Physics 207 – Lecture 20
Lecture 20
Goals:
• Chapter 14
Understand and use energy conservation in oscillatory
systems.
Understand the basic ideas of damping and resonance.
• Chapter 15
Understand pressure in liquids and gases
Use Archimedes’ principle to understand buoyancy
Understand the equation of continuity
Use an ideal-fluid model to study fluid flow.
Investigate the elastic deformation of solids and liquids
• Assignment
HW8, Due Wednesday, Apr. 8th
Tuesday: Read all of Chapter 15
Physics 207: Lecture 20, Pg 1
SHM So Far
The most general solution is x(t) = A cos(ωt + φ)
Velocity: v(t) = -ωA sin(ωt + φ)
Acceleration: a(t) = -ω2A cos(ωt + φ)
where A = amplitude
ω = (angular) frequency = 2π f = 2π/T
φ = phase constant
Hooke’s Law Spring:
ω=
k
m
Simple Pendulum:
ω=
g
L
Spring constant
Inertia
Physics 207: Lecture 20, Pg 2
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Physics 207 – Lecture 20
SHM So Far
The most general solution is x(t) = A cos(ωt + φ)
Velocity: v(t) = -ωA sin(ωt + φ)
Acceleration: a(t) = -ω2A cos(ωt + φ)
where A = amplitude
ω = (angular) frequency = 2π f = 2π/T
φ = phase constant
x(t)
Here φ = 0
T = 2π/ω
A
−π/ω
−π/ω
A
π/ω
2π
π/ω
time
Physics 207: Lecture 20, Pg 3
SHM So Far
For SHM without friction
The frequency does not depend on the amplitude !
The oscillation occurs around the equilibrium point
where the force is zero!
Mechanical Energy is constant, it transfers between
potential and kinetic energies.
Physics 207: Lecture 20, Pg 4
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Physics 207 – Lecture 20
The shaker cart
You stand inside a small cart attached to a heavy-duty spring, the
spring is compressed and released, and you shake back and forth,
attempting to maintain your balance. Note that there is also a
sandbag in the cart with you.
At the instant you pass through the equilibrium position of the
spring, you drop the sandbag out of the cart onto the ground.
What effect does jettisoning the sandbag at the equilibrium position
have on the amplitude of your oscillation?
A. It increases the amplitude.
B. It decreases the amplitude.
C. It has no effect on the amplitude.
Hint: At equilibrium, both the cart and the bag
are moving at their maximum speed. By
dropping the bag at this point, energy
(specifically the kinetic energy of the bag) is
lost from the spring-cart system. Thus, both the
elastic potential energy at maximum displacement
and the kinetic energy at equilibrium must decrease
Physics 207: Lecture 20, Pg 5
The shaker cart
Instead of dropping the sandbag as you pass through equilibrium, you
decide to drop the sandbag when the cart is at its maximum distance
from equilibrium.
What effect does jettisoning the sandbag at the cart’s maximum
distance from equilibrium have on the amplitude of your oscillation?
A. It increases the amplitude.
B. It decreases the amplitude.
C. It has no effect on the amplitude.
Hint: Dropping the bag at maximum
distance from equilibrium, both the cart
and the bag are at rest. By dropping the
bag at this point, no energy is lost from
the spring-cart system. Therefore, both the
elastic potential energy at maximum displacement
and the kinetic energy at equilibrium must remain constant.
Physics 207: Lecture 20, Pg 6
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Physics 207 – Lecture 20
The shaker cart
What effect does jettisoning the sandbag at the cart’s maximum
distance from equilibrium have on the maximum speed of the cart?
A. It increases the maximum speed.
B. It decreases the maximum speed.
C. It has no effect on the maximum speed.
Hint: Dropping the bag at maximum distance
from equilibrium, both the cart and the bag
are at rest. By dropping the bag at this
point, no energy is lost from the spring-cart
system. Therefore, both the elastic
potential energy at maximum displacement
and the kinetic energy at equilibrium must
remain constant.
Physics 207: Lecture 20, Pg 7
What about Vertical Springs?
For a vertical spring, if y is measured from
the equilibrium position
1
U = ky 2
2
Recall: force of the spring is the negative
derivative of this function:
FNET = −
j
k
dU
= −ky
dy
This will be just like the horizontal case:
d2y
-ky = ma = m
dt 2
y=0
m
Which has solution y(t) = A cos( ωt + φ) where ω =
F= -ky
k
m
Physics 207: Lecture 20, Pg 8
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Physics 207 – Lecture 20
Exercise Simple Harmonic Motion
A mass oscillates up & down on a spring. It’s position as a
function of time is shown below. At which of the points
shown does the mass have positive velocity and negative
acceleration ?
Remember: velocity is slope and acceleration is the curvature
y(t) = A cos( ωt + φ)
v(t) = -A ω sin( ωt + φ)
y(t)
a(t) = -A ω2 cos( ωt + φ)
(a)
(c)
t
(b)
Physics 207: Lecture 20, Pg 9
Home Exercise
A mass m = 2 kg on a spring oscillates (no friction) with
amplitude A = 10 cm. At t = 0 its speed is at a maximum, and
is v=+2 m/s
What is the angular frequency of oscillation ω ?
What is the spring constant k ?
General relationships E = K + U = constant, ω = (k/m)½
So at maximum speed U=0 and ½ mv2 = E = ½ kA 2
thus k = mv2/A2 = 2 x (2) 2/(0.1)2 = 800 N/m, ω = 20 rad / sec
k
m
x
Physics 207: Lecture 20, Pg 10
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Physics 207 – Lecture 20
Home Exercise Initial Conditions
A mass hanging from a vertical spring is lifted a distance d
above equilibrium and released at t = 0.
Which of the following describe its velocity and acceleration
as a function of time (upwards is positive y direction)?
(A) v(t) = - vmax sin( ωt )
a(t) = -amax cos( ωt )
(B) v(t) = vmax sin( ωt )
a(t) = amax cos( ωt )
(C) v(t) = vmax cos( ωt )
a(t) = -amax cos(ωt )
k
y
m
t=0
d
0
(both vmax and amax are positive numbers)
Physics 207: Lecture 20, Pg 11
Home Exercise Initial Conditions
A mass hanging from a vertical spring is lifted a
distance d above equilibrium and released at t = 0.
Which of the following describe its velocity and
acceleration as a function of time (upwards is positive y
direction):
(A) v(t) = - vmax sin( ωt )
(B) v(t) = vmax sin( ωt )
(C) v(t) = vmax cos( ωt )
a(t) = -amax cos( ωt )
a(t) = amax cos( ωt )
t=0
a(t) = -amax cos(ωt )
k
y
m
d
0
(both vmax and amax are positive numbers)
Physics 207: Lecture 20, Pg 12
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Physics 207 – Lecture 20
Exercise
Simple Harmonic Motion
You are sitting on a swing. A friend gives you a small push
and you start swinging back & forth with period T1.
Suppose you were standing on the swing rather than sitting.
When given a small push you start swinging back & forth with
period T2.
Which of the following is true recalling that ω = (g / L)½
(A) T1 = T2
(B) T1 > T2
(C) T1 < T2
T1
T2
Physics 207: Lecture 20, Pg 13
Exercise
Simple Harmonic Motion
You are sitting on a swing. A friend gives you a small
push and you start swinging back & forth with period T1.
Suppose you were standing on the swing rather than
sitting. When given a small push you start swinging back
& forth with period T2.
If you are standing, the center of mass moves towards the
pivot point and so L is less, ω is bigger, T2 is smaller
(A) T1 = T2
(B) T1 > T2
(C) T1 < T2
Physics 207: Lecture 20, Pg 14
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Physics 207 – Lecture 20
Energy in SHM
For both the spring and the pendulum, we can derive
the SHM solution using energy conservation.
The total energy (K + U) of a
system undergoing SMH will
always be constant!
U
E
K
U
-A
This is not surprising since
there are only conservative
forces present, hence energy is conserved.
0
A
x
Physics 207: Lecture 20, Pg 15
SHM and quadratic potentials
SHM will occur whenever the potential is quadratic.
For small oscillations this will be true:
For example, the potential between
H atoms in an H2 molecule looks
something like this:
U
E
U
x
K
U
-A
0
A
x
Physics 207: Lecture 20, Pg 16
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Physics 207 – Lecture 20
SHM and quadratic potentials
Curvature reflects the spring constant
or modulus (i.e., stress vs. strain or
force vs. displacement)
U
x
Measuring modular proteins with an AFM
See: http://hansmalab.physics.ucsb.edu
Physics 207: Lecture 20, Pg 17
What about Friction?
A velocity dependent drag force (A model)
d 2 x b dx k
+
+ x=0
dt 2 m dt m
dx
d 2x
− kx − b = m 2
dt
dt
We can guess at a new solution.
x (t ) = A exp ( − 2btm ) cos (ω t + φ ) and now ω02 k / m
Note
2
With,
k  b 
 b 
2
ω=
−
 = ωo − 

m  2m 
 2m 
2
Physics 207: Lecture 20, Pg 18
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Physics 207 – Lecture 20
What about Friction?
A damped exponential
x (t ) = A exp ( − b t ) cos (ω t + φ )
2m
if
ωo > b / 2 m
1.2
1
0.8
0.6
0.4
A
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
ωt
Physics 207: Lecture 20, Pg 19
Variations in the damping
Small damping time constant (m/b)
Low friction coefficient, b << 2m
Moderate damping time constant
(m/b)
Moderate friction coefficient (b < 2m)
Physics 207: Lecture 20, Pg 20
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Physics 207 – Lecture 20
Damped Simple Harmonic Motion
ω = ωo2 − (b / 2m) 2
A downward shift in the angular frequency
There are three mathematically distinct regimes
ωo > b / 2 m
underdamped
ωo = b / 2m
critically damped
ωo < b / 2 m
overdamped
Physics 207: Lecture 20, Pg 21
Exercise
Damped oscillations:
A can of coke is attached to a spring and is
displaced by hand (m = 0.25 kg & k = 25.0 N/m)
The coke can is released, and it starts oscillating
with an amplitude of A = 0.3 m.
How damped is the system?
A. Underdamped (multiple oscillations with an
exponential decay in amplitude)
B. Critically damped (simple decaying motion with
at most one overshoot of the system's resting
position)
C. Overdamped (simple exponentially decaying
motion, without any oscillations)
Physics 207: Lecture 20, Pg 22
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Physics 207 – Lecture 20
Driven SHM with Resistance
Apply a sinusoidal force, F0 cos (ωt), and now
consider what A and b do,
steady state amplitude
d 2 x b dx k
F
+
+
x
=
cos ω t
dt 2 m dt m
m
A=
b/m small
Not Zero!!!
F0 / m
(ω 2 − ω 02 ) 2 + (
bω 2
)
m
b/m middling
b large
ω ≅ ω0
ω
Physics 207: Lecture 20, Pg 23
Resonance-based DNA detection with nanoparticle probes
Change the mass of the
cantilever & change the
resonant frequency
Su et al., APL 82: 3562 (2003)
Physics 207: Lecture 20, Pg 24
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Physics 207 – Lecture 20
Stick - Slip Friction
How can a constant motion
produce resonant vibrations?
Examples:
Strings, e.g. violin
Singing / Whistling
Tacoma Narrows Bridge
…
Physics 207: Lecture 20, Pg 25
Dramatic example of resonance
In 1940, a steady wind set up a torsional vibration in the
Tacoma Narrows Bridge
Physics 207: Lecture 20, Pg 26
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Physics 207 – Lecture 20
A short clip
In 1940, a steady wind sets up a torsional vibration in the
Tacoma Narrows Bridge
Physics 207: Lecture 20, Pg 27
Dramatic example of resonance
Eventually it collapsed
Physics 207: Lecture 20, Pg 28
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Physics 207 – Lecture 20
Exercise
Resonant Motion
Consider the following set of pendulums all attached to the
same string
B
C
D
A
If I start bob D swinging which of the others will gain the most
mechanical energy (assuming virtually no friction) ?
(A)
(B)
(C)
Physics 207: Lecture 20, Pg 29
Chapter 15, Fluids
This is an actual photo of an iceberg, taken by a rig manager for
Global Marine Drilling in St. Johns, Newfoundland. The water was
calm and the sun was almost directly overhead so that the diver
Physics 207: Lecture 20, Pg 30
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Physics 207 – Lecture 20
Fluids (Ch. 15)
At ordinary temperature, matter exists
in one of three states
Solid - has a shape and forms a
surface
Liquid - has no shape but forms a
surface
Gas - has no shape and forms no
surface
What do we mean by “fluids”?
Fluids are “substances that
flow”…. “substances that take the
shape of the container”
Atoms and molecules are free to
move.
No long range correlation between
positions.
Physics 207: Lecture 20, Pg 31
Fluids
An intrinsic parameter of a fluid
Density
m
ρ =
V
kg/m3
units :
= 10-3 g/cm3
ρ(water) = 1.000 x 103 kg/m3
= 1.000 g/cm3
ρ(ice)
= 0.917 x 103 kg/m3
= 0.917 g/cm3
ρ(air)
= 1.29 kg/m3
= 1.29 x 10-3 g/cm3
ρ(Hg)
= 13.6 x103 kg/m3
= 13.6 g/cm3
Physics 207: Lecture 20, Pg 32
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Physics 207 – Lecture 20
Fluids
p=
Another parameter: Pressure
F
A
Any force exerted by a fluid is perpendicular to a surface
of contact, and is proportional to the area of that surface.
Force (a vector) in a fluid can be expressed in terms
of pressure (a scalar) as:
r
F = p A nˆ
n
A
Physics 207: Lecture 20, Pg 33
What is the SI unit of pressure?
A. Pascal
B. Atmosphere
C. Bernoulli
D. Young
E. p.s.i.
Units :
1 N/m2
1 bar
1 mbar
1 torr
= 1 Pa (Pascal)
= 105 Pa
= 102 Pa
= 133.3 Pa
1 atm = 1.013 x105 Pa
= 1013 mbar
= 760 Torr
= 14.7 lb/ in2 (=PSI)
Physics 207: Lecture 20, Pg 34
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Physics 207 – Lecture 20
Pressure vs. Depth
Incompressible Fluids (liquids)
When the pressure is much less
than the bulk modulus of the
fluid, we treat the density as
constant independent of
pressure:
incompressible fluid
p
0
y1
F1
y2
p
1
A
p
2
mg F2
For an incompressible fluid, the
density is the same everywhere,
but the pressure is NOT!
p(y) = p0 - y g ρ
Gauge pressure (subtract p0)
F2 = F1+ m g
= F1+ ρVg
F2 /A = F1/A + ρVg/A
p2 = p1 - ρg y
Physics 207: Lecture 20, Pg 35
Lecture 20
• Assignment
HW8, Due Wednesday, Apr. 8th
Tuesday: Read all of Chapter 15
Physics 207: Lecture 20, Pg 36
Page 18