SME
Mechanical Vibrations · Problem type
Springs in series and in parallel
Use this for “Springs of 3000 and 6000 N/m hold a 5 kg block. Find keq and ωₙ with the springs in parallel, then in series.”
Diagram
kₚ = k₁ + k₂ = 3,000 + 6,000
kₚ = 9,000 N/m
In parallel both stretch the same and their forces add: stiffer than either.
kₛ
=
k₁
k₂
k₁ + k₂
=
3,000
×
6,000
9,000
kₛ = 2,000 N/m
In series both carry the same force and their stretches add: softer than either.
Natural frequency in parallel ωₚ = 42.4264 rad/s
Natural frequency in series ωₛ = 20 rad/s
This is an example. Type your own values to start a new problem.
kₚ (N/m)
=
k₁ (N/m)
+
k₂ (N/m)
kₛ (N/m)
=
k₁ (N/m)
×
k₂ (N/m)
÷
(
k₁ (N/m)
+
k₂ (N/m)
)
m
Mass
example
ωₚ
Natural frequency in parallel
calculated · rad/s
ωₛ
Natural frequency in series
calculated · rad/s
Formulas
kₚ
=
k₁
+
k₂
9,000 = 3,000 + 6,000
kₛ
=
k₁
×
k₂
k₁
+
k₂
2,000
=
3,000
×
6,000
3,000 + 6,000
ωₚ
=
√(
kₚ
m
)
42.4264
=
√(
9,000
5
)
ωₛ
=
√(
kₛ
m
)
20
=
√(
2,000
5
)
Assumptions
•
In parallel both springs stretch the same and their forces add: kₚ = k₁ + k₂.
•
In series both carry the same force and their stretches add: 1 ÷ kₛ = 1 ÷ k₁ + 1 ÷ k₂.
•
Light linear springs and no damping, so each ωₙ = √(keq ÷ m).
Step-by-step
WE KNOW
k₁ = 3,000 N/m, k₂ = 6,000 N/m, m = 5 kg
FIND
parallel stiffness (kₚ), series stiffness (kₛ), natural frequency in parallel (ωₚ), natural frequency in series (ωₛ)
Step 1 · Find parallel stiffness (kₚ)
Use kₚ = k₁ + k₂
Side by side, both stretch x, so the forces k₁x and k₂x add.
kₚ = 3,000 + 6,000
kₚ
=
9,000
N/
m
Step 2 · Find series stiffness (kₛ)
Use kₛ = k₁ × k₂ ÷ (k₁ + k₂)
End
to
end,
the
stretches
F
k₁
and
F
k₂
add;
that
makes
the
product
over
the
sum.
kₛ
=
3,000
×
6,000
3,000 + 6,000
kₛ
=
3,000
×
6,000
9,000
kₛ
=
18,000,000
9,000
kₛ
=
2,000
N/
m
Step 3 · Find natural frequency in parallel (ωₚ)
Use ωₚ = √(kₚ ÷ m)
The pair acts as one spring kₚ on the mass.
ωₚ
=
√(
9,000
5
)
ωₚ
=
√
1
,800
ωₚ = 42.4264 rad/s
Step 4 · Find natural frequency in series (ωₛ)
Use ωₛ = √(kₛ ÷ m)
The pair acts as one softer spring kₛ on the mass.
ωₛ
=
√(
2,000
5
)
ωₛ
=
√
400
ωₛ = 20 rad/s
Check
Put the numbers back in. Both sides should match.
9,000 = 3,000 + 6,000 ✓
2,000
=
3,000
×
6,000
3,000 + 6,000
✓
42.4264
=
√(
9,000
5
)
✓
20
=
√(
2,000
5
)
✓
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Natural frequency from static deflection
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A compound pendulum: a rod swinging from its end
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