Transcript Monday, Dec. 1, 2003

PHYS 1443 – Section 003
Lecture #23
Monday, Dec. 1, 2003
Dr. Jaehoon Yu
1.
2.
3.
4.
5.
6.
7.
8.
Simple Harmonic Motion and Uniform Circular Motion
Damped Oscillation
Waves
Speed of Waves
Sinusoidal Waves
Rate of Wave Energy Transfer
Superposition and Interference
Reflection and Transmission
Monday, Dec. 1, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
1
Announcements
• Homework # 12
– Due at 5pm, Friday, Dec. 5
• The final exam
– On Monday, Dec. 8, 11am – 12:30pm in SH103.
– Covers: Chap. 10 not covered in Term #2 – Ch15.
• Need to talk to me? I will be around this week.
Monday, Dec. 1, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
2
Simple Harmonic and Uniform Circular Motions
Uniform circular motion can be understood as a
superposition of two simple harmonic motions in x and y axis.
y
y
A
P
y
f
O
x
P
y
w
A
q
O x Q
y
q
A a
q
O
Q
A
O vx Q
x
t=t
t=0
q=wt+f
When the particle rotates at a uniform angular
speed w, x and y coordinate position become
Since the linear velocity in a uniform circular
motion is Aw, the velocity components are
Since the radial acceleration in a uniform circular
motion is v2/A=w2A, the components are
Monday, Dec. 1, 2003
v
P
x
ax P
x  A cosq  Acoswt  f 
y  A sin q  Asin wt  f 
vx  v sin q   Aw sin wt  f 
v y  v cosq  Aw coswt  f 
a x  a cosq   Aw 2 coswt  f 
a y  a sin q   Aw 2 sin wt  f 
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
3
x
Example for Uniform Circular Motion
A particle rotates counterclockwise in a circle of radius 3.00m with a constant angular
speed of 8.00 rad/s. At t=0, the particle has an x coordinate of 2.00m and is moving to
the right. A) Determine the x coordinate as a function of time.
Since the radius is 3.00m, the amplitude of oscillation in x direction is 3.00m. And the
angular frequency is 8.00rad/s. Therefore the equation of motion in x direction is
x  A cosq  3.00mcos8.00t  f 
Since x=2.00, when t=0
2.00   3.00m  cos f ;
However, since the particle was
moving to the right f=-48.2o,
 2.00 

  48.2
 3.00 
f  cos 1 

x  3.00m cos 8.00t  48.2

Find the x components of the particle’s velocity and acceleration at any time t.
Using the
displcement
Likewise,
from velocity


 

dx
vx 
 3.00  8.00sin 8.00t  48.2   24.0m / s sin 8.00t  48.2
dt
dv
a x    24.0  8.00cos8.00t  48.2   192m / s 2 cos 8.00t  48.2
dt
Monday, Dec. 1, 2003

PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
4
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.
F
Let’s consider a system whose retarding force
is air resistance R=-bv (b is called damping
coefficient) and restoration force is -kx
The solution for the above 2nd order
differential equation is
The angular frequency w
for this motion is
w
x
 kx  bv  max
dx
d 2x
 kx  b
m 2
dt
dt
x
k  b 


m  2m 
e
2

b
t
2m
Acoswt  f 
Damping Term
This equation of motion tells us that when the retarding force is much smaller than restoration
force, the system oscillates but the amplitude decreases, and ultimately, the oscillation stops.
We express the
angular frequency as
Monday, Dec. 1, 2003
w
 b 
 w 2  

 2m 
2
Where as the natural
frequency w0
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
w0 
k
m
5
More on Damped Oscillation
The motion is called Underdamped when the magnitude of
the maximum retarding force Rmax = bvmax <kA
How do you think the damping motion would change as
retarding force changes?
 bvmax  kA
As the retarding force becomes larger, the amplitude reduces
more rapidly, eventually stopping at its equilibrium position
Under what condition this system
does not oscillate?
The system is Critically damped
w
 0 w 
b
2m
b  2mw  2 mk
What do you think happen?
Once released from non-equilibrium position, the object
would return to its equilibrium position and stops.
If the retarding force is larger
than restoration force
Rmax  bvmax  kA The system is Overdamped
Monday, Dec. 1, 2003
Once released from non-equilibrium position, the object would return
PHYSposition
1443-003,and
Fall 2003
to its equilibrium
stops, but a lot slower than before 6
Dr. Jaehoon Yu
Waves
• Waves do not move medium rather carry energy from
one place to another
• Two forms of waves
– Pulse
– Continuous or periodic wave
• Wave can be characterized by
– Amplitude
– Wave length
– Period
• Two types of waves
– Transverse Wave
– Longitudinal wave
• Sound wave
Monday, Dec. 1, 2003
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
7
Speed of Transverse Waves on Strings
How do we determine the speed of a transverse pulse traveling on a string?
If a string under tension is pulled sideways and released, the tension is responsible for
accelerating a particular segment of the string back to the equilibrium position.
The acceleration of the
particular segment increases
The speed of the wave increases.
So what happens when the tension increases?
Which means?
Now what happens when the mass per unit length of the string increases?
For the given tension, acceleration decreases, so the wave speed decreases.
Newton’s second law of motion
Which law does this hypothesis based on?
Based on the hypothesis we have laid out
above, we can construct a hypothetical
formula for the speed of wave
Is the above expression dimensionally sound?
Monday, Dec. 1, 2003
T
v

T: Tension on the string
: Unit mass per length
T=[MLT-2], =[ML-1]
(T/)1/2=[L2T-2]1/2=[LT-1]
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
8
Speed of Waves on Strings cont’d
v
q
T
Ds
Fr
q q
q
T
R
O
Let’s consider a pulse moving right and look at it in
the frame that moves along with the the pulse.
Since in the reference frame moves with the pulse,
the segment is moving to the left with the speed v,
and the centripetal acceleration of the segment is
Now what do the force components
look in this motion when q is small?
F
F
t
r
What is the mass of the segment when
the line density of the string is ?
Using the radial
force component
 T cosq  T cosq  0
 2T sin q  2q
m  Ds  R 2q  2 Rq
2
2
v
v
 Fr  ma  m R  2Rq R  2Tq
Therefore the speed of the pulse is
Monday, Dec. 1, 2003
v2
ar 
R
v
T

PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
9
Example for Traveling Wave
A uniform cord has a mass of 0.300kg and a length of 6.00m. The cord passes over a
pulley and supports a 2.00kg object. Find the speed of a pulse traveling along this cord.
5.00m
1.00m
Since the speed of wave on a string with line
density  and under the tension T is
The line density  is  
M=2.00kg
The tension on the string is
provided by the weight of the
object. Therefore
v
T

0.300kg
 5.00 10  2 kg / m
6.00m
T  Mg  2.00  9.80  19.6kg  m / s
Thus the speed of the wave is
v
Monday, Dec. 1, 2003
T


19.6
 19.8m / s
2
5.00  10
PHYS 1443-003, Fall 2003
Dr. Jaehoon Yu
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