SME
Mechanical Vibrations · Problem type
Tuned vibration absorber
Use this for “A pump runs at 1,500 rpm (157.08 rad/s) and shakes with a 150 N force. Size the spring for a 3 kg absorber, and find how far the absorber swings.”
Diagram
kₐ
=
mₐ
ω
2
=
2
×
188.5
2
kₐ = 71,060 N/m
Xₐ
=
F₀
kₐ
=
200
71,060
=
0.002814
m
Tuned to the running speed, the absorber’s spring pulls on the machine with kₐ xₐ = −F₀ sin ωt at every moment: the two cancel and the machine stands still.
This is an example. Type your own values to start a new problem.
kₐ (N/m)
=
mₐ (kg)
×
(
ω (rad/s)
)
2
Xₐ (mm)
=
F₀ (N)
÷
kₐ (N/m)
Formulas
kₐ
=
mₐ
×
ω
2
71,061.1517
=
2
×
188.4956
2
Xₐ
=
F₀
kₐ
0.002814
=
200
71,061.1517
Formulas are worked in m. The step-by-step shows the conversions.
Assumptions
•
The machine is driven by F₀ sin ωt at one steady speed ω; no damping.
•
The absorber is tuned to that speed, kₐ ÷ mₐ = ω², so the machine itself stands still.
•
The absorber then swings so its spring pushes back on the machine with kₐXₐ = F₀.
Step-by-step
WE KNOW
ω = 188.4956 rad/s, mₐ = 2 kg, F₀ = 200 N
FIND
absorber spring (kₐ), absorber swing (Xₐ)
Step 1 · Find absorber spring (kₐ)
Use kₐ = mₐ × ω²
Tune
the
absorber:
its
own
natural
frequency
√(
kₐ
mₐ
)
must
equal
the
driving
ω
.
kₐ
=
2
×
188.49556
2
kₐ = 2 × 35,530
kₐ = 71,061.15 N/m
Step 2 · Find absorber swing (Xₐ)
Use Xₐ = F₀ ÷ kₐ
With
the
machine
still,
the
absorber
spring
alone
holds
back
the
force:
kₐXₐ
=
F₀
.
Xₐ
=
200
71,061.15
Xₐ = 0.002814 m
Step 3 · Convert the answers
Change each answer back to the unit you chose.
Xₐ = 0.002814 m = 2.8145 mm (1 m = 1000 mm)
Check
Put the numbers back in. Both sides should match.
71,061.15
=
2
×
188.49556
2
✓
0.002814
=
200
71,061.15
✓
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Multi-degree-of-freedom systems
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Two masses on one spring, free at both ends
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