SME
Mechanical Vibrations
Multi-degree-of-freedom systems
Use this for “A 2 kg cart and a 1 kg cart sit between walls on springs of 300 N/m, 200 N/m and 100 N/m. Find both natural frequencies and the mode shapes.”
Diagram
det(K − ω²M) = 0: ω₁² = 131, ω₂² = 419, so ω₁ = 11.5 and ω₂ = 20.5 rad/s
Mode shapes x₂/x₁ = 1.19 (slow) and −1.69 (fast)
Started alone, m₁ hands its motion to m₂ and takes it back: the two modes beat.
This is an example. Type your own values to start a new problem.
m₁
Mass 1
example · kg
m₂
Mass 2
example · kg
k₁
Left wall spring
example
k₂
Middle spring
example
k₃
Right wall spring
example
ω₁
Slow mode
worked out (not typed)
ω₂
Fast mode
worked out (not typed)
r₁
Slow mode shape x₂ ÷ x₁
worked out (not typed)
r₂
Fast mode shape x₂ ÷ x₁
worked out (not typed)
Formulas
ω₁
=
√((
m₁
×
(
k₂
+
k₃
)
+
m₂
×
(
k₁
+
k₂
)
−
√((
m₁
×
(
k₂
+
k₃
)
+
m₂
×
(
k₁
+
k₂
))²
−
4
×
m₁
×
m₂
×
((
k₁
+
k₂
)
×
(
k₂
+
k₃
)
−
k₂
2
)))
÷
(2
×
m₁
×
m₂
))
11.4624
=
√((2
×
(200
+
100)
+
1
×
(300
+
200)
−
√((2
×
(200
+
100)
+
1
×
(300
+
200))²
−
4
×
2
×
1
×
((300
+
200)
×
(200
+
100)
−
200
2
)))
÷
(2
×
2
×
1))
ω₂
=
√((
m₁
×
(
k₂
+
k₃
)
+
m₂
×
(
k₁
+
k₂
)
+
√((
m₁
×
(
k₂
+
k₃
)
+
m₂
×
(
k₁
+
k₂
))²
−
4
×
m₁
×
m₂
×
((
k₁
+
k₂
)
×
(
k₂
+
k₃
)
−
k₂
2
)))
÷
(2
×
m₁
×
m₂
))
20.4601
=
√((2
×
(200
+
100)
+
1
×
(300
+
200)
+
√((2
×
(200
+
100)
+
1
×
(300
+
200))²
−
4
×
2
×
1
×
((300
+
200)
×
(200
+
100)
−
200
2
)))
÷
(2
×
2
×
1))
r₁
=
k₁
+
k₂
−
m₁
×
ω₁
2
k₂
1.1861
=
300 + 200 − 2 ×
11.4624
2
200
r₂
=
k₁
+
k₂
−
m₁
×
ω₂
2
k₂
−1.6861
=
300 + 200 − 2 ×
20.4601
2
200
Assumptions
•
Small motions along the line, no friction or damping. Springs: k₁ wall to m₁, k₂ between, k₃ m₂ to wall.
•
K = [k₁ + k₂, −k₂; −k₂, k₂ + k₃] and M = [m₁, 0; 0, m₂]; det(K − ω²M) = 0 is a quadratic in ω².
•
The smaller root is the slow mode ω₁, the larger the fast mode ω₂.
•
Mode shape r = x₂ ÷ x₁: positive, the masses move together; negative, opposite.
Step-by-step
WE KNOW
m₁ = 2 kg, m₂ = 1 kg, k₁ = 300 N/m, k₂ = 200 N/m, k₃ = 100 N/m
FIND
slow mode (ω₁), fast mode (ω₂), slow mode shape x₂ ÷ x₁ (r₁), fast mode shape x₂ ÷ x₁ (r₂)
Step 1 · Find slow mode (ω₁)
Use ω₁ = √((m₁ × (k₂ + k₃) + m₂ × (k₁ + k₂) − √((m₁ × (k₂ + k₃) + m₂ × (k₁ + k₂))² − 4 × m₁ × m₂ × ((k₁ + k₂) × (k₂ + k₃) − k₂²))) ÷ (2 × m₁ × m₂))
Expand
det(K
−
ω
2
M)
=
0:
2ω⁴
−
1,100ω²
+
110,000
=
0,
a
quadratic
in
ω
2
.
Its
smaller
root
is
the
slow
mode.
ω₁
=
√((1,100
−
√
1,
100
2
− 4 × 2 × 110,000
)
÷
(2
×
2))
ω₁
=
11.4624
rad/s
Step 2 · Find fast mode (ω₂)
Use ω₂ = √((m₁ × (k₂ + k₃) + m₂ × (k₁ + k₂) + √((m₁ × (k₂ + k₃) + m₂ × (k₁ + k₂))² − 4 × m₁ × m₂ × ((k₁ + k₂) × (k₂ + k₃) − k₂²))) ÷ (2 × m₁ × m₂))
The larger root of the same quadratic, 2ω⁴ − 1,100ω² + 110,000 = 0, is the fast mode.
ω₂
=
√((1,100
+
√
1,
100
2
− 4 × 2 × 110,000
)
÷
(2
×
2))
ω₂
=
20.4601
rad/s
Step 3 · Find slow mode shape x₂ ÷ x₁ (r₁)
Use r₁ = (k₁ + k₂ − m₁ × ω₁²) ÷ k₂
The
first
row
of
(K
−
ω
2
M)x
=
0
gives
x₂
x₁
at
the
slow
mode:
positive,
so
the
masses
move
together.
1
×
11.4624
2
≥
100:
mass
2
moves
at
least
as
much:
use
the
first
row
r₁
=
300 + 200 − 2 ×
11.4624
2
200
r₁
=
300 + 200 − 2 × 131.4
200
r₁
=
300 + 200 − 262.8
200
r₁
=
237.2
200
r₁
=
1.1861
Step 4 · Find fast mode shape x₂ ÷ x₁ (r₂)
Use r₂ = (k₁ + k₂ − m₁ × ω₂²) ÷ k₂
The
first
row
of
(K
−
ω
2
M)x
=
0
gives
x₂
x₁
at
the
fast
mode:
negative,
so
the
masses
move
opposite.
1
×
20.4601
2
≤
2
×
200
+
100:
mass
2
moves
at
least
as
much:
use
the
first
row
r₂
=
300 + 200 − 2 ×
20.4601
2
200
r₂
=
300 + 200 − 2 × 418.6
200
r₂
=
300 + 200 − 837.2
200
r₂
=
−
337.2
200
r₂
=
−1.6861
Check
Put the numbers back in. Both sides should match.
11.4624
=
√((2
×
(200
+
100)
+
1
×
(300
+
200)
−
√((2
×
(200
+
100)
+
1
×
(300
+
200))²
−
4
×
2
×
1
×
((300
+
200)
×
(200
+
100)
−
200
2
)))
÷
(2
×
2
×
1))
✓
20.4601
=
√((2
×
(200
+
100)
+
1
×
(300
+
200)
+
√((2
×
(200
+
100)
+
1
×
(300
+
200))²
−
4
×
2
×
1
×
((300
+
200)
×
(200
+
100)
−
200
2
)))
÷
(2
×
2
×
1))
✓
1.1861
=
300 + 200 − 2 ×
11.4624
2
200
✓
−1.6861
=
300 + 200 − 2 ×
20.4601
2
200
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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