Physics 116
ELECTROMAGNETISM AND
OSCILLATORY MOTION
Lecture 2
SHM and circular motion
Sept 30, 2011
R. J. Wilkes
Email: ph116@u.washington.edu
In case you were not
here yesterday:
First: I’m not Wilkes!
Your lecturer today: Victor Polinger
Substituting for R. J. Wilkes
(He will return on Monday)
Announcements
In case you were not
here yesterday:
(we’ll have a slide like this one most days)
- PHYS 116 Course home page:
Visit frequently for updated course info!
http://faculty.washington.edu/wilkes/116/
- Physics Dept’s general-information page for
100 courses:
http://www.phys.washington.edu/1xx/
- Webassign page (for homework and grades):
http://webassign.net/washington/login.html
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All HW Assignments and Grades on WebAssign
Homework solutions are submitted online.
No work is accepted past due date.
Click on assignment name to begin working.
Grades for homework are stored in lecture section.
NOTE: Lab grades stored separately, under their own sections.
In case you were not
here yesterday:
Other Class Resources
• Your Fellow Students
– You’re encouraged to study and discuss class together!
• Physics Study Center (A-wing Mezzanine, just under
our lecture room)
– Meeting place / work space for students
– Get guidance and help from TAs outside 116 office hours
– Look over other textbooks on the same subjects
• Physics Library (6th floor of C-wing)
– Best view from a comfy chair on campus!
• Class discussion board
–
https://catalyst.uw.edu/gopost/board/wilkes/23253/
• Your TA
– Kyle Armour (take a bow, Kyle)
– He will post office hours next week
– Email questions to him via class email, ph116@uw.edu
In case you were not
here yesterday:
“Clickers” are required
iCue / H-ITT Clickers (TX-3100)
• Required to enter answers in quizzes
– Be sure to get radio (RF), not infrared (IR)
• We will not use them this week
• We’ll practice using them next week, and begin using
them for pop quizzes thereafter
– Bring your clicker to class every day from Oct 3 onward
• Why must we torture you with pop quizzes?
– Motivation to attend class (physically and mentally)!
– Quizzes are designed to be easy IF you are paying attention
• Questions will be about something we just discussed!
Lecture Schedule
(up to exam 1)
Today
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summary from
yesterday:
Oscillation terminology
•
•
•
Period T = length of one full cycle (time between peaks)
Frequency f = number of cycles per unit time (units = 1 / time)
Amplitude A = maximum displacement (length)
•
Graph of oscillating object’s position vs time for T= 1 sec, f = 1 Hz, A = 2 m
position along the cycle is the phase of the wave:
( t / T ) = (φ / 2π) so φ = 2π ( t / T ) , or φ = 360° ( t / T )
2
f=1/T
Unit of
frequency =
1 cycle/sec
= 1 hertz (Hz)
T = 1 sec
1
0
0
0.25
0.5
0.75
1
1.25
A=2m
-1
-2
Distance,
meters
time, seconds
1.5
1.75
2
Examples / applications
1. Pulse rate is 70 beats / minute
• What is f in Hz ?
This is just conversion of units: 1 min = 60 sec so
f in Hz (cycles/sec) = (cycles/min)*(min/sec)
f = { 70 (1/min) }* { 1/60 (min/sec) } = 70/60 (1/sec=Hz) = 1.17 Hz
• What is period T in sec ?
connection between f and T : f = 1 / T
So in this case T = 1 / f = 1 / 1.17 Hz (cycles/sec) = 0.85 sec
2. Mass on a spring moves with amplitude A and period T
• How long does it take to move a net distance A ?
Definition of A = distance displaced in ¼ of a cycle (1 cycle takes t = T)
so it reaches x = A at time t = T/4
A
Note: it does NOT reach A/2 in T/8!
A/2
Speed is not constant: x = A sin(2p { t / T })
x=A/2 when sin(2p { t / T }) = ½
here t / T = fraction of a full cycle at time t,
2p { t / T } = fraction of 2p = phase of sine curve at t
sin-1( ½ ) = 0.52 radians = 0.083*2p
So x=A/2 when { t / T }=0.083, or t = 0.083T = T / 12
sin-1 (x) means “angle whose sine is x” (arcsine, or inverse sine)
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t=T
T/4
T/12
9
Position vs time in simple harmonic motion
•
•
Restoring force (spring-mass example): F = - k x
For Simple Harmonic Motion (SHM), k = constant
– force is proportional to displacement
•
•
Acceleration a = F/m = -(k/m)*x
Now comes magic (i.e. calculus, beyond our scope): as you know,
displacement
– a = rate of change of velocity v (“derivative” of v)
– v = rate of change of position x (“derivative” of x)
d ⎛ dx ⎞
d 2x
⎛k⎞
– In the language of calculus,
F = ma → − kx = m ⎜ ⎟ →
=
−
⎜ ⎟x
dt ⎝ dt ⎠
dt 2
⎝m⎠
– This is a “differential equation” for x vs time
phase, 0 - 2π
0
1
2
3
4
5
6
…whose solution (magic!) is
1.5
⎛ ⎛ t ⎞⎞
x(t ) = A cos⎜⎜ 2π ⎜ ⎟ ⎟⎟
1
⎝ ⎝ T ⎠⎠
0.5
Where A=displacement at t=0
0
And
m
T = 2π
-0.5
k
Graph’s lower axis shows time/T
⎛t⎞
Upper axis shows phase 2π ⎜ ⎟
⎝T ⎠
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-1.5
0
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0.2
0.4
0.6
0.8
1
t / T = fraction of period
10
SHM’s connection to circular motion
•
•
•
Circular motion is periodic
Coordinate of a point on a circle is given by sin/cos functions
So x or y coordinate of object moving in a circle has exactly the
same mathematical description as object in SHM
– Plot the x coordinate of a pin on a turntable – and we…
– Get the x coordinate of a mass on a spring with same period and
A=R
– Here’s a web applet that illustrates this:
http://www.ngsir.netfirms.com/englishhtm/SpringSHM.htm
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Angular velocity in SHM and circular motion
•
We can use the mathematical equivalence of circular motion
and SHM to find useful formulas about SHM
– The angular position of a peg on a turntable is just θ (t ) = ω t
– here θ is the total angle swept out, and ω is the angular velocity
• ω (omega) is the rate of change of angle with time
(this assumes we accumulate the total number of radians the peg sweeps out,
we do not reset to 0 radians whenever it passes its starting point)
– We found that the x coordinate of the peg has the same form as in SHM:
⎛1⎞
– Notice: 2π ⎜ ⎟ = 2π f
⎝T ⎠
⎛ ⎛ t ⎞⎞
x(t ) = A cos⎜⎜ 2π ⎜ ⎟ ⎟⎟
⎝ ⎝ T ⎠⎠
– So if we define ω = 2π f we can write x(t ) = A cos (ωt )
– ω = “angular frequency”
• number of radians per second of rotation, for peg on turntable
• Nothing actually rotating in SHM, but we use ω = 2 π f anyway
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Velocity in SHM and circular motion
•
Recall from PHYS 114:
– For uniform (constant ω ) circular motion
• peg’s speed is v = R ω
• x component of v points in – x direction for 0 < θ < π radians
• From triangle shown, we can see vx= - (v sin (θ) )
• so x component of velocity is vx= -Rω sin (ωt)
• So for SHM and circular motion we find:
⎛1⎞
2π ⎜ ⎟ = 2π f
⎝T ⎠
vy θ
vx
v
R
θ
ω = 2π f
y
x
θ (t ) = ω t
⎛ ⎛ t ⎞⎞
x(t ) = A cos ⎜ 2π ⎜ ⎟ ⎟ = A cos (ωt )
⎝ ⎝ T ⎠⎠
v(t ) = − Aω sin (ωt )
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Acceleration in SHM and circular motion
•
Again recall from PHYS 114:
– For uniform (constant ω ) circular motion
• Centripetal acceleration has magnitude a = v2 / R =(R ω)2/R = R ω2
• Centripetal acceleration points toward center
• x component of a points in – x direction, for 0 < θ < π/2 radians
• From triangle shown, we see x component ax= - a cos (θ)
• so x component of acceleration is ax(t)= R ω2 cos (ωt)
• So for both SHM and circular motion we find:
⎛1⎞
2π ⎜ ⎟ = 2π f
⎝T ⎠
ax
ω = 2π f
θ
ay
θ (t ) = ω t
x(t ) = A cos (ωt )
θ
a
R
x
v(t ) = − Aω sin (ωt )
a (t ) = − Aω 2 cos (ωt )
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Phase relationships and max values
•
•
Maximum values occur when sin/cos = 1, so
x(t ) = A cos (ωt )
v(t ) = − Aω sin (ωt )
max x = A
max v = Aω
a (t ) = − Aω 2 cos (ωt )
max a = Aω 2
x, v and a have phase relationships determined by their trig functions: cos, -sin
and -cos, all with the same value of ( ω t )
so at t = 0, x = +max, v = 0, and a = - max. After t=0,
x and v are 90 deg out of phase (¼ cycle shift)
x and a are 180 deg out of phase (opposite signs – ½ cycle shift)
1.5
1
x, v, a
0.5
x=Acos(wt)
0
v= - Awsin(wt)
a= - Aw^2cos(wt)
-0.5
-1
-1.5
0
0.25
0.5
0.75
1
t / T = fraction of period
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