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Physics 121C - MechanicsLecture 23 (Knight: 14.1 -
14.6)Simple Harmonic Motion
December 7, 2005 (29 Slides)
Physics 121C - MechanicsLecture 23 (Knight: 14.1 -
14.6)Simple Harmonic Motion
December 7, 2005 (29 Slides)
John G. CramerProfessor of Physics
B451 PABcramer@phys.washington.edu
December 7, 2005 Physics 121C - Lecture 23 2
AnnouncementsAnnouncementsTycho is Fixed! Homework Assignment #9 is due
on Midnight on Friday, December 9.
Tycho Survey: Please complete the “Survey” in the Homework section of Tycho as soon as possible. Its completion is required for this course.
The Final Exam will be held in this room on Monday, December 12 at 8:30 AM (Note: NOT 10:30 AM!).
Exam #3 will be returned on Friday. The Exam 3 grades and solutions are now on the web. If your Exam 3 scores are missing, see Laura Clement (C136 PAB). Exam #3 regrade requests will be accepted until noon on Monday, December 12.
December 7, 2005 Physics 121C - Lecture 23 3
20 30 40 50 60 70 80 90Score
0
5
10
15
20
rebmu
N
Exam 3
Exam 3 StatisticsExam 3 Statistics
Average = 57.83Std Dev = 13.91
1.7 2.7 3.7
High = 88Median = 58Low = 14
December 7, 2005 Physics 121C - Lecture 23 4
11
5-Dec-0522
3-D Rotation, Precession Supplemental
Angular Momentum
Make-up labs 7-Dec-05
23Simple Harmonic Motion 14.1 to 14.6
9-Dec-05 R4 Review & Extension-
12-Dec-05 FEFINAL EXAMINATION: Monday, December 12, 8:30 to 10:20 AM
Lecture Schedule (Part 4)
Lecture Schedule (Part 4)
We are here!
December 7, 2005 Physics 121C - Lecture 23 5
OscillationsOscillations
Oscillations (whether sinusoidal or otherwise) have some common characteristics:1. They take place around an equilibrium
position;2. The motion is periodic and repeats with each
cycle.
December 7, 2005 Physics 121C - Lecture 23 6
Frequency and PeriodFrequency and Period
1/ and 1/f T T f
is the frequency (units: Hz oscillations per second)f
is the period (units: s)T
December 7, 2005 Physics 121C - Lecture 23 7
Example: Radio Station Frequency and Period
Example: Radio Station Frequency and Period
What is the oscillation period of an FM radio station that broadcasts at 100 MHz?
8100 MHz 1.0 10 Hzf
88
11/ 1.0 10 s 10 ns
1.0 10 HzT f
Note that 1/Hz = s
December 7, 2005 Physics 121C - Lecture 23 8
Simple Harmonic MotionSimple Harmonic Motion
The special case of an oscillation that has smooth sinusiodal motion is called “Simple Harmonic Motion”, abbreviated SHM. The motion can be described by a period (the time required for one complete oscillation) and an amplitude (the distance from the equilibrium position to a turning point).
An example is a glider attached to a spring and moving on an air track.
December 7, 2005 Physics 121C - Lecture 23 9
Kinematics ofSimple Harmonic Motion
Kinematics ofSimple Harmonic Motion
2( ) cos cos 2
tx t A A f t
T
2(in rad/s) 2 (in Hz)f
T
( ) cosx t A t
max
( )( ) sin sinx
dx tv t A t v t
dt
max
22
Av A fA
T
December 7, 2005 Physics 121C - Lecture 23 10
Example: A System in SHM
Example: A System in SHM
An air-track glider is attached to a spring,pulled 20 cm to the right, and releasedat t-=0. It makes 15 completeoscillations in 10 s.
a. What is the period of oscillation?b. What is the object’s maximum speed?c. What is its position and velocity at t=0.80 s?
15 oscillations
10 s1.5 oscillations/s 1.5 Hz
f
1/ 0.667 sT f
max
2 2 (0.20 m)1.88 m/s
(0.667 s)
Av
T
2 2 (0.80 s)cos (0.20 m)cos 0.062 m 6.2 cm
(0.667 s)
tx A
T
max
2 2 (0.80 s)sin (1.88 m/s)sin 1.79 m/s 179 cm/s
(0.667 s)
tv v
T
December 7, 2005 Physics 121C - Lecture 23 11
Example: Finding the Time
Example: Finding the Time
A mass, oscillating in simple harmonic motion, starts at x=A and has period T. At what time, as a fraction of T, does the mass first pass through x=½A?
11 162
cos2 2 3
T Tt T
1
2
2cos
tx A A
T
December 7, 2005 Physics 121C - Lecture 23 12
SHM and Circular MotionSHM and Circular Motion
Uniform circular motion projected into one dimension is simple harmonic motion.
December 7, 2005 Physics 121C - Lecture 23 13
SHM and Circular MotionSHM and Circular Motion
cosx A
d
dt
0If 0 at 0, t t
( ) cos cosx t A A t
December 7, 2005 Physics 121C - Lecture 23 14
The Phase Constant 0The Phase Constant 0
0t
0
( ) cos ( )
cos
x t A t
A t
0
max 0
( ) sin
sin
v t A t
v t
0 0
0 0
cos
sinx
x A
v A
December 7, 2005 Physics 121C - Lecture 23 15
The Phase Constant 0The Phase Constant 0
December 7, 2005 Physics 121C - Lecture 23 16
Example:Using Initial Conditions
Example:Using Initial Conditions
A mass on a spring oscillates with a period of 0.80 s and an amplitude of 10 cm. At t=0 s it is 5.0 cm to the left of equilibrium and moving to the left. What are its position and direction of motion at t=2.0 s?
21 100 3
(0.05 m)cos cos 120
(0.10 m)
x
A
0The mass is moving to the left, so 120 .
2 27.854 rad/s
(0.80 s)T
0
2
3
( ) cos
(0.10 m)cos (7.854 rad/s)(2.0 s)
5.00 cm
x t A t
0
2
3
( ) sin
(0.10 m)(7.854 rad/s)sin (7.854 rad/s)(2.0 s)
68.1 cm/s ( >0 so moving to the right)
x
x
v t A t
v
December 7, 2005 Physics 121C - Lecture 23 17
Clicker Question 1Clicker Question 1
Shown are four harmonic oscillators at t=0. All are oscillating with amplitude A as shown. Their velocity directions are shown by the green arrows.
Which one has the phase constant ?
December 7, 2005 Physics 121C - Lecture 23 18
Energy and SHMEnergy and SHM1 12 22 2
constantxE K U mv kx
1 22
( )E x A U kA 1 2max2
( 0)E x K mv
1 12 2max2 2
mv kA max
kv A A
m
12
2
k k mf T
m m k
1 1 1 12 2 2 2max2 2 2 2
constantxE mv kx kA mv
December 7, 2005 Physics 121C - Lecture 23 19
Energy and SHMEnergy and SHM1 1 1 12 2 2 2
max2 2 2 2constantxE mv kx kA mv
2 2 2 2kv A x A x
m
222 20 00 0
mv vA x x
k
December 7, 2005 Physics 121C - Lecture 23 20
Example:Using Conservation of
Energy
Example:Using Conservation of
Energy A 500 g block on a spring is pulled a distance of 20 cm and released. The subsequent oscillations are measured to have a period of 0.80 s. At what position (or positions) is the speed of the block 1.0 m/s?
2 2 2 2kv A x A x
m
222 2 (1.0 m/s)
(0.20 m) 0.154 m 15.4 cm(7.85 rad/s)
vx A
2 20.80 s so 7.85 rad/s
(0.80 s)T
T
December 7, 2005 Physics 121C - Lecture 23 21
Clicker Question 2Clicker Question 2
Four springs have been compressed from their equilibrium positions at x=0 cm.
Which system has the largest maximum speed in its oscillation?
December 7, 2005 Physics 121C - Lecture 23 22
The Dynamics of SHMThe Dynamics of SHM
2( sin ) cosxx
dv da A t A t
dt dt
sinxv A t
( ) cosx t A t
December 7, 2005 Physics 121C - Lecture 23 23
The Dynamics of SHMThe Dynamics of SHM
2 cosxa A t
( ) cosx t A t
2xa x
December 7, 2005 Physics 121C - Lecture 23 24
The Equation of MotionThe Equation of Motion
sp( )xF k x
net sp x( ) ( )x xF F kx ma
2x
x 2
dv d xa
dt dt
2
2
d x kx
dt m
December 7, 2005 Physics 121C - Lecture 23 25
Solving the SHMEquation of MotionSolving the SHM
Equation of Motion2
2
d x kx
dt m
0( ) cosx t A t
0sindx
A tdt
2
202
cosd x
A tdt
20 0cos cos
kA t A t
m
2 The ( ) equation is a solution if k
x tm
What is x?
Let’s make a wild guess:
0Solution: ( ) cos with /x t A t k m
December 7, 2005 Physics 121C - Lecture 23 26
Example:Analyzing an Oscillator
Example:Analyzing an Oscillator
At t=0, a 500 g block oscillating on a spring is observed to be moving to the right at x=15 cm. It reaches a maximum displacement of 25 cm at t=0.300s.
a. Draw a position vs. time graph for one cycle of motion.
b. At what times during the cycle will the mass pass through 20 cm?
1 100
(15 cm)cos cos 0.927 rad
(25 cm)
x
A
0It moves to the right at t=0, so 0.927 rad.
max max 0 max 0cos so 0x A A t t
0 max/ ( 0.927 rad) /(0.300 s) 3.09 rad/st
22.03 sT
10
1
1cos
1 (20 cm)0.927 rad cos
3.09 rad/s (25 cm)
0.092 s and 0.508 s
xt
A
December 7, 2005 Physics 121C - Lecture 23 27
Vertical OscillationsVertical Oscillationssp( ) (because )yF k y k L y L
net sp
eq
For equilibrium ( ) ( )
0
y y yF F w
k L mg
eq eq. Now displace the mass from .mg
L Lk
net sp
eq
eq
( ) ( )
( )
( )
y y yF F w
k L y mg
k L mg ky
net( ) yF ky
December 7, 2005 Physics 121C - Lecture 23 28
Vertical OscillationsVertical Oscillations
0( ) cos( )y t A t
net( ) yF ky
December 7, 2005 Physics 121C - Lecture 23 29
Example:A Vertical Oscillation
Example:A Vertical Oscillation
A 200 g block hangs from a spring with spring constant 10 N/m. The block is pulled down to a point where the spring is 30 cm longer than it’s unstretched length, then released. Where is the block and what is its velocity 3.0 s later?
/ 0.196 m =19.6 cmL mg k
0( 0.30 m) ( 0.196 m) 0.104 m; =0A
/ (10 N/m) /(0.2 kg) 7.07 rad/sk m
0( ) cos( )
( 0.104 m)cos[(7.07 rad/s)(3.0 s)]
0.074 m
= 7.7 cm (above equilibrium position).
y t A t
0sin( ) 0.52 m/s 52 cm/sxv A t
December 7, 2005 Physics 121C - Lecture 23 30
End of Lecture 23End of Lecture 23Tycho is Fixed! Homework Assignment #9 is due
on Midnight on Friday, December 9.
Tycho Survey: Please complete the “Survey” in the Homework section of Tycho as soon as possible.
The Final Exam will be held in this room on Monday, December 12 at 8:30 AM (Note: NOT 10:30 AM!).
Exam #3 should be graded and returned by Friday. The Exam 3 solutions are on the web. Exam #3 regrade requests will be accepted until noon on Monday, December 12.