Download Oscillations and Periodic Motion

Oscillations and Periodic
Motion
24 June 2002
Dr. Gary Grim
Overview
• Observations & Definitions
• Simple Harmonic Motion (SHM)
• SHM Systems
– Mass & Spring
– Pendulum Systems
• Damped & Forced Oscillations
Observations & Definitions
• Equilibrium
– Natural position of rest when oscillations are not
present.
• Restoring force
– A force which always acts to return the system to its
equilibrium position.
• Amplitude
– Maximum displacement from equilibrium.
• Cycle
– One complete oscillation, i.e., a cycle is complete
when the position and velocity return to their initial
values.
Observations & Definitions
• Period
– Time to complete one cycle. (Inverse of
frequency.)
• Frequency
– Number of cycles per unit time. (Inverse of
period.)
• Hertz
– Unit of frequency: 1 Hz = 1 cycle/second.
Uniform Circular Motion
y
Initial conditions
X Displacement:
Y Displacement:
Speed =
R cos(φ )
R sin(φ )
vo
Projection onto X Axis
Amplitude = R
Period:
2πR
T =
vo
Angular velocity?
2π
2π
vo
ω≡
= 2πf =
=
2πR R
T
vo
yo
vo
R
φ
xo
x
Uniform Circular Motion
y
X & Y Motion
x (t ) = R cos(ωt + φ )
y (t ) = R cos(ω t + φ )
yo
vo
R
φ
xo
x
Observations & Definitions
• Angular Frequency (Velocity)
– Number of radians per unit time.
• Relationship to cycle frequency.
ω = 2πf
Simple Harmonic Motion
• SHM occurs when the restoring force is
proportional to the displacement.
• Ideal Spring (SHM Prototype)
– Hookes Law
F = −kx
2
d x
m 2 = −kx
dt
Simple Harmonic Motion
• SHM Equation of Motion:
2
d x
m 2 + kx = 0
dt
• Solution:
x (t ) = A cos( ωt + φ )
k
ω =
m
2
Simple Harmonic Motion
• Phase Angle:
ωt + φ
• Frequency:
ω
1
f =
=
2π 2π
• Period:
1
m
T = = 2π
f
k
k
m
Simple Harmonic Motion
• Initial Position:
xo = A cos(φ )
• Initial Velocity:
vo = −ωA sin(φ )
• Initial Phase:
− vo
φ = arctan(
)
ωxo
2
• Initial Amplitude: A = xo 2 + vo
ω2
Simple Harmonic Motion
• Total Energy (T+U) :
1 2 1 2
E = mv + kx
2
2
1 2
= kA
2
SHM Systems
• 1-D Systems
– Mass & Spring Horizontal
– Mass & Spring Vertical
• Same as horizontal, but with equilibrium point
shifted by gravity, such that: k∆x = mg
– Pendulum Systems
• Simple
• Physical
• Torsional
Example: P13-10
•
A 2.00 kg frictionless block is attached to an ideal spring with force constant
300 N/m. At t=0, the spring is neither stretched nor compressed and the
block is moving in the negative direction at 12.0 m/s. Find:
–
–
–
a) Amplitude
b) Phase
c) Write the equation for the position as a function of time.
300 N
k
m = 12.2rad / s
ω=
=
m
2.00 Kg
2
2
a) A = xo 2 + vo = 02 + 12 = 0.98m
2
2
ω
b)
φ = arctan(
12. 2
− vo
12
π
) = arctan( −
) = arctan( ∞) = ±
ω xo
12 .2 * 0
2
v o = −ωA sin(φ ) ⇒ φ > 0,
c)
π
x(t ) = 0. 98cos(12.2t + )
2
φ=
π
2
The Simple Pendulum
• Point mass suspended by a massless,
inextensible cable.
F = − mg sin θ
x
F ≈ − mgθ = − mg = kx
L
k
g
ω=
=
m
L
The Physical Pendulum
• A finite sized body, oscillating about a
frictionless, fixed point, within the body.
τ = −mgd sin θ ≈ −mgd θ
d 2θ
− mgdθ = Iα = I 2
dt
mgd
ω=
I
The Torsional Pendulum
• Rotating mass with torque proportional to
angular displacement, e.g. a watch spring.
Iα = −κθ
d 2θ
κ
=− θ
2
dt
I
κ
ω=
I
θ = Θ cos(ω t + φ )
Example P13-38
•
We want to support a thin hoop by a horizontal nail and have the hoop
make one complete small-angle oscillation every 2.0s What must the hoop
radius be?
A physical pendulum problem which means the angular frequency is given by:
ω=
mgd
,
I
where, the moment of inertia is taken about the edge of the ring, and the
distance from the pivot to the cg, is the radius R. From the parallel axis
theorem, the moment of inertia of the hoop about the nail
I = MR2 + MR2 = 2MR2, so
2π
mgR
=
⇒
2
T
2MR
2
2
1 T 
1  2 .0 
R= 
 g= 
 9.8 = 0 .497 m
2  2π 
2  6.28 
Damped Oscillations
• SHM:
d 2x
m 2 + kx = 0
dt
2
• Add velocity term:
d x
dx
m 2 + b + kx = 0
dt
dt
• Solution:
x (t ) = Ae
ω' =
− bt 2 m
cos(ω ' t + φ )
k  b 
−

m  2m 
2
Damped Oscillations
• Underdamped:
b < 2 km
• Critically Damped:
b = 2 km
• Overdamped:
b > 2 km
Oscillation Curves:
2.5
2
1.5
Displacment (m)
1
0.5
0
0
0.5
1
1.5
2
2.5
-0.5
-1
-1.5
Undamped
-2
Underdamped
Envelope
-2.5
Time (s)
Critically Dampled
Overdamped
3
Forced Oscillations & Resonance
• Unforced oscillations occur at the natural
frequency of the oscillator
• Forced oscillations have the period of the
driving force.
F (t ) = F cos(ω t )
max
d
• Amplitude of Forced Oscillations:
Fmax
A=
2 2
2
2
(k − mω d ) + b ω d
Forced Oscillations & Resonance
• Resonance:
k − mω d = 0,
2
k
ωd =
m