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Fluids
Monday, November 28, 11
Monday, November 28, 11
Density
m = ρV
Monday, November 28, 11
Density
m = ρV
Monday, November 28, 11
Density
m = ρV
Monday, November 28, 11
Density
m = ρV
Monday, November 28, 11
Pressure
The pressure of the
water behind each hole
pushes the water out.
The SI unit of pressure is 1 pascal = 1 Pa = 1 N/m2.
Monday, November 28, 11
Monday, November 28, 11
Monday, November 28, 11
Pressure in a Liquid Increases with Depth
Monday, November 28, 11
Pascal’s Law
Pressure is everywhere equal in a uniform fluid of equal depth.
Monday, November 28, 11
Pascal’s Law
Pressure is everywhere equal in a uniform fluid of equal depth.
Monday, November 28, 11
Pascal’s Law
Pressure is everywhere equal in a uniform fluid of equal depth.
F1 F2
=
A1 A2
Monday, November 28, 11
Monday, November 28, 11
Blaise Pascal (June 19, 1623 – August 19, 1662)
Pascal was a French
mathematician, physicist,
inventor, writer and Catholic
philosopher. He was a child
prodigy who was educated
by his father.
Pascal's earliest work was in
the study of fluids, and
clarified the concepts of
pressure and vacuum by
generalizing the work of
Evangelista Torricelli. Pascal
also wrote in defense of the
scientific method.
Monday, November 28, 11
Weight of air on your hand
Monday, November 28, 11
Weight of air on your hand
Your hand has an area of approximately 150 cm2,
atmospheric pressure is 103,000 Pa, what is the
mass of the air pressing down on your hand?
Monday, November 28, 11
Weight of air on your hand
Your hand has an area of approximately 150 cm2,
atmospheric pressure is 103,000 Pa, what is the
mass of the air pressing down on your hand?
F
150
P = ⇒ F = PA = 103000 ×
= 154.5N
A
10000
F 154.5
F = mg ⇒ m = =
≈ 15.8kg
g
9.8
Monday, November 28, 11
Weight of air on your hand
Your hand has an area of approximately 150 cm2,
atmospheric pressure is 103,000 Pa, what is the
mass of the air pressing down on your hand?
F
150
P = ⇒ F = PA = 103000 ×
= 154.5N
A
10000
F 154.5
F = mg ⇒ m = =
≈ 15.8kg
g
9.8
Why don’t you notice this?
Monday, November 28, 11
Weight of air on your hand
Your hand has an area of approximately 150 cm2,
atmospheric pressure is 103,000 Pa, what is the
mass of the air pressing down on your hand?
F
150
P = ⇒ F = PA = 103000 ×
= 154.5N
A
10000
F 154.5
F = mg ⇒ m = =
≈ 15.8kg
g
9.8
Why don’t you notice this?
The same pressure is also pushing up under your
hand.
Monday, November 28, 11
Monday, November 28, 11
Monday, November 28, 11
Absolute Pressure & Gauge Pressure
What is the pressure inside a flat tire?
Monday, November 28, 11
Absolute Pressure & Gauge Pressure
What is the pressure inside a flat tire?
Atmospheric pressure
Monday, November 28, 11
Absolute Pressure & Gauge Pressure
What is the pressure inside a flat tire?
Atmospheric pressure
So when we see a tire wall that says inflate to
32 lb/in2 (220 kPa), this means that the
difference between the pressure inside the tire
and outside the tire needs to be 32 lb/in2 greater
than atmospheric pressure.
So the total pressure inside the tire is
atmospheric pressure 14.7 lb/in2 (103 kPa) plus
32 lb/in2 (220 kPa), this gives a total pressure of
47 lb/in2 (323 kPa).
Monday, November 28, 11
Monday, November 28, 11
Finding absolute and gauge pressure
By Pascal’s law
Pressure to be
measured, p
Monday, November 28, 11
There are many clever ways to measure pressure
Monday, November 28, 11
There are many clever ways to measure pressure
Monday, November 28, 11
Monday, November 28, 11
Buoyancy and Archimedes Principle
Monday, November 28, 11
Buoyancy and Archimedes Principle
Monday, November 28, 11
Buoyancy and Archimedes Principle
Bouyant Force, FB = ρ f V f g
Monday, November 28, 11
Buoyancy and Archimedes Principle
• The buoyant force is equal to the weight of the displaced fluid.
Bouyant Force, FB = ρ f V f g
Monday, November 28, 11
Monday, November 28, 11
A floating object is in static equilibrium.
Monday, November 28, 11
A floating object is in static equilibrium.
Monday, November 28, 11
Measuring the density of a liquid
Monday, November 28, 11
Measuring the density of a liquid
Monday, November 28, 11
Measuring the density of a liquid
How a boat floats
Monday, November 28, 11
How big does a ballon need to be?
Monday, November 28, 11
Buoyancy and Archimedes Principle
Monday, November 28, 11
Buoyancy and Archimedes Principle
Volume of Statue
Monday, November 28, 11
Buoyancy and Archimedes Principle
Volume of Statue
Weight of displaced seawater
Monday, November 28, 11
Buoyancy and Archimedes Principle
Volume of Statue
Weight of displaced seawater
When at rest net force is zero
Monday, November 28, 11
Buoyancy and Archimedes Principle
Volume of Statue
Weight of displaced seawater
When at rest net force is zero
Monday, November 28, 11
Buoyancy and Archimedes Principle
Weight of displaced air
Monday, November 28, 11
Q14.1
The sphere on the right has twice the
mass and twice the radius of the
sphere on the left.
Compared to the sphere on the left,
the larger sphere on the right has
A. twice the density.
B. the same density.
C. 1/2 the density.
D. 1/4 the density.
E. 1/8 the density.
Monday, November 28, 11
mass m
radius R
mass 2m
radius 2R
A14.1
The sphere on the right has twice the
mass and twice the radius of the
sphere on the left.
Compared to the sphere on the left,
the larger sphere on the right has
mass m
radius R
mass 2m
radius 2R
A. twice the density.
B. the same density.
C. 1/2 the density.
D. 1/4 the density.
E. 1/8 the density.
Monday, November 28, 11
m
m
2m
ρ = , ρ1 =
, ρ2 =
4
4
V
3
πR
π (2R)3
3
3
m
4
3
π
R
ρ1
1
1
1
3
⇒
=
=
=
=
2m
2
1
1
ρ2
3
2
4
2
4
π (2R)3 2
3
Q14.2
A block of ice (density 920 kg/m3) and a block of iron
(density 7800 kg/m3) are both submerged in a fluid. Both
blocks have the same volume. Which block experiences the
greater buoyant force?
A. the block of ice
B. the block of iron
C. Both experience the same buoyant force.
D. The answer depends on the density of the fluid.
Monday, November 28, 11
A14.2
A block of ice (density 920 kg/m3) and a block of iron
(density 7800 kg/m3) are both submerged in a fluid. Both
blocks have the same volume. Which block experiences the
greater buoyant force?
A. the block of ice
B. the block of iron
C. Both experience the same buoyant force.
D. The answer depends on the density of the fluid.
Both displace same volume of water, hence same weight of water,
hence both have the same buoyant force.
Monday, November 28, 11
Q14.3
A cylinder is completely filled with water. The top of the
cylinder is sealed with a tight-fitting lid.
If you push down on the lid with a pressure of 1000 Pa, the
water pressure at the bottom of the cylinder
A. increases by more than 1000 Pa.
B. increases by 1000 Pa.
C. increases by less than 1000 Pa.
D. is unchanged.
E. The answer depends on the height of the cylinder.
Monday, November 28, 11
A14.3
A cylinder is completely filled with water. The top of the
cylinder is sealed with a tight-fitting lid.
If you push down on the lid with a pressure of 1000 Pa, the
water pressure at the bottom of the cylinder
A. increases by more than 1000 Pa.
B. increases by 1000 Pa.
C. increases by less than 1000 Pa.
D. is unchanged.
E. The answer depends on the height of the cylinder.
Monday, November 28, 11
Surface tension
• How is it that water striders can walk
on water (although they are more dense
than the water)?
The surface behaves like a membrane
Monday, November 28, 11
Surface tension
The cohesive forces among the liquid molecules are responsible for this
phenomenon of surface tension. In the bulk of the liquid, each molecule is
pulled equally in every direction by neighboring liquid molecules, resulting in a
net force of zero. The molecules at the surface do not have other molecules on
all sides of them and therefore are pulled inwards. This creates some internal
pressure and forces liquid surfaces to contract to the minimal area.
Monday, November 28, 11
Surface tension
• Surface tension also explains why falling
raindrops are spherical not tear drop shaped.
A falling drop of water, filmed at
3000 images per second.
Monday, November 28, 11
Fluid Flow
Fluid flow can be very complex, as can be seen in
the smoke from the match.
Monday, November 28, 11
Ideal Fluid
An Ideal Fluid is a fluid that is incompressible (its density does
not change) and has no internal friction (called viscosity).
Liquids are approximately incompressible in most situations.
We can treat gases as incompressible only if the pressure
change from one region to another is not too great.
Internal friction in a fluid causes shear stresses when two
adjacent layers of fluid move relative to each other, as when
fluid flows around an obstacle.
Monday, November 28, 11
An Ideal Fluid
Monday, November 28, 11
Fluid flow
Monday, November 28, 11
Fluid flow
Monday, November 28, 11
Fluid flow
Monday, November 28, 11
Monday, November 28, 11
Continuity Equation
The mass of a moving fluid does
not change as it flows, the mass
remains constant. This leads to an
important quantitative
relationship called the
continuity equation.
ρ A1v1dt = ρ A2 v2 dt or
A1v1 = A2 v2 is the continuity equation for incompressible flow
dV
Q=
= Av is the volume flow rate
dt
dV
ρ
is the mass flow rate
dt
ρ1 A1v1dt = ρ2 A2 v2 dt is the continuity equation for compressible flow
Monday, November 28, 11
Fluid flow
Flow speed changes through a tapered pipe
Monday, November 28, 11
Fluid flow
The speed of the water is
inversely proportional to the
diameter of the stream.
Monday, November 28, 11
Speed of water through a hose
Monday, November 28, 11
Deriving Bernoulli’s
The fluid is incompressible, so
dV = A1ds1 = A2 ds2
The work done on this fluid element during dt
F
dW = p1 A1ds1 − p2 A2 ds2 (as W=Fs, P= ⇒ F = PA)
A
dW = ( p1 − p2 )dV
As m = ρdV, the change in kinetic energy is
1
dK = ρdV (v22 − v12 )
2
The change in gravitational potential energy is
dU = ρdVg(y2 − y1 )
The energy equation tells us dW = dK + dU, so
1
2
2
( p1 − p2 )dV = ρdV (v2 − v1 ) + ρdVg(y2 − y1 )
2
1
2
2
p1 − p2 = ρ(v2 − v1 ) + ρ g(y2 − y1 ) this is Bernoulli's equation
2
Monday, November 28, 11
equation
Pressure difference associated
with change in speed of fluid
1
2
2
p1 − p2 = ρ(v2 − v1 ) + ρ g(y2 − y1 )
2
Pressure difference associated
caused by the fluid weight and the
difference in elevation at the two ends
Monday, November 28, 11
Pressure difference associated
with change in speed of fluid
1
2
2
p1 − p2 = ρ(v2 − v1 ) + ρ g(y2 − y1 )
2
Pressure difference associated
caused by the fluid weight and the
difference in elevation at the two ends
Or can write as:
Monday, November 28, 11
Pressure difference associated
with change in speed of fluid
1
2
2
p1 − p2 = ρ(v2 − v1 ) + ρ g(y2 − y1 )
2
Pressure difference associated
caused by the fluid weight and the
difference in elevation at the two ends
Or can write as:
Monday, November 28, 11
Pressure difference associated
with change in speed of fluid
1
2
2
p1 − p2 = ρ(v2 − v1 ) + ρ g(y2 − y1 )
2
Pressure difference associated
caused by the fluid weight and the
difference in elevation at the two ends
Or can write as:
Monday, November 28, 11
Water pressure in a home (Bernoulli’s Principle)
Find speed at bathroom, v2, using the continuity equation
Volume flow rate is
Find pressure at bathroom, p2, using the Bernoulli’s equation
Monday, November 28, 11
Speed of efflux (Bernoulli’s Equation)
Apply Bernoulli’s equation to points 1 & 2
y2=h
y1=0
Monday, November 28, 11
The Venturi meter (Bernoulli’s Equation)
Apply Bernoulli’s equation noting that y1=y2
We know that the pressure difference (p1-p2) is ρgh, so substituting this in the above expression and rearranging for v1
Monday, November 28, 11
Lift on an airplane wing
Monday, November 28, 11
Viscosity and turbulence
Viscosity is internal friction
in a fluid. Viscous forces
oppose the motion of one
portion of a fluid relative to
another.
Viscosity is the reason it
takes effort to paddle a canoe
through calm water, but it is
also the reason the paddle
works. Viscous effects are
important in the flow of fluids
in pipes, the flow of blood,
the lubrication of engine
parts, and many other
situations.
Monday, November 28, 11
Viscosity and turbulence
When we cease to treat fluids
as ideal, molecules can attract
or repel one another—they
can interact with container
walls and the result is
turbulence.
Monday, November 28, 11
A curve ball (Bernoulli’s equation applied to sports)
Bernoulli’s equation allows us to explain why a curve ball
would curve, and why a slider turns downward.
Slower higher pressure air
Slower higher pressure air
Faster lower pressure air
Monday, November 28, 11
Faster lower pressure air
A curve ball (Bernoulli’s equation applied to sports)
Bernoulli’s equation allows us to explain why a curve ball
would curve, and why a slider turns downward.
Slower higher pressure air
Slower higher pressure air
Faster lower pressure air
Monday, November 28, 11
Faster lower pressure air
Q14.4
An incompressible fluid
flows through a pipe of
varying radius (shown in
cross-section). Compared
to the fluid at point P, the
fluid at point Q has
P
radius R
radius 2R
A. greater pressure and greater volume flow rate.
B. greater pressure and the same volume flow rate.
C. the same pressure and greater volume flow rate.
D. lower pressure and the same volume flow rate.
E. none of the above
Monday, November 28, 11
Q
A14.4
An incompressible fluid
flows through a pipe of
varying radius (shown in
cross-section). Compared
to the fluid at point P, the
fluid at point Q has
P
radius R
radius 2R
A. greater pressure and greater volume flow rate.
B. greater pressure and the same volume flow rate.
C. the same pressure and greater volume flow rate.
D. lower pressure and the same volume flow rate.
E. none of the above
Monday, November 28, 11
Q
Q14.5
An incompressible fluid
flows through a pipe of
varying radius (shown in
cross-section). Compared
to the fluid at point P, the
fluid at point Q has
P
radius R
radius 2R
A. 4 times the fluid speed.
B. 2 times the fluid speed.
C. the same fluid speed.
D. 1/2 the fluid speed.
E. 1/4 the fluid speed.
Monday, November 28, 11
Q
A14.5
An incompressible fluid
flows through a pipe of
varying radius (shown in
cross-section). Compared
to the fluid at point P, the
fluid at point Q has
P
radius R
radius 2R
A. 4 times the fluid speed.
B. 2 times the fluid speed.
C. the same fluid speed.
D. 1/2 the fluid speed.
E. 1/4 the fluid speed.
A1 = π (2R)2 = 4π R 2 , A2 = π R 2 ⇒ A1 = 4A2
From continuity equation A1v1 = A2 v2 = 4A2 v1
v1 = 4v1
Monday, November 28, 11
Q
Blood flow in capillaries
Monday, November 28, 11
Blood flow in capillaries
Monday, November 28, 11
Blood flow in capillaries
Monday, November 28, 11
Simple Harmonic
Motion, Pendulums
Monday, November 28, 11
Introduction
If you look to the
right, you’ll see a
time-lapse photograph
of a simple pendulum.
It’s a great example of
the regular oscillatory
motion we’re about to
study.
Monday, November 28, 11
Oscillations about
Equilibrium
Monday, November 28, 11
Monday, November 28, 11
Periodic Motion
Period: time required for one cycle of periodic
motion
Frequency: number of oscillations per unit time
1
f =
T
This unit is called
the Hertz:
Monday, November 28, 11
Simple Harmonic Motion
A mass on a spring has a displacement as a
function of time that is a sine or cosine curve:
Here, A is called
the amplitude of
the motion.
Monday, November 28, 11
Simple Harmonic Motion
A spring exerts a restoring force
that is proportional to the
displacement from equilibrium:
Monday, November 28, 11
Simple Harmonic Motion
A spring exerts a restoring force
that is proportional to the
displacement from equilibrium:
Monday, November 28, 11
Simple harmonic motion
An ideal spring responds to
stretch and compression
linearly, obeying Hooke’s
Law.
For a real spring, Hookes’Law
is a good approximation.
Monday, November 28, 11
Linear Restoring Forces and Simple Harmonic
Motion
Monday, November 28, 11
Monday, November 28, 11
Equilibrium and Oscillation
Monday, November 28, 11
Monday, November 28, 11
Sinusoidal Relationships
Monday, November 28, 11
Mathematical Description of Simple Harmonic
Motion
Monday, November 28, 11
Simple Harmonic Motion
If we call the period of the motion T – this is the
time to complete one full cycle – we can write the
position as a function of time:
⎛ 2π ⎞
x = A cos ⎜
t⎟
⎝ T ⎠
It is then straightforward to show that the
position at time t + T is the same as the position
at time t, as we would expect.
Monday, November 28, 11
Connections between Uniform Circular Motion
and Simple Harmonic Motion
An object in simple
harmonic motion has the
same motion as one
component of an object in
uniform circular motion:
Monday, November 28, 11
Simple harmonic motion viewed as a projection
If you illuminate uniform circular motion (say by shining a
flashlight on a candle placed on a rotating lazy-Susan spice
rack), the shadow projection that will be cast will be
undergoing simple harmonic motion.
Monday, November 28, 11
Connections between Uniform Circular Motion
and Simple Harmonic Motion
Here, the object in circular motion has an angular
speed of
2π
ω=
T
where T is the period of motion of the object
in simple harmonic motion.
Monday, November 28, 11
Connections between Uniform Circular Motion
and Simple Harmonic Motion
The position as a function of time:
⎛ 2π ⎞
x = A cosθ = A cos(ω t) = A cos ⎜
t⎟
⎝ T ⎠
The angular frequency:
Monday, November 28, 11
SHM phase, position, velocity, and acceleration
SHM can occur with
various phase angles.
x = A cos(ω t + φ )
Monday, November 28, 11
For a given phase we can examine
position, velocity, and acceleration.
Connections between Uniform Circular Motion
and Simple Harmonic Motion
x = A cos(ω t)
The velocity as a function of time:
dx
v=
= −Aω sin(ω t)
dt
And the acceleration:
2
dv d x
2
a=
= 2 = −Aω cos(ω t)
dt dt
Both of these are found by taking components
of the circular motion quantities.
Monday, November 28, 11
The Period of a Mass on a Spring
The force on a mass on a spring is proportional to the displacement
(Hooke’s law), and to the acceleration (Newton’s second law)
F = ma = −kx
since x = A cos(ω t)
and a = −Aω cos(ω t)
2
m ⎡⎣ −Aω cos(ω t) ⎤⎦ = −k [ A cos(ω t)]
2
mω = k
k
2
ω =
m
2
ω=
Monday, November 28, 11
k
m
As
ω=
k
2π
= 2π f =
m
T
So
1
f =
2π
k
m
m
T = 2π
k
Monday, November 28, 11
Frequency and Period
The frequency of oscillation depends on physical properties of the
oscillator; it does not depend on the amplitude of the oscillation.
Monday, November 28, 11
The Period of a Mass on a Spring
Therefore, the period is
m
T = 2π
k
Monday, November 28, 11
Characteristics of SHM
Monday, November 28, 11
Characteristics of SHM
Monday, November 28, 11
X versus t for SHO then simple variations on a theme
Monday, November 28, 11
Energy in Simple Harmonic Motion
As a mass on a spring goes through its cycle of oscillation, energy is
transformed from potential to kinetic and back to potential.
Monday, November 28, 11
Energy Conservation in Oscillatory Motion
In an ideal system with no nonconservative forces,
the total mechanical energy is conserved. For a mass
on a spring:
1
1
E = K +U =
mv +
2
2
2
kx
2
Since we know the position and velocity as
functions of time, we can find the maximum kinetic
and potential energies:
U max
1 2
= kA , i.e. when x = A
2
As v = rω , and in this case rmax = A, and ω =
K max
k
m
1 2
1
1
1 2
2 2
2 k
= mv max = mA ω max = mA
= kA
2
2
2
m 2
Monday, November 28, 11
Energy Conservation in Oscillatory Motion
As a function of time,
1 2 1 2
E = K + U = mv + kx
2
2
2
2 2
2
v = −Aω sin(ω t) ⇒ v = A ω sin (ω t)
x = A cos(ω t) ⇒ x = A cos (ω t)
2
So the total energy
is constant; as the
kinetic energy
increases, the
potential energy
decreases, and
vice versa.
Monday, November 28, 11
ω=
2
2
k
2
⇒ mω = k
m
1
1 2
2 2
2
2
E = mA ω sin (ω t) + kA cos (ω t)
2
2
1 2 2
1 2
2
E = kA sin (ω t) + kA cos (ω t)
2
2
1 2
2
2
E = kA ⎡⎣sin (ω t) + cos (ω t) ⎤⎦
2
Energy Conservation in Oscillatory Motion
This diagram shows how the energy
transforms from potential to kinetic and back,
while the total energy remains the same.
Monday, November 28, 11
Energy in SHM
Monday, November 28, 11
Energy in SHM
• Energy is conserved during SHM and the forms (potential and
kinetic) interconvert as the position of the object in motion
changes.
Monday, November 28, 11
Monday, November 28, 11
The Pendulum
A physical pendulum is a
solid mass that oscillates
around its center of mass.
Examples:
Monday, November 28, 11
The Pendulum
A simple pendulum consists of a mass m (of
negligible size) suspended by a string or rod of
length L (and negligible mass).
The angle it makes with the vertical varies with
time as a sine or cosine.
Monday, November 28, 11
The Pendulum
Looking at the forces
on the pendulum bob,
we see that the
restoring force is
proportional to sin θ,
whereas the restoring
force for a spring is
proportional to the
displacement (which is
θ in this case).
Monday, November 28, 11
The Pendulum
However, for small angles, sin θ and θ in radians
are approximately equal.
Monday, November 28, 11
The Pendulum
However, for small angles, sin θ and θ in radians
are approximately equal.
Monday, November 28, 11
The Pendulum
For small angles θ in radians is approximately the sine of θ, θ ≈ sin θ
The net tangential force acting on m is the tangential component of its
weight, and is always direct towards the equilibrium point
F = mg sin θ ≈ mgθ
The arc length displacement of the mass from equilibrium is
s
⎛ mg ⎞
S = Lθ ⇒ θ = ⇒ F = mg sin θ ≈ mgθ = ⎜
s
⎟
⎝ L ⎠
L
For a spring F=kx
We can use this same form for a pendulum
if we let s=x, and k=mg/L
m
m
L
As T = 2π
= 2π
= 2π
k
mg L
g
So the period depends on the length of the pendulum
and the acceleration due to gravity
Monday, November 28, 11
Vibrations of molecules
Two atoms separated by their internuclear distance r can be
pondered as two balls on a spring.
Monday, November 28, 11
Vibrations of molecules
Two atoms separated by their internuclear distance r can be
pondered as two balls on a spring. The potential energy of such a
model is constructed many different ways. The Leonard–Jones
potential is sketched below. The atoms on a molecule vibrate as
shown below.
Monday, November 28, 11
Damped Oscillations
In most physical situations, there is a
nonconservative force of some sort, which will
tend to decrease the amplitude of the oscillation,
and which is typically proportional to the speed:
This causes the amplitude to decrease
exponentially with time:
Monday, November 28, 11
Damped Oscillations
This exponential decrease is shown in the figure:
Monday, November 28, 11
Damped Oscillations
The previous image shows a system that is
underdamped – it goes through multiple
oscillations before coming to rest. A critically
damped system is one that relaxes back to the
equilibrium position without oscillating and in
minimum time; an overdamped system will
also not oscillate but is damped so heavily
that it takes longer to reach equilibrium.
Monday, November 28, 11
Damping
Monday, November 28, 11
Damped oscillations
A person may not
wish for the object
they study to
remain in SHM.
Consider shock
absorbers and your
automobile.
Without damping
the oscillation,
hitting a pothole
would set your car
into SHM on the
springs that
support it.
Monday, November 28, 11
Driven Oscillations and Resonance
An oscillation can be driven by an oscillating driving
force; the frequency of the driving force may or may
not be the same as the natural frequency of the
system.
Monday, November 28, 11
A mass on a spring is driven by a large
geared motor apparatus, and exhibits
resonance at the appropriate frequency.
Below resonance, the driving motion is
in phase with the motion of the mass.
At resonance the mass is 90 degrees out
of phase with the driving motion, and
above resonance it is 180 degrees out of
phase.
Note the amplitude of motion difference
when the system is at resonance.
Monday, November 28, 11
Resonance
y = Ae
t
−
τ
The response curve becomes
taller and narrower as damping
is reduced
Monday, November 28, 11
Driven Oscillations and Resonance
If the driving frequency
is close to the natural
frequency, the amplitude
can become quite large,
especially if the damping
is small. This is called
resonance.
Monday, November 28, 11
Forced (driven) oscillations and resonance
• A force applied “in synch” with a motion already in progress will
resonate and add energy to the oscillation
• A singer can shatter a glass with a pure tone in tune with the
natural “ring” of a thin wine glass.
Monday, November 28, 11
Forced (driven) oscillations and resonance
The Tacoma Narrows Bridge suffered spectacular structural
failure after absorbing too much resonant energy
Monday, November 28, 11
Summary
• Period: time required for a motion to go through a
complete cycle
• Frequency: number of oscillations per unit time
• Angular frequency:
• Simple harmonic motion occurs when the
restoring force is proportional to the displacement
from equilibrium.
Monday, November 28, 11
Summary
• The amplitude is the maximum displacement from
equilibrium.
• Position as a function of time:
⎛ 2π ⎞
x = A cos ⎜
t ⎟ = A cos (ω t )
⎝ T ⎠
• Velocity as a function of time:
v = −Aω sin (ω t )
Monday, November 28, 11
Summary
© 2010 Pearson Education, Inc.
Monday, November 28, 11
Slide 14-32
Summary
• Acceleration as a function of time:
• Period of a mass on a spring:
• Total energy in simple harmonic motion:
Monday, November 28, 11
Summary
• Potential energy as a function of time:
• Kinetic energy as a function of time:
• A simple pendulum with small amplitude
exhibits simple harmonic motion
Monday, November 28, 11
Summary
• Period of a simple pendulum:
• Period of a physical pendulum:
Monday, November 28, 11
Summary
© 2010 Pearson Education, Inc.
Monday, November 28, 11
Slide 14-33
Summary
• Oscillations where there is a nonconservative
force are called damped.
• Underdamped: the amplitude decreases
exponentially with time:
• Critically damped: no oscillations; system
relaxes back to equilibrium in minimum time
• Overdamped: also no oscillations, but slower
than critical damping
Monday, November 28, 11
Summary
• An oscillating system may be driven by an
external force
• This force may replace energy lost to friction, or
may cause the amplitude to increase greatly at
resonance
• Resonance occurs when the driving frequency is
equal to the natural frequency of the system
Monday, November 28, 11
Quiz
1. The type of function that describes simple harmonic motion is
A.
B.
C.
D.
E.
linear
exponential
quadratic
sinusoidal
inverse
Monday, November 28, 11
Answer
1. The type of function that describes simple harmonic motion is
A.
B.
C.
D.
E.
linear
exponential
quadratic
sinusoidal
inverse
Monday, November 28, 11
Quiz
2. A mass is bobbing up and down on a spring. If you increase the
amplitude of the motion, how does this affect the time for one
oscillation?
A. The time increases.
B. The time decreases.
C. The time does not change.
Monday, November 28, 11
Answer
2. A mass is bobbing up and down on a spring. If you increase the
amplitude of the motion, how does this affect the time for one
oscillation?
A. The time increases.
B. The time decreases.
C. The time does not change.
Monday, November 28, 11
Reading Quiz
3. If you drive an oscillator, it will have the largest amplitude if you
drive it at its _______ frequency.
A. special
B. positive
C. resonant
D. damped
E. pendulum
Monday, November 28, 11
Answer
3. If you drive an oscillator, it will have the largest amplitude if you
drive it at its _______ frequency.
A. special
B. positive
C. resonant
D. damped
E. pendulum
Monday, November 28, 11
Additional Questions
Four 100 g masses are hung from four springs, each with unstretched
length 10 cm. The four springs stretch as noted in the following
diagram:
Now, each of the masses is lifted a small distance, released, and
allowed to oscillate. Rank the oscillation frequencies, from
highest to lowest.
A. a > b > c > d
B. d > c > b > a
C. a = b = c = d
Monday, November 28, 11
Answer
Four 100 g masses are hung from four springs, each with unstretched
length 10 cm. The four springs stretch as noted in the following
diagram:
Now, each of the masses is lifted a small distance, released, and
allowed to oscillate. Rank the oscillation frequencies, from
highest to lowest.
A. a > b > c > d
B. d > c > b > a
C. a = b = c = d
Monday, November 28, 11
Increasing k
increases f
Additional Questions
A typical earthquake produces vertical oscillations of the earth.
Suppose a particular quake oscillates the ground at a frequency 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
Monday, November 28, 11
Answer
A typical earthquake produces vertical oscillations of the earth.
Suppose a particular quake oscillates the ground at a frequency 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
1
1
1
T
f = ⇒T = =
= 6.67 ⇒ = 3.33
T
f 0.15
2
Time between highest and lowest point is T/2
Monday, November 28, 11
Answer
A typical earthquake produces vertical oscillations of the earth.
Suppose a particular quake oscillates the ground at a frequency 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
1
1
1
T
f = ⇒T = =
= 6.67 ⇒ = 3.33
T
f 0.15
2
Time between highest and lowest point is T/2
Monday, November 28, 11
Q13.1
An object on the end of a spring is oscillating in simple harmonic
motion. If the amplitude of oscillation is doubled, how does this affect
the oscillation period T and the object’s maximum speed vmax?
A. T and vmax both double.
B. T remains the same and vmax doubles.
C. T and vmax both remain the same.
D. T doubles and vmax remains the same.
E. T remains the same and vmax increases by a factor of
Monday, November 28, 11
.
A13.1
An object on the end of a spring is oscillating in simple harmonic
motion. If the amplitude of oscillation is doubled, how does this affect
the oscillation period T and the object’s maximum speed vmax?
A. T and vmax both double.
B. T remains the same and vmax doubles.
C. T and vmax both remain the same.
D. T doubles and vmax remains the same.
E. T remains the same and vmax increases by a factor of
Monday, November 28, 11
.
Q13.2
This is an x-t graph for
an object in simple
harmonic motion.
At which of the following times does the object have the most
negative velocity vx?
A. t = T/4
B. t = T/2
C. t = 3T/4
D. t = T
Monday, November 28, 11
A13.2
This is an x-t graph for
an object in simple
harmonic motion.
At which of the following times does the object have the most
negative velocity vx?
A. t = T/4
B. t = T/2
C. t = 3T/4
D. t = T
Monday, November 28, 11
Q13.3
This is an x-t graph for
an object in simple
harmonic motion.
At which of the following times does the object have the most
negative acceleration ax?
A. t = T/4
B. t = T/2
C. t = 3T/4
D. t = T
Monday, November 28, 11
A13.3
This is an x-t graph for
an object in simple
harmonic motion.
At which of the following times does the object have the most
negative acceleration ax?
A. t = T/4
B. t = T/2
C. t = 3T/4
D. t = T
Monday, November 28, 11
Q13.4
This is an ax-t graph for
an object in simple
harmonic motion.
At which of the following times does the object have the most
negative displacement x?
A. t = 0.10 s
B. t = 0.15 s
C. t = 0.20 s
D. t = 0.25 s
Monday, November 28, 11
A13.4
This is an ax-t graph for
an object in simple
harmonic motion.
At which of the following times does the object have the most
negative displacement x?
A. t = 0.10 s
B. t = 0.15 s
C. t = 0.20 s
D. t = 0.25 s
Monday, November 28, 11
Q13.5
This is an ax-t graph for
an object in simple
harmonic motion.
At which of the following times does the object have the most
negative velocity vx?
A. t = 0.10 s
B. t = 0.15 s
C. t = 0.20 s
D. t = 0.25 s
Monday, November 28, 11
A13.5
This is an ax-t graph for
an object in simple
harmonic motion.
At which of the following times does the object have the most
negative velocity vx?
A. t = 0.10 s
B. t = 0.15 s
C. t = 0.20 s
D. t = 0.25 s
Monday, November 28, 11
Q13.6
This is an x-t graph for
an object connected to
a spring and moving in
simple harmonic
motion.
At which of the following times is the potential energy
of the spring the greatest?
A. t = T/8
B. t = T/4
C. t = 3T/8
D. t = T/2
E. more than one of the above
Monday, November 28, 11
A13.6
This is an x-t graph for
an object connected to
a spring and moving in
simple harmonic
motion.
At which of the following times is the potential energy
of the spring the greatest?
A. t = T/8
B. t = T/4
C. t = 3T/8
D. t = T/2
E. more than one of the above
Monday, November 28, 11
Q13.7
This is an x-t graph for
an object connected to
a spring and moving in
simple harmonic
motion.
At which of the following times is the kinetic energy of
the object the greatest?
A. t = T/8
B. t = T/4
C. t = 3T/8
D. t = T/2
E. more than one of the above
Monday, November 28, 11
A13.7
This is an x-t graph for
an object connected to
a spring and moving in
simple harmonic
motion.
At which of the following times is the kinetic energy of
the object the greatest?
A. t = T/8
B. t = T/4
C. t = 3T/8
D. t = T/2
E. more than one of the above
Monday, November 28, 11
Q13.8
To double the total energy of a mass-spring system
oscillating in simple harmonic motion, the amplitude
must increase by a factor of
A. 4.
B.
C. 2.
D.
E.
Monday, November 28, 11
A13.8
To double the total energy of a mass-spring system
oscillating in simple harmonic motion, the amplitude
must increase by a factor of
A. 4.
B.
C. 2.
D.
E.
Monday, November 28, 11
Q13.9
A simple pendulum consists of a point mass
suspended by a massless, unstretchable string.
If the mass is doubled while the length of the string
remains the same, the period of the pendulum
A. becomes 4 times greater.
B. becomes twice as great.
C. becomes greater by a factor of
D. remains unchanged.
E. decreases.
Monday, November 28, 11
.
A13.9
A simple pendulum consists of a point mass
suspended by a massless, unstretchable string.
If the mass is doubled while the length of the string
remains the same, the period of the pendulum
A. becomes 4 times greater.
B. becomes twice as great.
C. becomes greater by a factor of
D. remains unchanged.
E. decreases.
Monday, November 28, 11
.