane, roofs are
“lifted” off a house by
pressure differences
The pressure
differences are small
but the force is
proportional to the area
of the roof
As air moves over a
hill, the streamlines
bunch together, so that
the air speeds up
This means there must
exist a zone of low
pressure at the crest of
the hill
Bernoulli’s Equation
We can find a numerical
relationship for pressure,
height and speed of a fluid
by applying conservation of
energy:
ΔK + ΔU = W
As a fluid moves through a
tube of varying widths,
parts of a segment of fluid
will lose energy that the
other parts of the fluid will
gain
The system moves out of
cylindrical volume V1 and
into V2
The system moves out of
cylindrical volume V1 and
into V2
The kinetic energies are
The net change in kinetic
energy is
The net change in kinetic
energy is
The net change in
gravitational potential
energy is
ΔU = U2 − U1
= ρ ΔVgy2 − ρ ΔVgy1
The positive and negative
work done are
W1 = F1 Δx1 = (p1 A1) Δx1
= p1 (A1 Δx1)
= p1 ΔV
The positive and negative
work done are
W1 = F1 Δx1 = (p1 A1) Δx1
= p1 (A1 Δx1)
= p1 ΔV
W2 = −F2 Δx2 = −( p2 A2) Δx2
= −p2 (A2 Δx2)
= −p2 ΔV
The net work done on the system is:
W = W1 + W2 = p1 ΔV − p 2 ΔV = (p1 − p 2) ΔV
We combine the equations for kinetic energy,
potential energy, and work done:
Rearranged, this equation is Bernoulli’s
equation, which relates ideal-fluid quantities at
two points along a streamline:
You can use Bernoulli’s equation to predict the
pressures and forces due to moving fluids
Viscosity
Viscosity is the measure
of a fluid’s resistance to
flow
A very viscous fluid
flows slowly when
poured
Real fluids (viscous fluids) require a pressure
difference in order to flow at a constant speed
Viscosity
The pressure difference needed to keep a fluid
moving is proportional to vavg and to the tube
length L, and inversely proportional to crosssection area A
η is the coefficient of viscosity with units of
N ⋅ s/m2 or Pa⋅ s
Viscosity
The viscosity
of many
liquids
decreases very
rapidly with
temperature
Poiseuille’s Equation
In an ideal fluid,
all fluid particles
move with the
same speed
For a viscous
fluid, the fluid
moves fastest in
the center of the
tube and
decreases speed
as you move
away from the
center, towards
the walls of the tube, where speed is 0
In average speed of a viscous fluid is
The volume flow rate for a viscous fluid is
This is called the Poiseuille’s Equation after the person
who first performed this calculation.
Springs & Restoring Forces
In Chapter 8, you
learned that a
stretched spring
exerts a restoring
force proportional
to the stretch:
Fsp = –kΔx
Springs & Restoring Forces
In Chapter 8, you
learned that a
stretched spring
exerts a restoring
force proportional
to the stretch:
Fsp = –kΔx
How does this linear restoring force lead to an
oscillation of the spring?
Stop To Think Review
A hanging spring has length 10 cm. A 100 g mass is hung from
the spring, stretching it to 12 cm. What will be the length of
the spring if this mass is replaced by a 200 g mass?
A.
B.
C.
D.
14 cm
16 cm
20 cm
24 cm
A.
B.
C.
D.
14 cm
16 cm
20 cm
24 cm
Remember:
Fsp = –kΔx
A.
B.
C.
D.
14 cm
16 cm
20 cm
24 cm
Remember:
Fsp = –kΔx
ΣFy = Fsp - w = 0
⇒ Fsp = w = mg
A.
B.
C.
D.
14 cm
16 cm
20 cm
24 cm
Remember:
Fsp = –kΔx
ΣFy = Fsp - w = 0
⇒ Fsp = w = mg
Same k different (double) m
⇒ double the displacement
A.
B.
C.
D.
14 cm
16 cm
20 cm
24 cm
Remember:
Fsp = –kΔx
ΣFy = Fsp - w = 0
⇒ Fsp = w = mg
Same k different (double) m
⇒ double the displacement
Equilibrium
Consider the marble
that is free to roll
inside a spherical
bowl
Equilibrium
Consider the marble
that is free to roll
inside a spherical
bowl
It has an equilibrium
position at the bottom
of the bowl where it
will rest with no net
force on it
Equilibrium
If pushed away from
equilibrium, the
marble’s weight leads
to a net force toward
the equilibrium
position
Equilibrium
If pushed away from
equilibrium, the
marble’s weight leads
to a net force toward
the equilibrium
position
Just like with the
spring, this force is the
restoring force
Equilibrium
When the marble is
released from the
side, it does not stop
at the bottom of the
bowl; it rolls up and
down each side of the
bowl, moving
through the
equilibrium position
Oscillations
When the marble is
released from the
side, it does not stop
at the bottom of the
bowl; it rolls up and
down each side of the
bowl, moving
through the
This type of motion or
oscillation is
characterized by a period
and frequency
Oscillations
Consider the girl
attached to the
bungee ropes
Oscillations
Consider the girl
attached to the
bungee ropes
When she moves
down, the springy
ropes pull up
Oscillations
Consider the girl
attached to the
bungee ropes
When she moves
down, the springy
ropes pull up
This restoring force produces an oscillation: one
bounce after another.
Frequency and Period
For an oscillation, the
time to complete one
full cycle is called the
period (T) of the
oscillation
period (T) of the
oscillation
The number of cycles
per second is called
the frequency (f ) of
the oscillation
period (T) of the
oscillation
The number of cycles
per second is called
the frequency (f ) of
the oscillation
The units of frequency
are hertz (Hz), or 1 s–1
A mass oscillates on a horizontal spring with period T = 2.0 s.
What is the frequency?
A.
B.
C.
D.
E.
0.50 Hz
1.0 Hz
2.0 Hz
3.0 Hz
4.0 Hz
A.
B.
C.
D.
E.
0.50 Hz
1.0 Hz
2.0 Hz
3.0 Hz
4.0 Hz
A.
B.
C.
D.
E.
0.50 Hz
1.0 Hz
2.0 Hz
3.0 Hz
4.0 Hz
f = 1/(2.0 s)
= 0.5 s-1
= 0.5 Hz
A.
B.
C.
D.
E.
0.50 Hz
1.0 Hz
2.0 Hz
3.0 Hz
4.0 Hz
f = 1/(2.0 s)
= 0.5 s-1
= 0.5 Hz
If the mass is pulled to the right and then released, how long
will it take for the mass to reach the leftmost point of its
motion?
A.
B.
C.
D.
E.
1.0 s
1.4 s
2.0 s
2.8 s
4.0 s
motion?
A.
B.
C.
D.
E.
1.0 s
1.4 s
2.0 s
2.8 s
4.0 s
Remember: The period is the time to complete one oscillation, which
means returning to the original position. In this case it is only half of an
oscillation.
motion?
A.
B.
C.
D.
E.
1.0 s
1.4 s
2.0 s
2.8 s
4.0 s
Remember: The period is the time to complete one oscillation, which
means returning to the original position. In this case it is only half of an
oscillation.
A typical earthquake produces vertical oscillations of the
earth. Suppose a particular quake oscillates the ground at a
frequency of 0.15 Hz. As the earth moves up and down, what
time elapses between the highest point of the motion and the
lowest point?
A.
B.
C.
D.
1s
3.3 s
6.7 s
13 s
lowest point?
A.
B.
C.
D.
1s
3.3 s
6.7 s
13 s
Half an oscillation
lowest point?
A.
B.
C.
D.
1s
3.3 s
6.7 s
13 s
Half an oscillation
T1 oscillation = 1/(0.15 Hz) ≈ 6.7 s
lowest point?
A.
B.
C.
D.
1s
3.3 s
6.7 s
13 s
Half an oscillation
T1/2 oscillation ≈ 3.3 s
lowest point?
A.
B.
C.
D.
1s
3.3 s
6.7 s
13 s
Half an oscillation
T1/2 oscillation ≈ 3.3 s
Oscillatory Motion
The graph of an oscillatory motion has the form
of a cosine function
Oscillatory Motion
A graph or a function that has the form of a sine
or cosine function is called sinusoidal.
Oscillatory Motion
The sand records
the motion of the
oscillating
pendulum
The sinusoidal
shape tells us that
this is simple
harmonic motion
Oscillatory Motion
A sinusoidal oscillation is called simple
harmonic motion (SHM)
All oscillations show a similar form
Oscillatory Motion
Linear Restoring Forces and Simple
Harmonic Motion (SHM)
If we displace a glider
attached to a spring from
its equilibrium position,
the spring exerts a
restoring force back
toward equilibrium
Linear Restoring Forces and Simple
Harmonic Motion (SHM)
This is a linear restoring
force, that is the net force is
toward the equilibrium
position and is proportional
to the distance from
equilibrium
Motion of Mass on a Spring
The amplitude A is the
object’s maximum
displacement from
equilibrium
Motion of Mass on a Spring
The amplitude A is the
object’s maximum
displacement from
equilibrium
Oscillation about an
with a linear restoring
force is always SHM
Vertical Mass on a Spring
For a hanging
weight, the
equilibrium
position of the
block is where it
hangs motionless
For a hanging
weight, the
equilibrium
position of the
block is where it
hangs motionless
The spring is
stretched by ∆L
The value of ∆L is
determined by
solving the staticequilibrium
problem
The value of ∆L is
determined by
solving the staticequilibrium
problem
Hooke’s Law says
Newton’s first law
for the block in
equilibrium is
Newton’s first law
for the block in
equilibrium is
Therefore the
length of the spring
at the equilibrium
position is
When the block is
above the equilibrium position, the
spring is still
stretched by an
amount ∆L – y
The net force on the
block is
The net force on the
block is
But k ∆L – mg = 0,
from Equation 14.4,
so the net force on
the block is
The role of gravity is to
determine where the
equilibrium position is, but
it doesn’t affect the restoring
force for displacement from
the equilibrium position
The role of gravity is to
determine where the
equilibrium position is, but
it doesn’t affect the restoring
force for displacement from
Due to the linearity of the restoring force, a mass
on a vertical spring oscillates with simple harmonic
motion.
The Pendulum
A pendulum is a mass
suspended from a pivot
point by a light string or rod
The Pendulum
The mass moves along a
circular arc
The Pendulum
The mass moves along a
circular arc
The net force is the
tangential component of the
weight:
The Pendulum
The equation is simplified
for small angles because
sin θ ≈ θ
The Pendulum
sin θ ≈ θ
This is called the smallangle approximation
The Pendulum
sin θ ≈ θ
This is called the smallangle approximation
Therefore the restoring force
is
The Pendulum
The force on a pendulum is
a linear restoring force for
small angles, so the
pendulum will undergo
simple harmonic motion
Describing Simple Harmonic Motion
1. The mass starts at its
maximum positive displacement, y = A. The velocity
is zero, but the acceleration is
negative because there is a net
downward force.
1. The mass starts at its
maximum positive displacement, y = A. The velocity
is zero, but the acceleration is
negative because there is a net
downward force.
2. The mass is now moving
downward, so the velocity is
negative. As the mass nears
equilibrium, the restoring
force—and thus the
magnitude of the acceleration
—decreases.
3. At this time the mass is
moving downward with
its maximum speed. It’s
at the equilibrium
position, so the net force
—and thus the
acceleration—is zero.
3. At this time the mass is
moving downward with
its maximum speed. It’s
at the equilibrium
position, so the net force
—and thus the
acceleration—is zero.
4. The velocity is still
negative but its
magnitude is decreasing,
so the acceleration is
positive.
5. The mass has reached the
lowest point of its motion, a
turning point. The spring is
at its maximum extension, so
there is a net upward force
and the acceleration is
positive.
5. The mass has reached the
lowest point of its motion, a
turning point. The spring is
at its maximum extension, so
there is a net upward force
and the acceleration is
positive.
6. The mass has begun moving
upward; the velocity and
acceleration are positive.
7. The mass is passing through
again, in the opposite
direction, so it has a positive
velocity. There is no net
force, so the acceleration is
zero.
7. The mass is passing through
again, in the opposite
direction, so it has a positive
velocity. There is no net
force, so the acceleration is
zero.
8. The mass continues moving
upward. The velocity is
positive but its magnitude
is decreasing, so the
acceleration is negative.
9. The mass is now back at its
starting position. This is
another turning point. The
mass is at rest but will soon
begin moving downward,
and the cycle will repeat.
The position-versus-time graph for oscillatory
motion is a cosine curve:
x(t) indicates that the position is a function of time
x(t) indicates that the position is a function of time
The cosine function can be written in terms of
frequency:
The velocity graph is an upside-down sine
function with the same period T:
The velocity graph is an upside-down sine
function with the same period T:
The restoring force causes an acceleration:
The accelerationversus-time graph
is inverted from the
position-versustime graph and can
also be written
The accelerationversus-time graph
is inverted from the
position-versustime graph and can
also be written
Example 14.2: Motion of a glider on
a spring
An air-track glider oscillates horizontally on a spring at a
frequency of 0.50 Hz. Suppose the glider is pulled to the right
of its equilibrium position by 12 cm and then released. Where
will the glider be 1.0 s after its release? What is its velocity at
this point?
a spring
PREPARE The
glider undergoes simple harmonic motion with
amplitude 12 cm. The frequency is 0.50 Hz, so the period is T = 1/f = 2.0 s. The glider is released at maximum extension
from the equilibrium position, meaning that we can take this
point to be t = 0.
a spring
1.0 s is exactly half the
period. As the graph of the motion
in the figure shows, half a cycle
brings the glider to its left turning
point, 12 cm to the left of the
equilibrium position. The velocity
at this point is zero.
SOLVE
a spring
1.0 s is exactly half the
period. As the graph of the motion
in the figure shows, half a cycle
brings the glider to its left turning
point, 12 cm to the left of the
equilibrium position. The velocity
at this point is zero.
SOLVE
ASSESS Drawing
a graph was an important step that helped us
make sense of the motion.
Connection to UCM
Circular motion and
simple harmonic
motion are motions
that repeat
Connection to UCM
Circular motion and
simple harmonic
motion are motions
that repeat
Uniform circular
motion projected onto
one dimension is
simple harmonic
motion
Connection to UCM
The x-component of the circular motion when the
particle is at angle ϕ is x = A cos ϕ
Connection to UCM
The angle at a later time is ϕ = ωt
Connection to UCM
The angle at a later time is ϕ = ωt
ω is the particle’s angular velocity: ω = 2πf
Connection to UCM
Therefore the particle’s x-component is expressed
x(t) = A cos(2πft)
Connection to UCM
Therefore the particle’s x-component is expressed
x(t) = A cos(2πft)
This is the same equation for the position of a
mass on a spring
Connection to UCM
The x-component of a particle in uniform circular
motion is simple harmonic motion
Connection to UCM
The x-component of a particle in uniform circular
motion is simple harmonic motion
The x-component of the velocity vector is
vx = −v sin ϕ = −(2πf )A sin(2πft)
Connection to UCM
This corresponds to simple harmonic motion if
we define the maximum speed to be
vmax = 2πfA
Connection to UCM
The x-component of the acceleration vector is
ax = −a cos ϕ = −(2πf )2A cos(2πft)
Connection to UCM
The x-component of the acceleration vector is
ax = −a cos ϕ = −(2πf )2A cos(2πft)
The maximum acceleration is thus
amax = (2πf )2A
Connection to UCM
For simple harmonic motion, if you know the
amplitude and frequency, the motion is
completely specified
Connection to UCM
A mass oscillates on a horizontal spring. It’s velocity is vx and
the spring exerts force Fx. At the time indicated by the arrow,
A. vx is + and Fx is +
B. vx is + and Fx is –
C. vx is – and Fx is 0
D. vx is 0 and Fx is +
E. vx is 0 and Fx is –
vx is 0
vx is 0
vx is 0
vx is -
vx is -
A block oscillates on a vertical spring. When the block is at
the lowest point of the oscillation, it’s acceleration ay is
A. Negative.
B. Zero.
C. Positive.
A. Negative.
B. Zero.
C. Positive.
A. Negative.
B. Zero.
C. Positive.
Example 14.3: Measuring the sway
of a tall building
The John Hancock Center in Chicago is 100 stories high.
Strong winds can cause the building to sway, as is the case
with all tall buildings. On particularly windy days, the top of
the building oscillates with an amplitude of 40 cm (≈16 in)
and a period of 7.7 s. What are the maximum speed and
acceleration of the top of the building?
of a tall building
The John Hancock Center in Chicago is 100 stories high.
Strong winds can cause the building to sway, as is the case
with all tall buildings. On particularly windy days, the top of
the building oscillates with an amplitude of 40 cm (≈16 in)
and a period of 7.7 s. What are the maximum speed and
acceleration of the top of the building?
PREPARE We
will assume that the oscillation of the building is
simple harmonic motion with amplitude A = 0.40 m. The
frequency can be computed from the period:
of a tall building
SOLVE We
can use the equations for maximum velocity and
acceleration in Synthesis 14.1.
of a tall building
SOLVE We
to compute:
vmax = 2πfA = 2π (0.13 Hz)(0.40 m) = 0.33 m/s
amax = (2πf )2A = [2π (0.13 Hz)]2(0.40 m) = 0.27 m/s2
In terms of the free-fall acceleration, the maximum
acceleration is amax = 0.027g.
of a tall building
SOLVE We
to compute:
vmax = 2πfA = 2π (0.13 Hz)(0.40 m) = 0.33 m/s
amax = (2πf )2A = [2π (0.13 Hz)]2(0.40 m) = 0.27 m/s2
In terms of the free-fall acceleration, the maximum
acceleration is amax = 0.027g.
ASSESS The acceleration is quite small, as you would expect;
if it were large, building occupants would certainly complain!
Even if they don’t notice the motion directly, office workers
on high floors of tall buildings may experience a bit of nausea
when the oscillations are large because the acceleration affects
the equilibrium organ in the inner ear.
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Things that are due
Homework #10
Due November 29, 2015 by 11:59 pm
Exam #4
November 24, 2015 at 5:00 pm
Reading Quiz #15
Due December 3, 2015 by 4:59 pm
EXAM #4
Covers Chapters 12-14 & Lectures 19-24
Tuesday, November 24, 2015
Bring a calculator and cheat sheet (turn in with
exam, one side of 8.5”x11” piece of paper)
Practice exam will be available on website
along with solutions
QUESTIONS?

Chapter 14 - Oscillations w./ QuickCheck Questions

Transcription

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