periodic motion
DESCRIPTION
Periodic Motion. Chapter 11: Sections 1-3, 5. Oscillation. A complete fluctuation of the value of a quantity above and below some median value. Simple Harmonic Motion. Equilibrium position - Central position where forces are balanced. - PowerPoint PPT PresentationTRANSCRIPT
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Periodic Motion
Chapter 11: Sections 1-3, 5
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Physics chapter 11 2
Oscillation
A complete fluctuation of the value of a quantity above and below some median value.
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Physics chapter 11 3
Simple Harmonic MotionEquilibrium position - Central position where
forces are balanced.Amplitude (A) - the maximum displacement
from equilibrium.(m)Period (T) - Time for one complete
oscillation.(s)Frequency (f)- Number of oscillations that
occur in a given unit of time. (cyc/s or Hz)Angular frequency (w) - the frequency given
in rad/s.
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Physics chapter 11 4
Simple Harmonic Motion
The displacement vector and the vector for the restoring force are always in opposite directions.
Hooke's LawF = kxThe force a spring exerts is proportional to
the distance that the spring is stretched or compressed.
k = F⁄x (N/m)
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Physics chapter 11 5
Simple Harmonic Motion
= 2f =2T
f =1
T
a = - k
mx
T = 2m
k
mass & springxa 2
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Physics chapter 11 6
Example
When a ball is suspended from a spring, the spring stretches by 70 mm.
If the ball oscillates up and down, what is its period?
What is its frequency?
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Physics chapter 11 7
Example
Write an equation for the force constant:
F kx
k Fx
mgx
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Physics chapter 11 8
Example
We don't know the mass of the ball.Write the equation for the period of
oscillation:T 2
mk
1k
xmg
T 2mxmg
2xg
20.07m
9.8m / s2 0.53s
f 1T1.89 Hz
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Physics chapter 11 9
Example 2
A piston undergoes simple harmonic motion in a vertical direction with an amplitude of 6 cm.
A coin is placed on top of the piston. What is the lowest frequency at which the
coin will be left behind by the piston on its downstroke?
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Physics chapter 11 10
Example 2
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Physics chapter 11 11
Example 2
The coin will leave the piston when the downward acceleration of the latter exceeds the acceleration of gravity g.
The maximum downward acceleration of piston occurs at the highest point of its motion.
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Physics chapter 11 12
Example 2
f 1
2gA
1
29.8m / s2
0.06m
= 2.0 Hz
amax 2A 42f 2A g
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Physics chapter 11 13
Simple Pendulum
All of the mass may be considered to be located at the end or bob.
2 =g
L
T = 2L
g
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Physics chapter 11 14
Example
A certain simple pendulum has a period on the earth of 0.500 s. What is its period on the surface of the moon, where g = 1.67 m/s2?
1.21s
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Physics chapter 11 15
Energy in SHM
2 21 1
2 2E mv kx
2max
1
2E mv
21
2E kA
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Physics chapter 11 16
Simple Harmonic Motion
x Acost
Simple Harmonic Motion
-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4 5 6 7 8
time (s)
po
siti
on
(m
)
x
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Physics chapter 11 17
Motion Equations
x = 0.5cos2πt (in standard units)Amplitude = 0.5 mAngular Frequency = 2π rad/sf = /2πT = 1/f
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Physics chapter 11 18
Motion Equations
x = Acostv = – Asinta = – 2Acost vmax = A= 2πfA where x = 0.
a = – 2xamax = 2A at x = ±A
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Physics chapter 11 19
x vs. t
x = Acoswt
Simple Harmonic Motion
-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4 5 6 7 8
time (s)
x
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Physics chapter 11 20
x and v vs. t
Simple Harmonic Motion
-2-1.5
-1-0.5
00.5
11.5
2
0 1 2 3 4 5 6 7 8
time (s)
x
v
v = – wAsinwtvmax = wA= 2πfA where x = 0.
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Physics chapter 11 21
x, v, and a vs. t
Simple Harmonic Motion
-3
-2
-1
0
1
2
3
0 1 2 3 4 5 6 7 8
time (s)
x
v
a
a = – w2Acoswta = w2xamax = w2A at x = ±A
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Physics chapter 11 22
More equations
km
m
kf
2
1
2
k
m
fT 2
1
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Physics chapter 11 23
ExampleAs shown on the next slide, a rotating
wheel drives a piston by means of a long connecting rod pivoted at both ends.
The wheel's radius is 20 cm and it is turning at 4.0 rev/s.
A) Write an equation for the displacement of the piston as a function of time.
B) Find the maximum speed and acceleration of the piston.
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Physics chapter 11 24
Example
•
x = ?
vmax = ?
amax = ?
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Physics chapter 11 25
Example
x = Acos t
A = 20 cm = 0.2 m = 4 rev/s x 2 rad/1 rev = 8 rad/s
x = 20cos8 t cmx = 0.2cos8 t m
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Physics chapter 11 26
Example
v = – Asintv = – 1.6sin8 t m/s
vmax = A= 2πfA
vmax = 1.6 m/s
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Physics chapter 11 27
Example
a = – 2Acosta = – 12.82cos8 t m/s2
amax = 2A
amax = 12.82 m/s2