Transcript PHYS 1443 – Section 501 Lecture #1

PHYS 1443 – Section 003
Lecture #21
Wednesday, Nov. 19, 2003
Dr. Mystery Lecturer
1.
2.
3.
4.
5.
Fluid Dymanics : Flow rate and Continuity Equation
Bernoulli’s Equation
Simple Harmonic Motion
Simple Block-Spring System
Energy of the Simple Harmonic Oscillator
Today’s Homework is #11 due on Wednesday, Nov. 26, 2003!!
Next Wednesday’s class is cancelled but there will be homework!!
Wednesday, Nov. 19, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
1
Flow Rate and the Equation of Continuity
Study of fluid in motion: Fluid Dynamics
If the fluid is water: Water
Hydro-dynamics
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 A2v2

t
t
Equation of Continuity
Wednesday, Nov. 19, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
2
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 A2v2
Since the air density is constant
A1v1  A2v2
Now let’s call the room as the
large section of the duct
A2 v2
A2l2 / t
V2
300
A1 



 0.11m 2
v1
v1
v1  t
3.0  900
Wednesday, Nov. 19, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
3
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 
Wednesday, Nov. 19, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
4
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
 A2 l2 v2   A1l1v12
2
2
 P1 A1l1  P2 A2 l2   A2 l2 gy2   A1l1 gy1
Wednesday, Nov. 19, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
5
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
1 2 1 2
 v2   v1  P1  P2   gy2   gy1
2
2
Reorganize P1 
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.
Wednesday, Nov. 19, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
6
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

Wednesday, Nov. 19, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu

7
Simple Harmonic Motion
What do you think a harmonic motion is?
Motion that occurs by the force that depends on displacement, and the
force is always directed toward the system’s equilibrium position.
What is a system that has such characteristics? A system consists of a mass and a spring
When a spring is stretched from its equilibrium position
by a length x, the force acting on the mass is
F  kx
It’s negative, because the force resists against the change of
length, directed toward the equilibrium position.
From Newton’s second law
F  ma  kx
we obtain
a

k
x
m
Condition for simple
k
This is a second order differential equation that can d 2 x


x
harmonic motion
be solved but it is beyond the scope of this class.
m
dt 2
Acceleration is proportional to displacement from the equilibrium
What do you observe
Acceleration is opposite direction to displacement
from this equation?
Wednesday, Nov. 19, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
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Equation of Simple Harmonic Motion
d 2x
k


x
2
dt
m
The solution for the 2nd order differential equation
x  A cos t   
Amplitude
Phase
Angular
Frequency
Phase
constant
Generalized
expression of a simple
harmonic motion
Let’s think about the meaning of this equation of motion
What happens when t=0 and =0?
What is  if x is not A at t=0?
x  A cos0  0  A
x  A cos   x'
  cos 1 x'
What are the maximum/minimum possible values of x?
Wednesday, Nov. 19, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
A/-A
An oscillation is fully
characterized by its:
•Amplitude
•Period or frequency
•Phase constant
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More on Equation of Simple Harmonic Motion
What is the time for full
cycle of oscillation?
The period
Since after a full cycle the position must be the same
x  Acos t  T      A cost  2   
T

How many full cycles of oscillation
does this undergo per unit time?
2

One of the properties of an oscillatory motion
f
1 
 
T 2
Frequency
Let’s now think about the object’s speed and acceleration.
What is the unit?
1/s=Hz
x  Acost   
dx
 Asin t    Max speed v
max  A
dt
Max acceleration
dv
2
2
Acceleration at any given time a    A cos  t      x a   2 A
dt
max
What do we learn
Acceleration is reverse direction to displacement
about acceleration?
Acceleration and speed are /2 off phase:
When v is maximum, a is at its minimum
Speed at any given time
Wednesday, Nov. 19, 2003
v

PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
10
Simple Harmonic Motion continued
Phase constant determines the starting position of a simple harmonic motion.
x  Acost   
x t 0  A cos 
At t=0
This constant is important when there are more than one harmonic oscillation
involved in the motion and to determine the overall effect of the composite motion
Let’s determine phase constant and amplitude
xi  A cos 
At t=0
vi  A sin 
 vi 

By taking the ratio, one can obtain the phase constant   tan  
 xi 
1
By squaring the two equation and adding them
together, one can obtain the amplitude

A cos   sin 
2
2
Wednesday, Nov. 19, 2003
2

2  x 2   vi 
A i   
xi2  A2 cos 2 
vi2   2 A2 sin 2 
2
A
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
 vi 
2
xi   
 
2
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Example for Simple Harmonic Motion
An object oscillates with simple harmonic motion along the x-axis. Its displacement from
the origin varies with time according to the equation; x  4.00mcost    where t is in seconds
and the angles is in the parentheses are in radians. a) Determine the amplitude,
frequency, and period of the motion.





4
.
00
m
cos

t




Acos

t




x


A  4.00m The angular frequency, , is   
From the equation of motion:
The amplitude, A, is
Therefore, frequency
and period are
T

2


2

 2s
f

1


1


 s 1
T

 2
b)Calculate the velocity and acceleration of the object at any time t.
Taking the first derivative on the
equation of motion, the velocity is
By the same token, taking the
second derivative of equation of
motion, the acceleration, a, is
Wednesday, Nov. 19, 2003

dx





4
.
00


sin

t

v

m / s

dt

d 2 x  4.00   2 cos t   m / s 2


a 2


dt
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
12
Simple Block-Spring System
A block attached at the end of a spring on a frictionless surface experiences
acceleration when the spring is displaced from an equilibrium position.
a
k
x
m
This becomes a second
order differential equation
If we    k
d 2x
k
  x denote
2
m
dt
m
2
d
x

The resulting differential equation becomes



x
2
dt
Since this satisfies condition for simple
x  A cost   
harmonic motion, we can take the solution
Does this solution satisfy the differential equation?
Let’s take derivatives with respect to time dx  A d cost      sin t   
Now the second order derivative becomes
dt
dt
d
d 2x
sin t      2  cost      2 x




2
dt
dt
Whenever the force acting on a particle is linearly proportional to the displacement from some
equilibrium position and is in the opposite direction, the particle moves in simple harmonic motion.
Wednesday, Nov. 19, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
13
More Simple Block-Spring System
How do the period and frequency of this harmonic motion look?
Since the angular frequency  is
The period, T, becomes
So the frequency is
Special case #1
x   cost
f
T



2


 2
1

1


T
2
2
k
m
m
k
k
m
What can we learn from these?
•Frequency and period do not
depend on amplitude
•Period is inversely proportional
to spring constant and
proportional to mass
Let’s consider that the spring is stretched to distance A and the block is let
go from rest, giving 0 initial speed; xi=A, vi=0,
2
dx
d
x
v
  sin t a  2   2  cos t ai   2   kA/ m
dt
dt
This equation of motion satisfies all the conditions. So it is the solution for this motion.
Suppose block is given non-zero initial velocity vi to positive x at the
instant it is at the equilibrium, xi=0
Is this a good

vi 


1 
1
  tan     
 tan  
x   cos t    A sin t  solution?
x 


Special case #2


i

Wednesday, Nov. 19, 2003


PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
14