spring SHM
F  -kx
k
a x 0
m
a  ω2 x  0
x  A sin ( ωt  δ)
v  A ω cos ( ωt  δ)
a = -A ω 2 sin ( ωt + δ)
Pendulum
g
ω
L
11-0
A=xmax
v max = A ω
amax = A ω2
k
ω
m
2π
ω=2πf =
T
1
f
T
Most mechanical systems- displace from equilibrium- 2 possibilities
restoring force 1st approx F=-kx
oscillates about equilibrium
with natural frequency
 potential energy 1st
approx
U=1/2 kx2
 repulsive force - runs away
Small displacement Simple Harmonic Motion Dynamics
Examples: Spring, Pendulum…
3 Points to note
 hit (tickle) system – it “rings” at a natural frequency (fo)
 friction makes natural ring die of exponentially with time
 vibrate system with frequency (f)- it responds with f but close
to (or at) fo get sharp, big “resonant” response
11-1
Harmonic Motion
Cyclic Transfer Between Potential Energy and Kinetic Energy
Kinetic Energy
-xmax x = 0
velocity
 Potential Energy
position
+vmax
+xmax
Galileo
-clock
-vmax
-xmax
T  period, time required for one complete
cycle [seconds/cycle = s]
Frequency, f
f =
1 [ Hz = cycle/s]
T
11-2
Spring: Simple Harmonic Motion Dynamics (Newton & Hook were bitter rivals)
F
Equilibrium (no stretch/compression)
x=0
F  -kx
compressed
F always acts toward x=0
F
stretched
F
more stretched
Newton says
F  ma
11-1
ma  -kx
k
a x 0
m
k
ω
m
simple harmonic motion
a  ω2 x  0
x  A sin ( ωt  δ)
v  A ω cos ( ωt  δ)
ω2πf
angular freq.
(rad./sec.)
Hooks Law
frequency
(Hz = sec.-1)
1
f
T
period (sec.)
a   A ω 2 sin ( ωt  δ)
11-3
Spring: Simple Harmonic Motion
x = A sin( ωt + δ)
a = -A ω2 sin(ωt + δ)
v = A ω cos(ωt + δ)
11-4
uniform circular motion- simple harmonic motion analogy
 = t
vx = v cos()
x
v
v ωA
a
a = v2/A
A
 = t
a = 2/A
x  A sin ( ωt )
11-5
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=148
cyclic transfer :: kinetic energy  potential energy
A
1
1
2
2
kA  mv max
2
2
v max
k
A
m
Energy
A
1
2
x = 0 E  mv max
2
1
E

kA 2
x= A
2
x= -A E  1 kA 2
2
1
1 2
2
E  mv  kx
2
2
11-6
Example Problem: Mass on a Spring
If m=1.0kg , k= 100 N/m , and vmax = 0.5m/s then
k
100N / m
rad


 10
m
1kg
sec
Energy Conservation:
T
a max =Aω2 =(0.05m)(10
2
1
 0.628sec;f f = 11=
.6Hz
1.6Hz
T

T
1
1 2
2
mv max = kx max
2
2
v max 0.5 m/s
x max =A=
=
= 0.05m=A
ω 10 rad/s
find A, T, f, xmax, amax
x 2max
x max 
mv 2max

k
m
v max
v max 
k

rad 2 m
) =5 2
s
s
rad
t)
Also if x=0 at t=0 then
sec
rad
t)
But if x=+A at t=0 then x(t)=A cos(ωt)=A cos(10
sec
x(t)=A sin(ωt)=A sin(10
11-7
another approach to same problem
m = 1.0kg k= 100N/m
maximum speed vmax = 0.5m/s.
find the A, T, f, xmax, amax
SHM Problem
k
100N / m
rad


 10
m
1kg
sec
2
1
T
 0.628sec;f   1.6Hz

T
rad
x(t )  A sin(t )  A sin(10
t)
sec
0.5m / s
vmax  A  A 
 0.05m
10rad / s
rad 2
m
2
amax  A  (0.05m)(10
) 5 2
s
s
11-7a
Problem/experiment
x  A sin ( ωt  δ)
A = 0.2 m
Find everything
given
t=0
v=0
x=xmax
x=0.2m
x=0
m=0.25 kg
given
T=2.2 s
given
11-8
http://phet.colorado.edu/sims/mass-spring-lab/mass-spring-lab_en.html
Problem/experiment
x  A sin ( ωt  δ)
A = 0.2 m
f=
1
1
=
= 0.45 Hz
T 2.2 s
ω = 2 π f = 2 (3.14) .45 = 2.85
Find everything
given
t=0
v=0
x=xmax
x=0.2m
rad
s
x=0
m=0.25 kg
given
T=2.2 s
given
11-8a
http://phet.colorado.edu/new/admin/gen-flashpage.php?flash=http://phet.colorado.edu/sims/massspring-lab/mass-spring-lab.swf&title=Masses & Springs
Problem/experiment
x  A sin ( ωt  δ)
A = 0.2 m
f=
1
1
=
= 0.45 Hz
T 2.2 s
ω = 2 π f = 2 (3.14) .45 = 2.85
Find everything
given
t=0
v=0
x=xmax
x=0.2m
rad
s
x=0
k
2
ω =  k =mω
m
k = 0.25 [kg] (2.85)2 [1/s 2 ] = 8.12 [kg/s 2 ]
2
m=0.25 kg
given
k = 8.12 [(mkg)/(s 2 )][1/m]
k = 8.12 [N/m]
T=2.2 s
given
11-8b
http://phet.colorado.edu/new/admin/gen-flashpage.php?flash=http://phet.colorado.edu/sims/massspring-lab/mass-spring-lab.swf&title=Masses & Springs
Problem/experiment
x  A sin ( ωt  δ)
A = 0.2 m
f=
1
1
=
= 0.45 Hz
T 2.2 s
ω = 2 π f = 2 (3.14) .45 = 2.85
Find everything
given
t=0
v=0
x=xmax
x=0.2m
rad
s
x=0
m=0.25 kg
given
vmax = A ω = 0.2 (2.85) [m/s]
v max = 0.57 m/s
T=2.2 s
given
11-8c
http://phet.colorado.edu/new/admin/gen-flashpage.php?flash=http://phet.colorado.edu/sims/massspring-lab/mass-spring-lab.swf&title=Masses & Springs
Problem/experiment
x  A sin ( ωt  δ)
A = 0.2 m
f=
1
1
=
= 0.45 Hz
T 2.2 s
ω = 2 π f = 2 (3.14) .45 = 2.85
Find everything
given
t=0
v=0
x=xmax
x=0.2m
rad
s
x=0
m=0.25 kg
given
amax = A ω2 = 0.2 (2.85)2 [m/s2 ]
amax = 1.62 m/s2
T=2.2 s
given
11-8d
http://phet.colorado.edu/new/admin/gen-flashpage.php?flash=http://phet.colorado.edu/sims/massspring-lab/mass-spring-lab.swf&title=Masses & Springs
Problem/experiment
x  A sin ( ωt  δ)
A = 0.2 m
f=
1
1
=
= 0.45 Hz
T 2.2 s
ω = 2 π f = 2 (3.14) .45 = 2.85
Find everything
given
t=0
v=0
x=xmax
x=0.2m
rad
s
x=0
k
2
ω =  k =mω
m
k = 0.25 [kg] (2.85)2 [1/s 2 ] = 8.12 [kg/s 2 ]
2
m=0.25 kg
given
k = 8.12 [(mkg)/(s 2 )][1/m]
k = 8.12 [N/m]
vmax = A ω = 0.2 (2.85) [m/s]
v max = 0.57 m/s
amax = A ω = 0.2 (2.85) [m/s ]
amax = 1.62 m/s2
2
2
2
T=2.2 s
given
11-8sum
http://phet.colorado.edu/en/simulation/
mass-spring-lab
Problem/experiment
203 NOT REQUIRED
δ=?
x  A sin ( ωt  δ)
Find everything
given
t=0
x=xmax=A
v=0
x=0.2m
a= is -
t=0
x = A sin ( 0 + δ) = A
sin (δ) = 1
δ=

x=0
m=0.25 kg
given
2
T=2.2 s
given
11-8
Extra
Role of friction --- damped oscillations
• F=-kx-bv
(bv is a frictional force and causes dampening)
E~ ½ k A2
A  A 0e

bt
2m
friction leads to loss of mechanical energy
11-9
http://phet.colorado.edu/en/simulation/mass-spring-lab
Driven harmonic motion – resonance
• F=-kx-bv +Fext sin(2 f t)
friction
Resonant frequency 2  f0 = k/m
driving frequency f
11-10
Simulations of approach to – passage through resonance
http://www.walter-fendt.de/ph14e/resonance.htm
11-10a
http://www.physics.rutgers.edu/~croft/tacnarr.mpg
Tacoma Narrows Bridge Collapse
11-10b
Pendulum

-F = ma
T
L
F
x
xL
F= mg sin()
-mg sin() = ma
  sin() small angle app
-g = a
 mg
A  L θ max
g
a+ x  0 ω  g
L
L
x  A sin ( ωt )
11-11
simple
harmonic
motion
An equivalent rotation approach
Pendulum

I = mL2
T
F= mg sin()
F
L
x
xL
 mg
=I
-FL = (mL2) 
-mg sin() L = (mL2) 
-g sin() = L 
full soln.
g
Euler
α  sin( θ )  0
L
Integrals
  sin() small angle app.
α
g
θ0
L
θ  θmax sin ( ωt )
11-11a
ω
simple
harmonic
g motion
L
Appendix I : check of solution with differential equation (not needed 203)
Spring: Simple Harmonic – x – v – a – check of motion eq.
x  A sin ( ωt  δ)
v
dx
d[sin ( ωt  δ)]
A
 A ω cos ( ωt  δ)
dt
dt
v  A ω cos ( ωt  δ)
a
a   A ω 2 sin ( ωt  δ)
dv
d[cos ( ωt  δ)]
 Aω
  A ω 2 sin ( ωt  δ)
dt
dt
a  ω2 x  0
d2x
2

ω
x0
2
dt
 A ω sin ( ωt  δ)  ω A sin ( ωt  δ)  0
2
2
11-12
Appendix II : determination of phase factor (not needed 203)
OR
x  A cos ( ωt  φ)
(1) x  A sin ( ωt  φ)
amplitude
phase
determine by initial conditions
(2) v  ω A cos ( ωt  φ)  v ocos ( ωt  φ)
example 1
t = 0 xo=0.05 m vo=0
0.05 m
A = 0.05
(1)  0.05 = A sin ()  0.05 = A sin (± /2)
 = + /2
(2)  0 = A cos ()  0 = cos ()   = ± /2
x = 0.05 sin ( t + /2)
example 2 t = 0 xo=0.03 m vo= 0.2 m/s
Let = 5 rad/s
(1)  0.03 = A sin ()
0.03 = A sin ()  tan () = 3/4
(2)  0.2 = A 5 cos ()
0.2 = A 5 cos ()
 = tan-1(.75)
0.03 = A sin (.64) = A .6
 = .64 rad
A = 0.05
11-13