PHYS 1441 – Section 501
Lecture #17
Monday, Aug. 2, 2004
Dr. Jaehoon Yu
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•
•
•
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Flow Rate and Equation of Continuity
Bernoulli’s Equation
Simple Harmonic Motion
Simple Block Spring System
Energy of a Simple Harmonic Oscillator
Final term exam next Wednesday, Aug. 11!
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
1
Announcements
• Final term exam Wednesday, Aug. 11
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Time: 6:00 – 8:00 pm
Location: SH125
Covers: Section 8.6 – Ch. 11
Review on Monday, Aug. 9
Mixture of multiple choice and free style
• Please do not miss the exam
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
2
Flow Rate and the Equation of Continuity
Study of fluid in motion: Fluid Dynamics
If the fluid is water: Water dynamics?? Hydro-dynamics
•Streamline or Laminar flow: Each particle of the fluid
Two main
types of flow follows a smooth path, a streamline
•Turbulent flow: Erratic, small, whirlpool-like circles called
eddy current or eddies which absorbs a lot of energy
Flow rate: the mass of fluid that passes a given point per unit time m / t
m1
1V1 1 A1l1


 1 A1v1
t
t
t
since the total flow must be conserved
m1 m2
1 A1v1   2 A2 v2

t
t
Equation of Continuity
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
3
Example for Equation of Continuity
How large must a heating duct be if air moving at 3.0m/s along it can
replenish the air every 15 minutes, in a room of 300m3 volume?
Assume the air’s density remains constant.
Using equation of continuity
1 A1v1  2 A2 v2
Since the air density is constant
A1v1  A2 v2
Now let’s imagine the room as
the large section of the duct
A2l2 / t
V2
A2 v2
300



A1 
 0.11m 2
v1
v1  t
v1
3.0  900
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
4
Bernoulli’s Equation
Bernoulli’s Principle: Where the velocity of fluid is high, the
pressure is low, and where the velocity is low, the pressure is high.
Amount of work done by the force, F1,
that exerts pressure, P1, at point 1
W1  F1l1  P1 A1l1
Amount of work done on the other
section of the fluid is
W2   P2 A2 l2
Work done by the gravitational force to
move the fluid mass, m, from y1 to y2 is
W3  mg  y2  y1 
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
5
Bernoulli’s Equation cont’d
The net work done on the fluid is
W  W1 W2 W3  P1 A1l1  P2 A2 l2 mgy2  mgy1
From the work-energy principle
1
1
2
mv2  mv12  P1 A1l1  P2 A2 l2 mgy2  mgy1
2
2
Since mass, m, is contained in the volume that flowed in the motion
A1l1  A2 l2
Thus,
and
m   A1l1   A2 l2
1
1
2
2
 A2 l2 v2   A1l1v1
2
2
 P1 A1l1  P2 A2 l2   A2 l2 gy2   A1l1 gy1
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
6
Bernoulli’s Equation cont’d
Since
1
1
 A2 l2 v22   A1l1v12  P1 A1l1  P2 A2 l2   A2 l2 gy2   A1l1 gy1
2
2
We
obtain
Reorganize P1 
1 2 1 2
 v2   v1  P1  P2   gy2   gy1
2
2
1 2
1 2
Bernoulli’s
 v1   gy1  P2   v2   gy2 Equation
2
2
Thus, for any two
points in the flow
1 2
P1   v1   gy1  const.
2
For static fluid P2  P1   g  y1  y2   P1   gh
1
2
2
P

P


v

v

2
1
1
2
For the same heights
2
Pascal’s
Law
The pressure at the faster section of the fluid is smaller than slower section.
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
7
Example for Bernoulli’s Equation
Water circulates throughout a house in a hot-water heating system. If the water is
pumped at a speed of 0.5m/s through a 4.0cm diameter pipe in the basement under
a pressure of 3.0atm, what will be the flow speed and pressure in a 2.6cm diameter
pipe on the second 5.0m above? Assume the pipes do not divide into branches.
Using the equation of continuity, flow speed on the second floor is
2
A1v1  r12 v1
 0.020 

v2 

0.5

 1.2m / s
2


A2
 r2
 0.013 
Using Bernoulli’s equation, the pressure in the pipe on the second floor is


1
P2  P1   v12  v22   g  y1  y2 
2
1
5
 3.0 10  1103 0.52  1.22  1103  9.8   5 
2
 2.5 105 N / m 2

Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu

8
Vibration or Oscillation
What are the things
that vibrate/oscillate?
•
•
•
•
•
So what is a vibration or oscillation?
Tuning fork
A pendulum
A car going over a bump
Building and bridges
The spider web with a prey
A periodic motion that repeats over the same path.
A simplest case is a block attached at the end of a coil spring.
When a spring is stretched from its equilibrium
position by a length x, the force acting on the mass is
F  kx
The sign is negative, because the force resists against the
change of length, directed toward the equilibrium position.
Acceleration is proportional to displacement from the equilibrium
Acceleration is opposite direction to displacement
Monday, Aug. 2, 2004
This system is doing a simple harmonic motion (SHM).
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
9
Vibration or Oscillation Properties
The maximum displacement from
the equilibrium is
Amplitude
One cycle of the oscillation
The complete to-and-fro motion from an initial point
Period of the motion, T
The time it takes to complete one full cycle
Unit?
s
Frequency of the motion, f
The number of complete cycles per second
Unit?
s-1
Relationship between
period and frequency?
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
f

1
T
or
T

1
f
10
Vibration or Oscillation Properties
Amplitude?
•
•
•
•
•
Monday, Aug. 2, 2004
A
When is the force greatest?
When is the velocity greatest?
When is the acceleration greatest?
When is the potential energy greatest?
When is the kinetic energy greatest?
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
11
Example 11-1
Car springs. When a family of four people with a total mass of 200kg step into their
1200kg car, the car’s springs compress 3.0cm. (a) What is the spring constant of the car’s
spring, assuming they act as a single spring? (b) How far will the car lower if loaded with
300kg?
(a) What is the force on the spring?
From Hooke’s law
F  mg  200  9.8  1960 N
F  kx  k  0.03  mg  1960 N
Fg mg 1960
k 


 6.5  104 N / m
x
x
0.03
(b) Now that we know the spring constant, we can solve for x in the force equation
F  kx  mg  300  9.8
mg 300  9.8
2
x


4.5

10
m
4
k
6.5 10
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
12
Energy of the Simple Harmonic Oscillator
How do you think the mechanical energy of the harmonic oscillator look without friction?
Kinetic energy of a
1

mv 2
KE
harmonic oscillator is
2
1 2

kx
The elastic potential energy stored in the spring
2
1 2 1 2
Therefore the total mechanical energy

mv  kx
E  KE  PE
2
2
of the harmonic oscillator is
PE
Total mechanical energy of a simple harmonic oscillator is
proportional to the square of the amplitude.
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
13
Energy of the Simple Harmonic Oscillator cont’d
1
1
2
KEmax  mvmax  k 2
2
2
k
vmax 
A
m
Maximum KE is
when PE=0
Maximum speed
The speed at any given
point of the oscillation
E
v
1
1 2 1
2
2

mv

kx

k

 KE  PE
2
2
2

 k m A x
2
2

  vmax
KE/PE
-A
Monday, Aug. 2, 2004
x
1  
 A
2
E=KE+PE=kA2/2
A
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
x
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Example 11-3
Spring calculations. A spring stretches 0.150m when a 0.300-kg mass is hung from it.
The spring is then stretched an additional 0.100m from this equilibrium position and
released.
(a) Determine the spring constant.
From Hooke’s law
F  kx  mg  0.300  9.8 N
mg 0.300  9.8
k

 19.6 N / m
x
0.150
(b) Determine the amplitude of the oscillation.
Since the spring was stretched 0.100m from
equilibrium, and is given no initial speed, the
amplitude is the same as the additional stretch.
A  0.100m
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
15
Example cont’d
(c) Determine the maximum velocity vmax.
k
A
m
vmax 
19.6
0.100  0.808m / s
0.300
(d) Determine the magnitude of velocity, v, when the mass is 0.050m from equilibrium.
vv
max
 x
1  
 A
2
 0.050 
 0.808 1  

 0.100 
2
 0.700m / s
(d) Determine the magnitude of the maximum acceleration of the mass.
Maximum acceleration is at the point where the mass is stopped to return.
F  ma  kA
Solve for a
kA 19.6  0.100

 6.53m / s 2
a 
m
0.300
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
16
Example for Energy of Simple Harmonic Oscillator
A 0.500kg cube connected to a light spring for which the force constant is 20.0 N/m oscillates on a
horizontal, frictionless track. a) Calculate the total energy of the system and the maximum speed of
the cube if the amplitude of the motion is 3.00 cm.
k  20.0N / m
A  3.00cm  0.03m
From the problem statement, A and k are
The total energy of
the cube is
E
 KE  PE 
1 2 1
2
kA  20.0 0.03  9.00 10 3 J
2
2
Maximum speed occurs when kinetic energy is the same as the total energy
KEmax
vmax
Monday, Aug. 2, 2004
1 2
 mvmax
2
E
1 2
 kA
2
k
20.0
A
 0.03
 0.190m / s
m
0.500
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
17
Example for Energy of Simple Harmonic Oscillator
b) What is the velocity of the cube when the displacement is 2.00 cm.
velocity at any given
displacement is
v

 k m A2  x 2



 20.0  0.032  0.02 2 / 0.500  0.141m / s
c) Compute the kinetic and potential energies of the system when the displacement is
2.00 cm.
Kinetic
energy, KE
Potential
energy, PE
Monday, Aug. 2, 2004
KE 
1 2
mv
2
PE 
1
2
 0.500  0.141  4.97 10 3 J
2
1 2 1
kx  20.0  0.022  4.00 10 3 J
2
2
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
18
Sinusoidal Behavior of SHM
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
19
The Period and Sinusoidal Nature of SHM
Consider an object moving on a circle with a constant angular speed w
v0
A2  x 2
x
 1  
A
 A
v
sin   
v0
Since it takes T to
complete one full
circular motion
From an energy
relationship in a
spring SHM
Thus, T is
v0 
2
x
v  v0 1   
 A
2 A
 2 Af
T
1
1
mv02  kA2
2
2
T  2
m
k
f 
T 
2 A
v0
v0 
k
A
m
1
1

T 2
k
m
Natural Frequency
If you look at it from the side, it looks as though it is doing a SHM
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
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20
Example 11-5
Car springs. When a family of four people with a total mass of 200kg step into their
1200kg car, the car’s springs compress 3.0cm. The spring constant of the spring is
6.5x104N/m. What is the frequency of the car after hitting the bump? Assume that the
shock absorber is poor, so the car really oscillates up and down.
T  2
1400
m
 2
 0.92s
4
6.5 10
k
1
1
f  
2
T
Monday, Aug. 2, 2004
k
1

m 2
6.5 10
 1.09 Hz
1400
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
4
21
Example 11-6
Spider Web. A small insect of mass 0.30 g is caught in a spider web of negligible mass.
The web vibrates predominantly with a frequency of 15Hz. (a) Estimate the value of the
spring constant k for the web.
f  1
2
k
 15Hz
m
Solve for k
k  4 mf  4  3  10  15   2.7 N / m
2
2
2
4
2
(b) At what frequency would you expect the web to vibrate if an insect of mass 0.10g were
trapped?
1
f 
2
Monday, Aug. 2, 2004
k
1

m 2
2.7
 26 Hz
4
110
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
22
The SHM Equation of Motion
The object is moving on a circle with a constant angular speed w
How is x, its position at any given time expressed with the known quantities?
v0
x  A cos since    t and   2 f
x  A cos t  A cos 2 ft
k
How about its velocity v at any given time?
v0 
m
A
v  v0 sin   v0 sin  t   v0 sin  2 ft 
How about its acceleration a at any given time?
From Newton’s
Monday, Aug. 2, 2004
2nd
law a 
F  kx
 kA 




 cos  2 ft   a0 cos  2 ft 
m
m
 m
kA
a0 
m
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
23
Sinusoidal Behavior of SHM
x  A cos  2 ft 
v  v0 sin  2 ft 
a  a0 cos  2 ft 
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
24
The Simple Pendulum
A simple pendulum also performs periodic motion.
The net force exerted on the bob is
 Fr  T  mg cos A  0
F
t
 mg sin  A  mg
Since the arc length, x, is
x  L
mg
F   Ft  
x
L
Satisfies conditions for simple harmonic motion!
It’s almost like Hooke’s law with. k  mg
L
The period for this motion is
T  2
m
 2
k
mL
 2
mg
L
g
The period only depends on the length of the string and the gravitational acceleration
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
25
Example 11-8
Grandfather clock. (a) Estimate the length of the pendulum in a grandfather clock that
ticks once per second.
Since the period of a simple
pendulum motion is
T  2
The length of the pendulum
in terms of T is
T 2g
L
4 2
Thus the length of the
pendulum when T=1s is
L
g
T 2 g 1 9.8
L

 0.25m
2
2
4
4
(b) What would be the period of the clock with a 1m long pendulum?
T  2
Monday, Aug. 2, 2004
L
1.0
 2
 2.0s
g
9.8
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
26
Damped Oscillation
More realistic oscillation where an oscillating object loses its mechanical
energy in time by a retarding force such as friction or air resistance.
How do you think the
motion would look?
Amplitude gets smaller as time goes on since its energy is spent.
Types of damping
A: Overdamped
B: Critically damped
C: Underdamped
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
27
Forced Oscillation; Resonance
When a vibrating system is set into motion, it oscillates with its natural
frequency f0.
1
k
f0 
2
m
However a system may have an external force applied to it that has
its own particular frequency (f), causing forced vibration.
For a forced vibration, the amplitude of vibration is found to be dependent
on the different between f and f0. and is maximum when f=f0.
A: light damping
B: Heavy damping
The amplitude can be large when
f=f0, as long as damping is small.
This is called resonance. The natural
frequency f0 is also called resonant frequency.
Monday, Aug. 2, 2004
PHYS 1441-501, Summer 2004
Dr. Jaehoon Yu
28