SME
Control Systems · Problem type
Mass–spring–damper model
Use this for “Find ωₙ, ζ and the DC gain of G(s) = 1/(2s² + 8s + 50) for a 2 kg mass.”
Diagram
ωₙ
=
√(
k
m
)
=
√(
50
2
)
=
5
rad/s
ζ
=
b
2
√
km
ζ
=
8
2
√
50 × 2
=
0.4
ζ
<
1,
underdamped:
ωd
=
ωₙ
√
1 −
ζ
2
=
4.58
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.
G(s)
=
1
s²
+
s
+
ωₙ
Natural frequency
calculated · rad/s
ζ
Damping ratio
calculated
K
DC gain
calculated
Formulas
ωₙ
=
√(
k
m
)
5
=
√(
50
2
)
ζ
=
b
2 ×
√
k
×
m
0.4
=
8
2 ×
√
50 × 2
K
=
1
k
0.02
=
1
50
Assumptions
•
G(s) = X(s) ÷ F(s) = 1 ÷ (ms² + bs + k) stands for mx″ + bx′ + kx = f, starting from rest.
•
Matching ms² + bs + k to m(s² + 2ζωₙ s + ωₙ²) gives ωₙ and ζ; below ζ = 1 the mass rings.
•
The damping b is in kg/s (the same as N·s/m), and the DC gain 1 ÷ k is in metres per newton.
Step-by-step
WE KNOW
m = 2 kg, b = 8 N·s/m, k = 50 N/m
FIND
natural frequency (ωₙ), damping ratio (ζ), DC gain (K)
Step 1 · Find natural frequency (ωₙ)
Use ωₙ = √(k ÷ m)
Divide
the
denominator
by
m
:
its
constant
term
k
m
is
ωₙ
2
.
ωₙ
=
√(
50
2
)
ωₙ
=
√
25
ωₙ = 5 rad/s
Step 2 · Find damping ratio (ζ)
Use ζ = b ÷ (2 × √(k × m))
Match
the
s
terms:
b
m
=
2ζωₙ,
so
ζ
=
b
2mωₙ
=
b
2
√
km
.
ζ
=
8
2 ×
√
50 × 2
ζ
=
8
2 ×
√
100
ζ
=
8
2 × 10
ζ
=
8
20
ζ
=
0.4
Step 3 · Find DC gain (K)
Use K = 1 ÷ k
Put s = 0: the mass and damper drop out, and only the spring holds the force.
K
=
1
50
K
=
0.02
m/N
Check
Put the numbers back in. Both sides should match.
5
=
√(
50
2
)
✓
0.4
=
8
2 ×
√
50 × 2
✓
0.02
=
1
50
✓
Topic check
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