Transcript 2 - cloudfront.net

Chapter 13
Oscillations About
Equilibrium
FOCUSED LEARNING TARGET
GIVEN VIBRATIONS AND OSCILLATIONS CAUSED
BY SPRING AND PENDULUM , I WILL BE ABLE TO
CALCULATE FORCE OF SPRING (FS) , TOTAL
ENERGY IN TERMS OF ELASTIC POTENTIAL
ENERGY (US) AND KINETIC ENERGY (K),
FREQUENCY (f) AND PERIOD (t) USING THE
FOLLOWING EQUATIONS
FS = -kx ; K= ½ mv2 ;
Ki + Ui = Kf +Uf
US = ½ kx2= ½ kA2 ; E = K + U ; T= 1/f ; f = 1/T
f = 1/2π √k/m ; T = 2π√m/k ; vmax = A√k/m
HOMEWORK :
• CHAPTER 13. 1- 13.7 SUMMARIES
• 1 EXAMPLE FOR EACH SECTION
• 2 HW PROBLEMS(ODD ) FOR EACH SECTION
• 2 PROBLEMS IN THE COLLEGE BOARD
•
https://apstudent.collegeboard.org/apcourse/
ap-physics-b/exam-practice
Experiment :PENDULUM
1. HYPOTHESIS :
WHAT WILL HAPPEN
TO THE NUMBER OF
VIBRATIONS OF THE
PENDULUM WHEN
THE MASS OF THE
PENDULUM
INCREASES?
________________
2. HYPOTHESIS :
WHAT WILL HAPPEN
TO THE NUMBER OF
VIBRATIONS OF THE
PENDULUM WHEN
THE STRING OF THE
PENDULUM
INCREASES?
________________
CW :Pendulum
Part 1:Half of the String’s Length
3. Length of the String
_______cm
4. How many complete
cycles in one minute ?
Do
3 trials and get the
average
Cycles = _______
Time = 1 minute
5. Frequency in cycles
per minute
= ___________
6.Frequency in cycles
per second
=_____________
7. Period =_____sec
Pendulum
Part 2 : Double the Length of the String
8. Length of the String
_______cm
9. How many complete
cycles in one minute ?
Do
3 trials and get the
average
Cycles = ______
Time = 1 minute
10. Frequency in cycles
per minute
= ___________
11.Frequency in cycles
per second
=_____________
12. Period =
________sec
Pendulum
Part 3 : Double the Mass and Use Half of the
String’s Length
13. Length of the String
_______cm
14. How many
complete cycles in one
minute ? Do
3 trials and get the
average
Cycles = ______
Time = 1 minute
15. Frequency in
cycles per minute
= ___________
16.Frequency in
cycles per second
=_____________
17. Period =
____sec
Conclusion
15. WHAT HAPPENS TO THE NUMBER OF
VIBRATIONS OF THE PENDULUM WHEN
THE STRING OF THE PENDULUM
INCREASES?
16. WHAT HAPPENS TO THE NUMBER OF
VIBRATIONS OF THE PENDULUM WHEN
THE MASS OF THE PENDULUM
INCREASES?
Read p 444-446
Vocabulary :
1. Oscillation
2. Propagation
3. Wave pulse
4. Hooke’s Law
5. Simple Harmonic
Motion
6. Displacement
7. Amplitude
8. Period
9. Frequency
10. Hertz
Equations Necessary :
1. Periodic Motion
• Motion that repeats itself over a fixed and
reproducible period of time.
• The revolution of a planet about its sun is an
example of periodic motion. The highly
reproducible period (T) of a planet is also
called its year.
• 2. Mechanical devices on earth can be
designed to have periodic motion. These
devices are useful timers. They are called
oscillators.
3. Simple Harmonic Motion
You attach a weight to a spring, stretch the spring past
its equilibrium point and release it. The weight bobs
up and down with a reproducible period, T.
• Plot position vs. time to get a graph that resembles
a sine or cosine function. The graph is "sinusoidal",
so the motion is referred to as simple harmonic
motion.
• Springs and pendulums undergo simple harmonic
motion and are referred to as simple harmonic
oscillators.
4. Crest
5. Troughs
6. Amplitude- Maximum displacement from
equilibrium. Related to energy.
7. Equilibrium
7. Period(T)
Length of time required for one oscillation.
8. Frequency
• How fast the oscillator is oscillating.
• f = 1/T
• Unit: Hz or s-1
9. Springs
Springs are a common type of simple harmonic
oscillator.
Our springs are "ideal springs", which means
• They are massless.
• They are both compressible and extensible.
They will follow Hooke's Law.
• Fs = -kx
11. F=-kx ( Fs opposite with X)
FS= 0
FS
X>0
FS
X< 0
12.Fs = -kx
The acceleration of the block is equal to
a = Fs / m
13.Another way to describe the block’s motion is
the energy it transfers.
Us = ½ k x2
When you pull the block, you are increasing the
elastic potential energy.
14. Releasing the block , potential energy
becomes kinetic energy as the block moves. As it
passes through equilibrium Us =0 , so all energy
is K.
15. As it passes again through equilibrium , it
compresses the spring , K –kinetic becomes Uselastic potential
11. F=-kx ( Fs opposite with X)
FS= 0
FS
Us = 0
K is maximized
FS
Us =maximized
K =0
V=0
X< 0
X>0
Us =maximized
K =0
V=0
• 13. A 12 cm long spring has a force constant
(k) of 400 N/m . How much force is required to
stretch the spring to a length of 14cm.
• 13. A 12 cm long spring has a force constant
(k) of 400 N/m . How much force is required to
stretch the spring to a length of 14cm.
• F = -kx
• F = - 400N/m ( .14m -.12m) = - 8 N
Conservation of Energy
Springs and pendulums obey conservation of
energy.
• The equilibrium position has high kinetic
energy and low potential energy.
• The positions of maximum displacement have
high potential energy and low kinetic energy.
• Total energy of the oscillating system is
constant.
14. A block of mass m = 2 kg is attached to an
ideal spring of force constant k = 500N/m . The
amplitude of the resulting oscillations is 8 cm .
Determine the total energy of the oscillator and
the speed of the block when it is 4 cm from
equilibrium.
14. A block of mass m = 2 kg is attached to an
ideal spring of force constant k = 500N/m . The
amplitude of the resulting oscillations is 8 cm .
Determine the total energy of the oscillator and
the speed of the block when it is 4 cm from
equilibrium.
E = Us + K = ½ kx2 + 0 = ½ (500N/m)(.08m)2 =1.6J
1.6J= ½ kx2 + ½ mv2 = ½ (500N/m)(.04m)2 + ½ (2kg) v2
v = 1.1 m/s
14. A block of mass m = 0.05 kg oscillates on a
spring whose force constant k is 500 N/m. The
amplitude of the oscillations is 4 cm . Calculate
the maximum speed of the block .
14. A block of mass m = 0.05 kg oscillates on a
spring whose force constant k is 500 N/m. The
amplitude of the oscillations is 4 cm . Calculate
the maximum speed of the block .
Us = ½ kx2  K = ½ mv2
½ kx2 = ½ mv2
v
= √ kX2 /m
v =√ 500N/m ( .04m)2/ 0.05kg
v
= 4m/s
15. A block of mass m = 8kg is attached to an
ideal spring of force constant k = 500N/m . The
block is at rest at its equilibrium position. An
impulsive force acts on a block , giving it an
initial speed of 2m/s . Find the amplitude of the
resulting oscillations?
15. A block of mass m = 8kg is attached to an ideal
spring of force constant k = 500N/m . The block is at
rest at its equilibrium position. An impulsive force
acts on a block , giving it an initial speed of 2m/s .
Find the amplitude of the resulting oscillations?
Ei = Ef
Ki + Ui = Kf + Uf
½ mv2 + 0 = 0 + ½ kx2
8kg (2m/s)2= 500N/m X2
x = 0.25 m
CW :
• A mass of 0.5 kg is connected to a massless
spring with a force constant k of 50N/m . The
system is oscillating on a frictionless horizontal
surface . If the amplitude of the oscillations is
2cm , what is the total energy of the system ?
• 16. A block oscillating on the end of a spring
moves from its position of a maximum spring
stretch to maximum spring compression in
0.25sec . Determine the period and frequency
of this motion.
• 16. A block oscillating on the end of a spring
moves from its position of a maximum spring
stretch to maximum spring compression in
0.25sec . Determine the period and frequency
of this motion.
• 16. A block oscillating on the end of a spring
moves from its position of a maximum spring
stretch to maximum spring compression in
0.25sec . Determine the period and frequency
of this motion.
• For whole cycle T = 0.5sec
• f = 1/T = 1/0.5s = 2 Hertz
CW:
. A student observing an oscillating block counts
45.5 cycles of oscillations in one minute .
Determine its frequency in hertz and period in
seconds.
CW
. A student observing an oscillating block counts
45.5 cycles of oscillations in one minute .
Determine its frequency in hertz and period in
seconds.
f= 45.5 cycles/min X 1min/60sec = 0.758
cycles/sec = 0.758 Hz
T = 1/f = 1/ 0.758Hz= 1.32 sec
17. A block of mass m = 2 kg is attached to a
spring whose force constant k , is 300 N/m .
Calculate the frequency and period of the
oscillations of this spring –block system.
17. A block of mass m = 2 kg is attached to a
spring whose force constant k , is 300 N/m .
Calculate the frequency and period of the
oscillations of this spring –block system.
f = 1/2π √k/m
f = 1/2π√ (300N/m) / 2kg
f = 1.9 Hz
T = 1/f = 1/ 1.9Hz = 0.51 sec
• 18. A block is attached to a spring and set into
oscillatory motion and its frequency is
measured . If this block were removed and
replaced by a second block with ¼ the mass of
the first block , how would the frequency of
the oscillations compare to the first block ?
• 18. A block is attached to a spring and set into
oscillatory motion and its frequency is
measured . If this block were removed and
replaced by a second block with ¼ the mass of
the first block , how would the frequency of
the oscillations compare to the first block ?
f = 1/2π √ k/m
= 1/2π √ k / (1/4)m
f = 1/2π√ 4k/m = 1/2π (2) √k/m
f increased by a factor of 2
CW:
Calculate the period of a 300-g mass
attached to an ideal spring with a
force constant of 25 N/m.
CW
A 300-g mass attached to a spring
undergoes simple harmonic motion
with a frequency of 25 Hz. What is
the force constant of the spring?
CW : An 80-g mass attached to a spring
hung vertically causes it to stretch 30 cm
from its unstretched position. If the mass
is set into oscillation on the end of the
spring, what will be the period?
Sample Problem
You wish to double the force
constant of a spring. You
• A. Double its length by connecting
it to another one just like it.
• B. Cut it in half.
• C. Add twice as much mass.
• D. Take half of the mass off.
Sample Problem
You wish to double the force
constant of a spring. You
• A. Double its length by connecting
it to another one just like it.
• B. Cut it in half.
• C. Add twice as much mass.
• D. Take half of the mass off.
CW : Sample problem.
A 2.0-kg mass attached to a spring oscillates with an
amplitude of 12.0 cm and a frequency of 3.0 Hz.
What is its total energy?
19. Pendulums
A simple pendulum consists of a weight of mass
m attached to a string or a massless rod that
swings without friction, about the vertical
equilibrium position .
The pendulum can be thought of as a simple
harmonic oscillator.
The displacement needs to be small for it to
work properly.
21. RESTORING FORCE
• FRESTORING = mg Sinθ
• The restoring force is provided by gravity.
• Displacement is zero at equilibrium.
• At the endpoints of the oscillation region , the
restoring force and tangential acceleration at
have the greatest magnitudes, the speed of the
pendulum is zero , potential energy is maximized.
• As the pendulum passes through the equilibrium
position, its kinetic energyand speed are
maximized.
22. A simple pendulum has a period of 1s on
earth. What would be its period on the Moon(
where g = 1/6 of the earth )
T= 2π√ L/g
T = increased by √6 = 1sec X √6
Sample problem
Predict the period of a pendulum consisting of a 500
gram mass attached to a 2.5-m long string.
Sample problem
Suppose you notice that a 5-kg weight tied to a string
swings back and forth 5 times in 20 seconds. How
long is the string?
Sample problem
The period of a pendulum is observed to be T.
Suppose you want to make the period 2T. What do
you do to the pendulum?
Conservation of Energy
Pendulums also obey conservation of energy.
• The equilibrium position has high kinetic
energy and low potential energy.
• The positions of maximum displacement have
high potential energy and low kinetic energy.
• Total energy of the oscillating system is
constant.