Mechanical and Electrical Oscillations (§ 2.5)
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Mechanical Oscillations
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Finding the Spring Constant
General Mechanical Oscillations
The RLC Electrical Circuit
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General Electrical Oscillations
Remark:
Different physical systems may be mathematically identical.
Mechanical and Electrical Oscillations (§ 2.5)
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Mechanical Oscillations
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I
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Finding the Spring Constant
General Mechanical Oscillations
The RLC Electrical Circuit
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General Electrical Oscillations
Remark:
Different physical systems may be mathematically identical.
Mechanical Oscillations
A spring with rest length l and a
mass m hangs from a ceiling.
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(l + ∆l) is called equilibrium
position of the spring loaded
with a mass m.
The coordinate y measures
vertical deviations from the
equilibrium position.
0
∆l
m
y (t)
y
Forces acting on the system:
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Weight: Fg = mg .
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Spring: Fs = −k(∆l + y ), with k is the spring constant.
(Hooke’s Law, accurate for small oscillations.)
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Damping: Fd (t) = −d y 0 (t). Fluid Resistance.
Newton’s Law: m y 00 (t) = Fg + Fs (t) + Fd (t).
Mechanical and Electrical Oscillations (§ 2.5)
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Mechanical Oscillations
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Finding the Spring Constant
General Mechanical Oscillations
The RLC Electrical Circuit
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General Electrical Oscillations
m
Finding the Spring Constant
Remark: The equilibrium displacement ∆l, the spring constant k,
the mass m, and the acceleration of gravity g , are all related.
Theorem (Spring Constant)
A spring-weight system with spring constant k, body mass m, at
rest with spring deformation ∆l on the Earth surface satisfies
mg = k ∆l.
Proof: Newton’s law says m y 00 (t) = Fg + Fs (t) + Fd (t).
Recall: Fg = mg , Fs = −k(∆l + y ), Fd (t) = −d y 0 (t). That is,
m y 00 (t) = mg − k(∆l + y (t)) − d y 0 (t).
When the spring is at rest, this is at y = 0, with y 0 = 0, and
y 00 = 0. Therefore, mg = k ∆l.
Finding the Spring Constant
Example (Finding a Spring Constant)
Find the spring constant of a spring that deforms 10 cm when a
mass of 500 gr is attached. Use that g = 1000 cm/sec2 .
Solution: The Spring at Rest Theorem says that
mg = k ∆l
⇒
k=
mg
.
∆l
In our case ∆l = 10 cm and m = 500 gr. Then
k=
(10)(1000)
500
⇒
k = 20
gr
.
sec2
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Mechanical and Electrical Oscillations (§ 2.5)
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Mechanical Oscillations
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Finding the Spring Constant
General Mechanical Oscillations
The RLC Electrical Circuit
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General Electrical Oscillations
General Mechanical Oscillations
Theorem (Mechanical Oscillations)
The vertical displacement y of a spring-weight with spring constant
k > 0, body mass m > 0, and damping constant d > 0, satisfies
m y 00 (t) + d y 0 (t) + k y (t) = 0.
(1)
The roots of the characteristic polynomial of Eq. (1) are
r
q
d
k
r± = −ωd ± ωd2 − ω02 , ωd =
, ω0 =
,
2m
m
where ωd and ω0 are the damping and natural frequencies. Then
(a) If ωd > ω0 , then ygen = c+ e r+ t + c- e r- t , (overdamped).
(b) If ωd = ω0 , then ygen = (c+ + c- t) e −ω0 t , (critically damped).
dt
(c) If ωd < ω0 , then ygen = A e −ω
q cos(βt − φ), (underdamped),
with system frequency β = ω02 − ωd2 , amplitude A > 0, and
phase φ ∈ (−π, π]. The case ωd = 0 is undamped.
General Mechanical Oscillations
Proof: Recall: Fg = mg , Fs = −k(∆l + y ), Fd (t) = −d y 0 . So
m y 00 = Fg + Fs + Fd .
⇒
m y 00 = mg − k(∆l + y ) − d y 0 .
Since k ∆l = mg , we get
m y 00 = −k y − d y 0
⇒
m y 00 + d y 0 + k y = 0.
To solve for the function y , we need the characteristic equation
p
1 2
−d ± d 2 − 4mk .
m r + d r + k = 0 ⇒ r± =
2m
r
r
2
d
k
d
d
k
, ω0 =
,
r± = −
±
− , ωd =
2m
m
2m
2m
m
q
so we get r± = −ωd ± ωd2 − ω02 .
General Mechanical Oscillations
Recall:
m y 00
+d
y0
q
+ k y = 0, and r± = −ωd ± ωd2 − ω02 .
From § 2.3 we know many things, such as
(a) Overdamped: ωd > ω0 . Two distinct real roots r± :
ygen (t) = c1 e r+ t + c2 e r- t .
(b) Critically Damped: ωd = ω0 . Repeated real root r+ = r- = ω0 :
ygen (t) = (c1 + c2 t) e ω0 t .
(c) Underdamped:
q ωd < ω0 . Complex roots r± = −ωd ± iβ,
where β = ω02 − ωd2 , so the general solution is
ygen (t) = c+ cos(βt) + c- sin(βt) e −ωd t .
General Mechanical Oscillations
Recall the case (c), underdamped:
y (t) = c+ cos(βt) + c- sin(βt) e −ωd t ,
q
β = ω02 − ωd2 .
The next result shows that the solution can be written as
y (t) = A e −ωd t cos(βt − φ),
with
A > 0,
φ ∈ (−π, π].
Theorem (Amplitude-Phase Transformation)
For every c+ , c- , β ∈ R, A > 0, and φ ∈ (−π, π] holds
c+ cos(βt) + c- sin(βt) = A cos(βt − φ),
with c+ = A cos(φ) and c- = A sin(φ).
Proof: cos(βt + (−φ)) = cos(βt) cos(−φ) − sin(βt) sin(−φ),
A cos(βt − φ) = [A cos(φ)] cos(βt) + [A sin(φ)] sin(βt).
Hence c+ = A cos(φ) and c- = A sin(φ).
General Mechanical Oscillations
Example (First Spring)
Find the movement of a 5kg mass attached to a spring with
constant k = 5kg/secs2 moving in a medium with damping
constant d = 5kg/secs, with initial conditions
√
y (0) = 3 and y 0 (0) = 0.
Solution: The equation is: my 00 + dy 0 + ky = 0, with m = 5,
k = 5, d = 5. The characteristic roots are
r
q
d
1
k
r± = −ωd ± ωd2 − ω02 , ωd =
= , ω0 =
= 1.
2m
2
m
r
√
1
1
1
3
r± = − ±
−1=− ±i
. Underdamped oscillations.
2
4
2
2
√3
y (t) = A cos
t − φ e −t/2 .
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General Mechanical Oscillations
Example (First Spring)
Find the movement of a 5kg mass attached to a spring with
constant k = 5kg/secs2 moving in a medium with damping
constant d = 5kg/secs, with initial conditions
√
y (0) = 3 and y 0 (0) = 0.
√3
Solution: Recall: y (t) = A cos
t − φ e −t/2 . We now find the
2
amplitude A and the phase φ from the initial conditions. Hence,
√
√3
√3
3
1
0
−t/2
y (t) = −
A sin
t −φ e
− A cos
t − φ e −t/2 .
2
2
2
2
√
√
3 = y (0) = A cos(φ) ⇒ A cos(φ) = 3.
√
3
1
0 = y 0 (0) =
A sin(φ) − A cos(φ).
2
2
√
√
3
1
3
A sin(φ) = A cos(φ) =
⇒ A sin(φ) = 1.
2
2
2
General Mechanical Oscillations
Example (First Spring)
Find the movement of a 5kg mass attached to a spring with
constant k = 5kg/secs2 moving in a medium with damping
constant d = 5kg/secs, with initial conditions
√
y (0) = 3 and y 0 (0) = 0.
√3
Solution: Recall: y (t) = A cos
t − φ e −t/2 , with
2
√
A cos(φ) = 3, and A sin(φ) = 1. (cos(φ) > 0, sin(φ) > 0.)
Therefore, A2 cos2 (φ) + A2 sin2 (φ) = A2 = 3 + 1, hence A = 2.
sin(φ)
1
π
π
5π
tan(φ) =
= √ . So either φ = or φ = − π = − .
cos(φ)
6
6
6
3
π
Since cos(φ) > 0, sin(φ) > 0, then φ = .
6
√3
π −t/2
Then we conclude y (t) = 2 cos
t−
e
.
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General Mechanical Oscillations
Example (Second Spring)
Find the movement of a 5kg mass attached to a spring with
constant k = 5kg/secs2 moving in a medium with damping
constant d = 5kg/secs, with initial conditions
√
y (0) = − 3 and y 0 (0) = 0.
√3
Solution: Recall: y (t) = A cos
t − φ e −t/2 . We now find the
2
amplitude A and the phase φ from the initial conditions. Hence,
√
√3
√3
3
1
0
−t/2
y (t) = −
A sin
t −φ e
− A cos
t − φ e −t/2 .
2
2
2 √
√2
− 3 = y (0) = A cos(φ) ⇒ A cos(φ) = − 3.
√
3
1
0 = y 0 (0) =
A sin(φ) − A cos(φ).
2
2
√
√
3
1
3
A sin(φ) = A cos(φ) = −
⇒ A sin(φ) = −1.
2
2
2
General Mechanical Oscillations
Example (Second Spring)
Find the movement of a 5kg mass attached to a spring with
constant k = 5kg/secs2 moving in a medium with damping
constant d = 5kg/secs, with initial conditions
√
y (0) = − 3 and y 0 (0) = 0.
√3
Solution: Recall: y (t) = A cos
t − φ e −t/2 , with
2
√
A cos(φ) = − 3, and A sin(φ) = −1. (cos(φ) < 0, sin(φ) < 0.)
Therefore, A2 cos2 (φ) + A2 sin2 (φ) = A2 = 3 + 1, hence A = 2.
sin(φ)
1
π
π
5π
tan(φ) =
= √ . So either φ = or φ = − π = − .
cos(φ)
6
6
6
3
5π
Since cos(φ) < 0, sin(φ) < 0, then φ = − .
6
√3
5π −t/2
Then we conclude y (t) = 2 cos
t+
e
.
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Mechanical and Electrical Oscillations (§ 2.5)
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Mechanical Oscillations
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Finding the Spring Constant
General Mechanical Oscillations
The RLC Electrical Circuit
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General Electrical Oscillations
The RLC Electrical Circuit
Consider an electric circuit with
resistance R, non-zero capacitor
C , and non-zero inductance L, as
in the figure.
R
C
L
I (t) = electric current
Kirchhoff’s Law: The electric current flowing in the circuit satisfies:
Z t
1
L I 0 (t) + R I (t) +
I (s) ds = 0.
C t0
Derivate both sides above: L I 00 (t) + R I 0 (t) +
00
Divide by L: I (t) + 2
Introduce ωd =
R 2L
I 0 (t) +
1
I (t) = 0.
C
1
I (t) = 0.
LC
R
1
and ω0 = √ , then I 00 + 2ωd I 0 + ω02 I = 0.
2L
LC
Mechanical and Electrical Oscillations (§ 2.5)
I
Mechanical Oscillations
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I
Finding the Spring Constant
General Mechanical Oscillations
The RLC Electrical Circuit
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General Electrical Oscillations
General Electrical Oscillations
Theorem (RLC Circuit)
The electric current function I in an RLC circuit with resistance
R > 0, capacitance C > 0, and inductance L > 0, satisfies
I 00 + 2ωd I 0 + ω02 I = 0,
ωd =
R
,
2L
1
ω0 = √ ,
LC
where ωd is the damping frequency and ω0 is the natural frequency
of the circuit. Furthermore, the results for the spring-weight system
parts (a)-(c), hold with the roots of the characteristic polynomial
q
r± = −ωd ± ωd2 − ω02 .
General Electrical Oscillations
Example (RLC Circuit)
Find real-valued fundamental solutions to I 00 + 2ωd I 0 + ω02 I = 0,
where ωd = R/(2L), ω02 = 1/(LC ), in the cases (a) (b) below.
Solution: The characteristic polynomial is p(r ) = r 2 + 2ωd r + ω02 .
The roots are:
q
q
1
2
r± = −2ωd ± 4ωd − 4ω02
⇒ r± = −ωd ± ωd2 − ω02 .
2
Case (a) R = 0. This implies ωd = 0, so r± = ±iω0 . Therefore,
I+ (t) = cos(ω0 t),
I- (t) = sin(ω0 t).
Remark: When the circuit has no resistance, the current oscillates
without dissipation. Undamped solution.
General Electrical Oscillations
Example (RLC Circuit)
Find real-valued fundamental solutions to I 00 + 2ωd I 0 + ω02 I = 0,
where ωd = R/(2L), ω02 = 1/(LC ), in the cases (a) (b) below.
q
Solution: Recall: r± = −ωd ± ωd2 − ω02 .
p
Case (b) R < 4L/C . This implies
4L
R <
C
2
⇔
R2
1
<
4L2
LC
⇔
ωd2 < ω02 .
q
Therefore, r± = −ωd ± i ω02 − ωd2 . The fundamental solutions are
I+ (t) = e −ωd t cos(βt),
q
with β = ω02 − ωd2 .
I- (t) = e −ωd t sin(βt).
General Electrical Oscillations
Example (RLC Circuit)
Find real-valued fundamental solutions to I 00 + 2ωd I 0 + ω02 I = 0,
where ωd = R/(2L), ω02 = 1/(LC ), in the cases (a) (b) below.
q
Solution: Given β = ω02 − ωd2 , the fundamental solutions are
I+ (t) = e −ωd t cos(βt),
I+
I- (t) = e −ωd t sin(βt).
I-
e −ωd t
t
−e −ωd t
The resistance R damps the
current oscillations. R is in
R
ωd =
, and ωd is the real
2L
part of the roots
q
r± = −ωd ± i ω02 − ωd2 .