@kul 2 shm - circular - energi
TRANSCRIPT
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SHM and Circular Motion
Uniform circularmotion projected onto
one dimension issimple harmonic
motion.
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SHM and Circular Motion
x(t) Acos
d
dt
t
x(t) Acos t
Start with the x-component ofposition of the particle in UCM
End with the same result asthe spring in SHM!
Notice it started at angle zero
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Initial conditions:
t 0
We will not always start ourclocks at one amplitude.
x(t) Acos t 0 v x (t) Asin t 0 v x (t) v max sin t 0
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Phase Shifts:
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An object on a spring oscillates with a period of 0.8s andan amplitude of 10cm. At t=0s, it is 5cm to the left of
equilibrium and moving to the left. What are its positionand direction of motion at t=2s?
x(t) Acos t 0
x 0 5cm Acos 0 Initial conditions:
0 cos 1 x 0
A
cos
1 5cm10cm
120
23
rads
From the period we get: 2T
20.8s
7.85rad/s
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An object on a spring oscillates with a period of 0.8s andan amplitude of 10cm. At t=0s, it is 5cm to the left of
equilibrium and moving to the left. What are its positionand direction of motion at t=2s?
x(t) Acos t 0
7.85rad /s0 2
3 rads
A 0.1m
t 2s
x(t) 0.1cos 7.85 2 23
x(t) 0.05m
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We have modeled SHM mathematically.Now comes the physics.
Total mechanical energy is conservedfor our SHM example of a spring with
constant k, mass m, and on africtionless surface.
E K U 12
mv 2 12
kx 2
The particle has all potential energyat x=A and x= A, and the particlehas purely kinetic energy at x=0.
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At turning points:
E U 12
kA 2
At x=0:
E k 12
mv max2
From conservation:12
kA 2 12
mv max2
Maximum speed as related toamplitude:
v max k
mA
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From energy considerations:
From kinematics:
Combine these:
v max k
mA
v max A
k
m
f 12
k m
T 2 m
k
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a 500g block on a spring is pulled a distance of 20cm andreleased. The subsequent oscillations are measured to
have a period of 0.8s. at what position or positions is theblocks speed 1.0m/s?
The motion is SHM and energy is conserved.
1
2mv 2
1
2kx 2
1
2kA 2
kx 2 kA 2 mv 2
x A 2 mk
v 2
x A2 v
2
2
2
T
2
0.8s
7.85rad/s
x 0.15m
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8/12/2019 @Kul 2 SHM - Circular - Energi
12/31Dynamics of SHM
Acceleration is at a maximum when the particle is atmaximum and minimum displacement from x=0.
a x dv x (t)dt d Asin t dt
2Acos t
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13/31Dynamics of SHM
Acceleration isproportional to the
negative of thedisplacement.
a x 2Acos t
a x 2x
x Acos t
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14/31Dynamics of SHM
As we found with energyconsiderations:
a x 2x
F ma x kxma x kx
a x k mx
According to Newtons 2 nd Law: a x d
2xdt 2
Acceleration is notconstant:
d 2xdt 2
k m
x
This is the equation ofmotion for a mass on aspring. It is of a general
form called a second orderdifferential equation.
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2 nd -Order Differential Equations:
Unlike algebraic equations, their solutions are notnumbers, but functions.
In SHM we are only interested in one form so we canuse our solution for many objects undergoing SHM.
Solutions to these diff. eqns. are unique (there is onlyone). One common method of solving is guessing the
solution that the equation should have
d 2x
dt2
k
mx
Fromevidence, we
expect thesolution:
x Acos t 0
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-Order Differential Equations:
Lets put this possible solution into our equation andsee if we guessed right!
d2
xdt 2
k m
x
IT WORKS. Sinusoidal oscillation ofSHM is a result of Newtons laws!
x Acos t 0
d 2xdt 2
2Acos t
dx
dt
Asin t
2Acos t k m Acos t 2 k
m
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What about vertical oscillations ofa spring-mass system??
Fnet k L mg 0Hanging at rest:
k L mg
L mk
g
this is the equilibriumposition of the system.
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Now we let the systemoscillate. At maximum:
But:
Fnet k L y mg
Fnet k L mg ky
k L mg 0So:
Fnet ky
Everything that we have learned abouthorizontal oscillations is equally valid for
vertical oscillations!
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The Pendulum
Fnet t mg sin ma td 2sdt 2 gsin
Equation of motionfor a pendulum
s L
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Small Angle Approximation:
d 2sdt 2
gsin
When is about0.1rad or less, h
and s are about thesame.
sin cos 1tan sin 1
d 2sdt 2
g
sL
Fnet t md 2sdt 2
mgsL
f
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The Pendulum
Equation ofmotion for a
pendulum
d2
sdt 2
gsL
gL
(t) max cos t 0 x(t) Acos t 0
h l h d l ll h d f l
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A Pendulum Clock
What length pendulum will have a period of exactly 1s?
g
L
T 2 L
g
g T
2
2
L
L 9.8m/s2 1s
2
2
0.248m
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Conditions for SHM
Notice that all objects thatwe look at are described
the same mathematically.
Any system with a linear restoring
force will undergo simpleharmonic motion around theequilibrium position.
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A Physical Pendulum
d 2
dt 2 mgl
I
I mgd mglsin
when there ismass in the
entire pendulum,not just the bob.
Small Angle Approx.
mglI
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Damped Oscillations
All real oscillators are dampedoscillators. these are any that slow
down and eventually stop.a model of drag force for
slow objects:
Fdrag bv
b is the damping
constant (sort of like acoefficient of friction).
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Damped Oscillations
F Fs Fdrag kx bv ma
kx bdxdt
md 2xdt 2
0
Another 2 nd -order diff eq.
Solution to 2 nd -order diff eq:
x(t) Ae bt / 2m
cos t 0
k m
b 2
4m 2
02 b2
4m 2
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Damped Oscillations
x(t) Ae bt / 2m cos t 0
A slowly changing linethat provides a border to
a rapid oscillation is
called the envelope ofthe oscillations.
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http://www.youtube.com/watch?v=IqK2r5bPFTM&feature=related -
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DrivenOscillations
Not all oscillating objects are disturbed from restthen allowed to move undisturbed.
Some objects may be subjected to a periodicexternal force.
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DrivenOscillations
All objects have a natural frequency at whichthey tend to vibrate when disturbed.
Objects may be exposed to a periodic force witha particular driving frequency .
If the driven
frequency matchesthe naturalfrequency of an
object, RESONANCE occurs
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THE
END