SME
Differential Equations
Second-order linear ODEs
Use this for “A 1 kg mass on a 10 N/m spring with damping 2 N·s/m starts 1 m out, moving outward at 2 m/s. Where is it after 1 s?”
Diagram
ωₙ
=
√(
k
m
)
=
√(
10
1
)
=
3.16
rad/s
ζ
=
c
2
√
km
ζ
=
2
2
√
10 × 1
=
0.316
ζ
<
1,
underdamped:
ωd
=
ωₙ
√
1 −
ζ
2
=
3
rad/s,
less
than
ωₙ;
the
crests
shrink
inside
±X
exp(−ζωₙt).
This is an example. Type your own values to start a new problem.
(
m (kg)
)y″
+
(
c (kg/s)
)y′
+
(
k (N/m)
)y
=
0
y(0)
=
y₀ (m)
,
y′(0)
=
v₀ (m/s)
y(
t (s)
)
=
y(t) (m)
Δ
Discriminant c² − 4mk
worked out (not typed)
regime
Damping regime
worked out (not typed)
α
Real part of the roots
worked out (not typed) · s⁻¹
β
Offset of the roots from α
worked out (not typed) · s⁻¹
Formulas
Δ
=
c
2
−
4
×
m
×
k
−36
=
2
2
−
4
×
1
×
10
regime
=
undamped
if
c
=
0,
underdamped
if
Δ
<
0,
critically
damped
if
Δ
=
0,
else
overdamped
−36 < 0 and 2 > 0, so underdamped
α
=
−
c
2 ×
m
−1
=
−
2
2 × 1
β
=
(
1
2 ×
m
)
×
√
|
Δ
|
3
=
(
1
2 × 1
)
×
√
|−36|
y(
t
)
=
e
α
×
t
×
(
y₀
×
cos(
β
×
t
)
+
v₀
−
α
×
y₀
β
×
sin(
β
×
t
))
−0.3123
=
e
−1 × 1
×
(1
×
cos(3
×
1)
+
2 − (−1) × 1
3
×
sin(3
×
1))
Assumptions
•
A linear spring and damper: my″ + cy′ + ky = 0. Trying y = ert gives mr² + cr + k = 0, with roots α ± β or α ± βi.
•
c² < 4mk oscillates (underdamped), c² = 4mk is critically damped, and c² > 4mk creeps back without swinging (overdamped).
•
C₁ and C₂ are fitted to the start: y(0) = y₀ and y′(0) = v₀. Angles are in radians.
•
c is in kg/s, the same unit as N·s/m; time is in s, so α and β are per second.
Step-by-step
WE KNOW
m = 1 kg, c = 2 kg/s, k = 10 N/m, y₀ = 1 m, v₀ = 2 m/s, t = 1 s
FIND
discriminant c² − 4mk (Δ), damping regime (regime), real part of the roots (α), offset of the roots from α (β), position at t (y(t))
Step 1 · Find discriminant c² − 4mk (Δ)
Use Δ = c² − 4 × m × k
Put
y
=
e
rt
into
the
equation:
e
rt
is
never
0,
so
mr²
+
cr
+
k
=
0.
Its
discriminant
sorts
the
motion.
r
2
+
2r
+
10
=
0
Δ
=
2
2
−
4
×
1
×
10
=
−36
Δ
=
−36
Step 2 · Find damping regime (regime)
Use regime = undamped if c = 0, underdamped if Δ < 0, critically damped if Δ = 0, else overdamped
Δ
<
0:
the
roots
are
complex,
so
the
mass
swings
back
and
forth
inside
a
shrinking
envelope.
−36 < 0 and 2 > 0: underdamped
regime = undamped if 2 = 0, underdamped if −36 < 0, critically damped if −36 = 0, else overdamped
regime = underdamped
Step 3 · Find real part of the roots (α)
Use α = −c ÷ (2 × m)
The
quadratic
formula
r
=
−
c
± √
Δ
2
m
splits
into
−
c
2
m
,
the
same
in
every
regime,
and
the
±
part.
α
=
−
2
2 × 1
α
=
−1
s⁻¹
Step 4 · Find offset of the roots from α (β)
Use β = (1 ÷ (2 × m)) × √(|Δ|)
With
Δ
<
0
the
square
root
is
imaginary:
the
roots
are
α
±
βi,
and
β
is
how
fast
the
mass
swings
(rad/s).
β
=
(
1
2 ×
m
)
×
√
−
Δ
β
=
(
1
2 × 1
)
×
√(−(−36))
r
=
−2 ±
√
−36
2
=
−1
±
3i
β
=
3
s⁻¹
Step 5 · Find position at t (y(t))
Use y(t) = eα × t × (y₀ × cos(β × t) + (v₀ − α × y₀) ÷ β × sin(β × t))
Underdamped:
y
=
e
αt
(C₁
cos
βt
+
C₂
sin
βt).
At
t
=
0
it
gives
C₁
=
y₀
;
its
slope
there
gives
C₂
=
v₀
−
αy₀
β
.
−36 < 0: underdamped
y″ + 2y′ + 10y = 0
r = −1 ± 3i
C₁
=
y₀
=
1;
C₂
=
v₀
−
αy₀
β
=
2 − (−1) × 1
3
=
1
y(
t
)
=
e
−
t
(cos(
3
t
)
+
sin(
3
t
))
y(1)
=
e
-1
(cos(3)
+
sin(3))
=
0.36787944
×
(−0.84887249)
=
−0.3123
y(
t
)
=
−0.3123
m
Check
Put the numbers back in. Both sides should match.
−36
=
2
2
−
4
×
1
×
10
✓
−36 < 0 and 2 > 0, so underdamped ✓
−1
=
−
2
2 × 1
✓
3
=
(
1
2 × 1
)
×
√
|−36|
✓
−0.3123
=
e
−1 × 1
×
(1
×
cos(3
×
1)
+
2 − (−1) × 1
3
×
sin(3
×
1))
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.