SME
Mechanical Vibrations
Forced vibration and resonance
Use this for “A 10 kg machine on a 4000 N/m mount with ζ = 0.1 is shaken by 100 N at 15 rad/s. Find the steady amplitude and the phase lag.”
Diagram
r
=
ω
ωₙ
=
15
20
=
0.75
X
=
(
F₀
k
)
÷
√(
(1 −
r
2
)
2
+
(2ζr)
2
)
X = 0.025 × 2.16 = 0.0541 m
tan
φ
=
2ζ
r
1 −
r
2
:
φ
=
18.9°
Near r = 1 the response peaks; far above it the block barely moves and lags by nearly 180°.
This is an example. Type your own values to start a new problem.
ωₙ (rad/s)
=
k (N/m)
÷
m (kg)
r
=
ω (rad/s)
÷
ωₙ (rad/s)
X (mm)
=
(
F₀ (N)
÷
k (N/m)
)
÷
(1
−
r
2
)
2
+
(2
×
×
)
2
φ
Phase lag
calculated
Formulas
ωₙ
=
√(
k
m
)
20
=
√(
4,000
10
)
r
=
ω
ωₙ
0.75
=
15
20
X
=
(
F₀
k
)
÷
√(
(1 −
r
2
)
2
+
(2 ×
ζ
×
r
)
2
)
0.05405
=
(
100
4,000
)
÷
√(
(1 −
0.75
2
)
2
+
(2 × 0.1 × 0.75)
2
)
φ
=
atan2(2
×
ζ
×
r
,
1
−
r
2
)
0.3303
=
atan2(2
×
0.1
×
0.75,
1
−
0.75
2
)
Formulas are worked in m, rad. The step-by-step shows the conversions.
Assumptions
•
Steady state under F₀ sin ωt: the start-up motion has died away, so the block swings at ω.
•
Viscous damping with ratio ζ, a linear spring and one degree of freedom.
•
The response lags the force by φ, from 0° (slow) through 90° at r = 1 to nearly 180°.
Step-by-step
WE KNOW
m = 10 kg, k = 4,000 N/m, ω = 15 rad/s, ζ = 0.1, F₀ = 100 N
FIND
natural frequency (ωₙ), frequency ratio (r), steady amplitude (X), phase lag (φ)
Step 1 · Find natural frequency (ωₙ)
Use ωₙ = √(k ÷ m)
How fast the mass rings on its own: the square root of stiffness over mass.
ωₙ
=
√(
4,000
10
)
ωₙ
=
√
400
ωₙ = 20 rad/s
Step 2 · Find frequency ratio (r)
Use r = ω ÷ ωₙ
The frequency ratio: how fast the force drives against how fast the mass rings.
r
=
15
20
r
=
0.75
Step 3 · Find steady amplitude (X)
Use X = (F₀ ÷ k) ÷ √((1 − r²)² + (2 × ζ × r)²)
The
static
stretch
F₀
k
,
magnified
near
r
=
1
where
the
root
is
smallest.
X
=
(
100
4,000
)
÷
√(
(1 −
0.75
2
)
2
+
(2 × 0.1 × 0.75)
2
)
X
=
100
4,000
÷
√(
(1 − 0.5625)
2
+
(0.2 × 0.75)
2
)
X
=
100
4,000
÷
√
0.4375
2
+
0.15
2
X
=
100
4,000
÷
√
0.1914 + 0.0225
X
=
100
4,000
÷
√
0.2139
X
=
100
4,000
÷
0.4625
X
=
0.025
0.4625
X
=
0.05405
m
Step 4 · Find phase lag (φ)
Use φ = atan2(2 × ζ × r, 1 − r²)
tan
φ
=
2
ζ
r
1 −
r
2
;
atan2
keeps
φ
past
90°
once
r
passes
1.
φ
=
atan2(2
×
0.1
×
0.75,
1
−
0.75
2
)
φ
=
0.3303
rad
Step 5 · Convert the answers
Change each answer back to the unit you chose.
X = 0.05405 m = 54.0541 mm (1 m = 1000 mm)
φ = 0.3303 rad = 18.9246° (180° = π rad)
Check
Put the numbers back in. Both sides should match.
20
=
√(
4,000
10
)
✓
0.75
=
15
20
✓
0.05405
=
(
100
4,000
)
÷
√(
(1 −
0.75
2
)
2
+
(2 × 0.1 × 0.75)
2
)
✓
0.3303
=
atan2(2
×
0.1
×
0.75,
1
−
0.75
2
)
✓
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Transmissibility: how much force a mount lets through
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