Practice Test
Simple harmonic motion
Solve. x(t) is the distance of an object O from a fixed point P. Find
x(t) in each case:
Math 216
Differential Equations
Kenneth Harris
kaharri@umich.edu
1
4x 00 + 24x 0 + 100x = 0.
2
4x 00 + 40x 0 + 100x = 0.
3
4x 00 + 48x 0 + 100x = 0
What is the long term behavior of x(t) in each case?
(a) O oscillates through P. (periodic motion)
Department of Mathematics
University of Michigan
(b) O slows to a halt without oscillating through P.
(c) O slows to a halt while oscillating through P. (pseudoperiodic
motion)
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(d) O drifts infinitely far from P.
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Review
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Review
Simple harmonic motion
Case 1
Case 1: x0 > 0 and v0 = 0
x(t) = x0 cos(ωt)
Simple harmonic motion of a mass m on a spring of elasticity k :
mx 00 + kx = 0,
x(0) = x0 , x 0 (0) = v0
The solution to this problem is
x(t) = x0 cos(ωt) +
v0
sin(ωt)
ω
Ω
C
O
Comp
where ω =
q
m
m
Str
k
m.
There are four cases when one of x0 = 0 or v0 = 0.
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Review
Review
Case 2
Case 3
Case 2: x0 < 0 and v0 = 0
Case 3: x0 = 0 and v0 > 0
x(t) = x0 cos(ωt) = −x0 cos(ωt − π)
x(t) =
v0
v0
π
sin(ωt) =
cos(ωt − )
ω
ω
2
C
C
O
O
Comp
m
Comp
Str
Str
Π
m
m
2
Ω
m
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Review
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Ω
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Simple Harmonic Oscillators: Putting it together
Case 4
Example
Case 4: x0 = 0 and v0 < 0
Problem. A 3 kg mass is attached to a spring of stiffness 48 N/m.
When the mass is stretched 12 m the velocity is 2 m/s away from the
equillibrium point. Find the equation for displacement.
v0
v0
3π
x(t) =
sin(ωt) = − cos(ωt −
)
ω
ω
2
Ω
m
Answer. The initial value problem is given by
x 00 + 16x = 0,
O
Comp
x(t) = A cos(4t) + B sin(4t).
Str
m
C
In this case, A = x0 =
1
2
and B =
x(t) =
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1 0
, x (0) = 2.
2
The general solution is
3Π
2
x(0) =
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v0
ω
= 12 ,
1
1
cos(4t) + sin(4t).
2
2
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Simple Harmonic Oscillators: Putting it together
Simple Harmonic Oscillators: Putting it together
Graph of solution
Circular motion
At time t = 0 the shadow mass m is π4 radians from P. The mass m
π
seconds to reach its maximal displacement at P
will take 16
π
The mass first reaches maximal stretch in 16
seconds (time lag), then
π
oscillates with period 2 seconds and amplitude √12 meters.
1
1
2
cosH4tL +
1
2
2
1
cosH4tL+ sinH4tL at t=0
2
sinH4tL
x
0.6
0.4
0.2
m
O
Q
1
2
3
4
5
6
-
t
Π
P
4
C
-0.2
Ω
-0.4
-0.6
m
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Conversion
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Conversion
Goal
Euler’s formula
Goal. Convert a periodic motion
The ♥ is Euler’s formula
ea+bi = ea cos b + i sin b
x(t) = A cos(ωt) + B sin(ωt)
for converting polar coordinates (reiθ )
to a circular motion
x(t) = C cos(ωt − α)
r = ea ,
where
θ=b
to rectangular coordinates (x + iy )
â C is the amplitude
â ω is angular velocity or circular frequency,
x = ea cos b,
â α is phase angle.
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y = ea sin b
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Conversion
Conversion
Trig identities
Converting to cos
eix + e−ix
eix − e−ix
sin x =
2
2i
Derivation of sin x using ♥ Euler’s formula:
eiωt + e−iωt eiωt − e−iωt A cos(ωt) + B sin(ωt) = A
+B
2
2i
eiωt + e−iωt eiωt − e−iωt = A
+ −iB
2
2
iωt
−iωt
e
e
+ (A + iB)
= (A − iB)
2
2
cos x =
eix − e−ix
2i
1
cos x + i sin x − cos(−x) + i sin(−x)
2i
1
=
cos x + i sin x − cos(x) − i sin(x)
2i
1
=
2i sin x
2i
= sin x
=
Convert A + iB and A − iB to polar coordinates:
A + iB = Ceiα
A − iB = Ce−iα .
We will compute C and α shortly.
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Conversion
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Conversion
Converting to cos
Computing amplitude to Phase angle
Recap,
A cos(ωt) + B sin(ωt) = C cos(ωt − α)
Let A + iB = Ceiα , so A − iB = Ce−iα .
A cos(ωt) + B sin(ωt) =
=
=
=
=
Kenneth Harris (Math 216)
where A + iB = Ceiα . So,
eiωt
e−iωt
(A − iB)
+ (A + iB)
2
2
−iωt
iωt
e
e
+ Ceiα
Ce−iα
2
2
i(ωt−α)
i(α−ωt)
e
e
C
+C
2
2
ei(ωt−α) + e−i(ωt−α) C
2
C cos(ωt − α)
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C=
p
A2 + B 2
α = arctan
|B|
.
|A|
(α is the angle A + iB makes with the x-axis.)
However, 0 ≤ α < π2 , so we must determine the quadrant of A + iB:


α
if A, B > 0



π − α
if A < 0, B > 0
α=

π+α
if A < 0, B < 0



2π − α if A > 0, B < 0
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Conversion
Back to the example
Determining quadrant
Putting the pieces together
8
α
>
>
>
<π − α
α=
>π + α
>
>
:
2π − α
if A, B > 0
if A < 0, B > 0
if A < 0, B < 0
if A > 0, B < 0
Simple harmonic motion with x0 = 12 m and v0 = 2m/s,
imaginary HBL
x(t) =
A+iB
A+iB
1
1
cos(4t) + sin(4t).
2
2
q
C = 14 + 14 = √1 and α = arctan(1) =
2
first quadrant). So,
Α
Α
Α
Α
real HAL
A+iB
where
â Amplitude is
√1
2
≈ 0.7 meters,
â Period is ≈ 1.6 seconds per oscillation,
â Frequency is π2 ≈ 0.64 oscillations per second,
π
â Time lag is 16
≈ 0.2 seconds (between start and maximum
displacement).
A+iB
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Back to the example
2
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Dampening
Graph of solution
1
(since x0 , v0 > 0, α is in the
π 1
π 1
x(t) = √ cos 4t − ) = √ cos 4(t −
) ,
4
16
2
2
π
2
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π
4
Forces in Mass-Spring Oscillators
cosH4tL +
1
2
sinH4tL =
1
cosH4t -
2
Π
4
L
There are three forces in mass-spring oscillators
1
x
1.0 Π
Π
16
2
The spring elasticity, which by Hooke’s law, is proportional to the
displacement x(t) in the opposite direction:
Fspring = −kx
1
2
2
0.5
Dampening forces, primarily friction, which is often proportional to
the velocity (but opposing the direction of motion):
Fdampening = −cx 0
1
2
3
4
5
t
6
3
External forces, such as gravity: Fext .
By Newton’s second law,
-0.5
-
1
mx 00 = −kx − cx 0 + Fext
2
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equivalently
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mx 00 + cx 0 + kx = Fext .
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Dampening
Critically damped motion
Damped Ocillatory Motion
Critical dampening
â Crically damped motion occurs when c 2 − 4mk = 0.
Section 3.4 considers damped oscillatory motion free of external forces
â There is only one real root:
mx 00 + cx 0 + kx = 0.
r=
(Section 3.6 considers motions with external forces.)
−c
2m
â The general solution to mx 00 + cx 0 + kx = 0 is
The characteristic equation is mr 2 + cr + k = 0 (m, c, k > 0), and the
roots are
−c
1 p 2
±
c − 4mk
r=
2m 2m
There are three cases, depending on whether the discriminant
c 2 − 4mk is positive, zero, or negative.
−c
x(t) = A + Bt e 2m t
â The mass eventually comes to a stop:
lim x(t) = lim
t→∞
t→∞
A + Bt
c
e 2m t
=0
(Use L’Hôpital’s rule to verify this.)
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Critically damped motion
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Critically damped motion
Critical dampening: non-oscillatory
Examples of critically damped motion
Equation: x 00 + 2x + 1 = 0, Solutions:
â The general solution to mx 00 + cx 0 + kx = 0 in the critical damping
case is
−c
x(t) = A + Bt e 2m t
x0 = .3, v0 = −1.3
(0.3 − t)e−t
x0 = .3, v0 = 1.3
(t − 0.3)e−t
x0 = .3, v0 = −0.15
(0.3 + 0.15)e−t
x
â The x 0 (t) vanishes at most one value of t (provided A or B is
nonzero)
−c
c
c
A−
Bt e 2m t
x 0 (t) = B −
2m
2m
â Therefore, the mass m does not oscillate.
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Overdamped Motion
Overdamped Motion
Overdamped motion
Overdampening: non-oscillatory
â Overdamped motion occurs when c 2 − 4mk > 0, so the effect of
dampening is large compared to the mass and spring elasticity.
â There are two distinct real roots:
−c
1 p 2
−c
1 p 2
r1 =
+
c − 4mk r2 =
−
c − 4mk
2m 2m
2m 2m
Since
c2
>
c2
â The general solution to mx 00 + cx 0 + kx = 0 in the overdamping
case is
x(t) = Aer1 t + Ber2 t
where r1 , r2 < 0.
â The x 0 (t) vanishes at most one value of t (provided A or B is
nonzero)
x 0 (t) = Ar1 er1 t + Br2 er2 t = er1 t Ar1 + Br2 e(r1 −r2 )t
− 4mk , both r1 and r2 are negative.
â The general solution to mx 00 + cx 0 + kx = 0 is
x(t) = Aer1 t + Ber2 t
â Therefore, the mass m does not oscillate.
where x(t) → 0 as t → ∞
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Underdamped motion
â The general solution to mx 00 + cx 0 + kx = 0 in the underdamping
case is
Ceρt cos ωt − α .
â There are complex conjugate roots ρ ± iω where
ρ=−
c
2m
ω=
1 p
4mk − c 2
2m
c
where ρ = − 2m
< 0.
â cos ωt − α oscillates between −Ceρt and Ceρt , so that x(t) → 0
as t → ∞.
â The general solution to mx 00 + cx 0 + kx = 0 is
â The motion is not actually periodic, since the amplitude diminishes
over time. However, motion periodically reverses:
c
Time-varying amplitude Ceρt = Ce− 2m t ,
√ 2π
pseudoperiod 2π
,
ω =
2
x(t) = eρt A cos(ωt) + B sin(ωt) .
â We can express x(t) in the alternative form
Ceρt cos ωt − α .
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Underdampening
â Underdamped motion occurs when c 2 − 4mk < 0. The
dampening is too weak to prevent oscillatory behavior.
√
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Underdamped motion
Underdampening
where C =
Math 216 Differential Equations
A2 + B 2 and tan α =
pseudofrequency
ω
2π .
4mk −c
B
A.
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Underdamped motion
Example of underdamped motion
Underdamping
x
-Ρt
Ce
2Π
Ω
t
-Ρt
-Ce
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