PHYS 1443 – Section 004
Lecture #23
Thursday, Nov. 20, 2014
Dr. Jaehoon Yu
•Bernoulli’s Principle
•Simple Block-Spring System
•Solution for SHM
•Energy of the Simple Harmonic
Oscillator
Thursday, Nov. 20, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
1
Announcements
• Last Quiz
– Beginning of the class coming Tuesday, Nov. 25
– Covers CH14.1 through what we finish today (CH15.1?)
– BYOF
• Final comprehensive exam
–
–
–
–
In this room, 9:00 – 10:30am, Thursday, Dec. 11
Covers CH 1.1 through what we finish Tuesday, Dec. 2 plus the math refresher
Mixture of multiple choice and free response problems
Bring your calculator but DO NOT input formula into it!
• Your phones or portable computers are NOT allowed as a replacement!
– You can prepare a one 8.5x11.5 sheet (front and back) of handwritten
formulae and values of constants for the exam
• None of the parts of the solutions of any problems
• No derived formulae, derivations of equations or word definitions!
– Do NOT Miss the exam!
Tuesday, Nov. 4, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
2
Bernoulli’s Principle
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 the work done by the force,
F1, that exerts pressure, P1, at point 1
W1  F × Dl = P1 A1l1
1
1
Amount of the work done by the force
on the other section of the fluid is
W2  F2 × Dl2 =  P2 A2 l2
Ü F2
F1 Þ
Thursday, Nov. 20, 2014
Work done on the gravitational force to
move the fluid mass, m, from y1 to y2 is
W3  mg  y2  y1 
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
3
Bernoulli’s Equation cont’d
The total amount of the work done on the fluid is
W  W1 W2 W3  P1 A1l1  P2 A2 l2 mgy2  mgy1
From the work- kinetic energy principle
1
1
2
mv2  mv12  P1 A1l1  P2 A2 l2 mgy2  mgy1
2
2
Since the mass m is contained in the volume that flowed in the motion
and
m   A1l1   A2 l2
A1l1  A2 l2
Thus,
1
1
2
2
 A2 l2 v2   A1l1v1
2
2
 P1 A1l1  P2 A2 l2   A2 l2 gy2   A1l1 gy1
Thursday, Nov. 20, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
4
Since A1l1
Bernoulli’s
Equation
cont’d
A

l

2
2
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 
Bernoulli’s
1 2
1 2
 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
For the same heights P2  P1    v12  v22 
2
Result of Energy
conservation!
Pascal’s
Law
The pressure at the faster section of the fluid is smaller than slower section.
Thursday, Nov. 20, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
5
Ex. for Bernoulli’s Equation
Water circulates throughout a house in a hot-water heating system. If the water is
pumped at the 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

Thursday, Nov. 20, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu

6
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
Buildings 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.
Thursday, Nov. 20, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
7
Simple Harmonic Motion
A motion that is caused by a force that depends on
displacement, and the force is always directed toward
the system’s equilibrium position.
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
k
This is a second order differential equation that can d 2 x
Condition for simple


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 the displacement.
from this equation?
This system is doing a simple harmonic motion (SHM).
Thursday, Nov. 20, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
8
Equation of Simple Harmonic Motion
d 2x
k
=- x
2
m
dt
The solution for the 2nd order differential equation
x = Acos (w t + j )
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 φ if x is not A at t=0?
x  A cos0  0  A
x = Acos (j ) x' An oscillation is fully
j = cos
-1
What are the maximum/minimum possible values of x?
Thursday, Nov. 20, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
( x' A)
A/-A
characterized by its:
•Amplitude
•Period or frequency
•Phase constant
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?
sec
Frequency of the motion, f
The number of complete cycles per second
Unit?
s-1
Relationship between
period and frequency?
Thursday, Nov. 20, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
f

1
T
or
T
=
1
f
10
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 (w ( t + T ) + j )= Acos (w t + 2p + j )
T
=
How many full cycles of oscillation
does this undergo per unit time?
2p
w
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 (w t + j )
dx
= -w Asin (w t + j ) Max speed vmax  A
dt
Max acceleration
dv
2
2
Acceleration at any given time a  = -w Acos (w t + j )  x
amax = w 2 A
dt
What do we learn
Acceleration is reverse direction to displacement
about acceleration?
Acceleration and speed are /2 off phase of each other:
When v is maximum, a is at its minimum
Speed at any given time
Thursday, Nov. 20, 2014
v

PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
11
Simple Harmonic Motion continued
Phase constant determines the starting position of a simple harmonic motion.
x = Acos (w t + j )
x t =0 = Acosj
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
At t=0
xi = Acosj
vi = -w Asinj
æ vi ö
By taking the ratio, one can obtain the phase constant j = tan ç è w xi ÷ø
-1
xi2 = A2 cos2 j
By squaring the two equations and adding
them together, one can obtain the amplitude
(
)
v
A2 cos2 j + sin2 j = A2 = xi2 + æç i ö÷
èwø
Thursday, Nov. 20, 2014
vi2 = w 2 A2 sin 2 j
2
æ vi ö
2
A = xi + ç ÷
èwø
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
2
12
Sinusoidal Behavior of SHM
What do you think the trajectory will look if the
oscillation was plotted against time?
Thursday, Nov. 20, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
13
Sinusoidal Behavior of SHM
x  A cos  2 ft 
v  v0 sin  2 ft 
a  a0 cos  2 ft 
Thursday, Nov. 20, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
14
Example for A 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 æçè p t + p4 ö÷ø where t is in
seconds and the angles in the parentheses are in radians. a) Determine the amplitude,
frequency, and period of the motion.
From the equation of motion:
pö
æ
=
4.00m
cos
p
t
+
=
Acos
w
t
+
j
) çè
x
(
) (
÷
4ø
The amplitude, A, is A = 4.00m The angular frequency, ϖ, is w = p
Therefore, frequency
2 2
1

1
w

 =

 s 1

 2s
f
and period are

T

2p  2
T
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
pö
dx
æ
v  = - ( 4.00 ´ p ) sin ç p t + ÷ m / s
è
dt
4ø
By the same token, taking the
second derivative of equation of
motion, the acceleration, a, is
pö
d2x
æ
2
2
a = 2 = - 4.00 ´ p cos ç p t + ÷ m / s
è
4ø
dt
Thursday, Nov. 20, 2014
(
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
)
15
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.
d2x
k
If we    k
= - x denote
2
m
dt
m
d2x
2
The resulting differential equation becomes
=
w
x
2
dt
This becomes a second
order differential equation
Fspring  ma
 kx
k
a=- x
m
x = Acos (w t + j )
Since this satisfies condition for simple
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 (w t + j ))= -wAsin (w t + j )
Now the second order derivative becomes
dt
dt
d
d2x
2
=
w
A
sin
w
t
+
j
=
w
Acos (w t + j )= -w 2 x
(
)
(
)
2
dt
dt
Whenever the force acting on an object is linearly proportional to the displacement from some
equilibrium position and is in the opposite direction, the particle moves in simple harmonic motion.
Thursday, Nov. 20, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
16
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 = Acosw t
f
T
=
=
w
2p
w
=
= 2p
1
w
1
=
=
T
2p
2p
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 a distance A, and the block is
let go from rest, giving 0 initial speed; xi=A, vi=0,
2
dx
d
x
v=
= -wAsin w t a  2 = -w 2 Acos w t ai = -w 2 A = -kA / m
dt
dt
This equation of motion satisfies all the SHM conditions. So it is the solution for this motion.
Suppose a 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
pö
vi ö
p
æ
-1 æ
-1
j = tan ç - ÷ = tan ( -¥) = - x = Acos ç w t - ÷ = Asin (w t ) solution?
è
2
è wx ø
2ø
Special case #2
i
Thursday, Nov. 20, 2014
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
17
Example for Spring Block System
A car with a mass of 1300kg is constructed so that its frame is supported by four
springs. Each spring has a force constant of 20,000N/m. If two people riding in the
car have a combined mass of 160kg, find the frequency of vibration of the car after it is
driven over a pothole in the road.
Let’s assume that mass is evenly distributed to all four springs.
The total mass of the system is 1460kg.
Therefore each spring supports 365kg each.
From the frequency relationship
based on Hook’s law
Thus the frequency for
vibration of each spring is
f =
f
1
2p
=

1
1

=
2
T
2p
k
1
=
m
2p
k
m
20000
= 1.18s -1 = 1.18 Hz
365
How long does it take for the car to complete two full vibrations?
The period is
Thursday, Nov. 20, 2014
T =
1
= 2p
f
m
= 0.849s For two cycles
k
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
2T  1.70s
18
Example for Spring Block System
A block with a mass of 200g is connected to a light spring for which the force constant is
5.00 N/m and is free to oscillate on a horizontal, frictionless surface. The block is displaced
5.00 cm from equilibrium and released from reset. Find the period of its motion.
From the Hook’s law, we obtain

X=0
X=0.05
k

m

5.00
 5.00s 1
0.20
As we know, period does not depend on the
amplitude or phase constant of the oscillation,
therefore the period, T, is simply
T

2


2
 1.26 s
5.00
Determine the maximum speed of the block.
From the general expression of the
simple harmonic motion, the speed is
Thursday, Nov. 20, 2014
dx
= -w Asin (w t + j )
dt
= w A  5.00  0.05  0.25m/ s
vmax 
PHYS 1443-004, Fall 2014
Dr. Jaehoon Yu
19