SME
Classical Mechanics · Problem type
Normal modes of a two-mass chain
Use this for “A wall holds a 300 N/m spring, then a 2 kg block, then a 150 N/m spring and a 1 kg block with its far end free. Find both normal-mode frequencies and each mode’s shape.”
Diagram
det(K − ω²M) = 0: ω₁² = 58.6, ω₂² = 341, so ω₁ = 7.65 and ω₂ = 18.5 rad/s
Mode shapes x₂/x₁ = 2.41 (slow) and −0.414 (fast)
This is an example. Type your own values to start a new problem.
+
−
ω²
−
−
−
ω²
=
0
(k
in
N/m,
m
in
kg)
ω₁
Slow-mode ω
worked out (not typed) · rad/s
ω₂
Fast-mode ω
worked out (not typed) · rad/s
r₁
Slow mode shape x₂ ÷ x₁
worked out (not typed)
r₂
Fast mode shape x₂ ÷ x₁
worked out (not typed)
Formulas
ω₁
=
√((
m₂
×
(
k₁
+
k₂
)
+
m₁
×
k₂
−
√((
m₂
×
(
k₁
+
k₂
)
+
m₁
×
k₂
)²
−
4
×
m₁
×
m₂
×
k₁
×
k₂
))
÷
(2
×
m₁
×
m₂
))
7.6537 = √((1 × (200 + 100) + 1 × 100 − √((1 × (200 + 100) + 1 × 100)² − 4 × 1 × 1 × 200 × 100)) ÷ (2 × 1 × 1))
ω₂
=
√((
m₂
×
(
k₁
+
k₂
)
+
m₁
×
k₂
+
√((
m₂
×
(
k₁
+
k₂
)
+
m₁
×
k₂
)²
−
4
×
m₁
×
m₂
×
k₁
×
k₂
))
÷
(2
×
m₁
×
m₂
))
18.4776 = √((1 × (200 + 100) + 1 × 100 + √((1 × (200 + 100) + 1 × 100)² − 4 × 1 × 1 × 200 × 100)) ÷ (2 × 1 × 1))
r₁
=
k₁
+
k₂
−
m₁
×
ω₁
2
k₂
2.4142
=
200 + 100 − 1 ×
7.6537
2
100
r₂
=
k₂
k₂
−
m₂
×
ω₂
2
−0.4142
=
100
100 − 1 ×
18.4776
2
Assumptions
•
Small motions along the line, no friction; the far end is free. K = [k₁ + k₂, −k₂; −k₂, k₂] and M = [m₁, 0; 0, m₂].
•
det(K − ω²M) = 0 is m₁m₂ω⁴ − (m₂(k₁ + k₂) + m₁k₂)ω² + k₁k₂ = 0: the smaller root in ω² is the slow mode, the larger the fast.
•
Each mode’s shape is its amplitude ratio r = x₂ ÷ x₁ = (k₁ + k₂ − m₁ω²) ÷ k₂: positive, the blocks move together; negative, opposite.
Step-by-step
WE KNOW
m₁ = 1 kg, m₂ = 1 kg, k₁ = 200 N/m, k₂ = 100 N/m
FIND
slow-mode ω (ω₁), fast-mode ω (ω₂), slow mode shape x₂ ÷ x₁ (r₁), fast mode shape x₂ ÷ x₁ (r₂)
Step 1 · Find slow-mode ω (ω₁)
Use ω₁ = √((m₂ × (k₁ + k₂) + m₁ × k₂ − √((m₂ × (k₁ + k₂) + m₁ × k₂)² − 4 × m₁ × m₂ × k₁ × k₂)) ÷ (2 × m₁ × m₂))
det(K
−
ω
2
M)
=
0
is
m₁
m₂
ω
4
−
(
m₂
(
k₁
+
k₂
)
+
m₁
k₂
)
ω
2
+
k₁
k₂
=
0,
a
quadratic
in
ω
2
;
the
smaller
root
is
the
slow
mode.
ω₁
=
√((400
−
√
400
2
− 4 × 1 × 20,000
)
÷
(2
×
1))
ω
4
−
400ω²
+
20,000
=
0
ω₁
=
7.6537
rad/s
Step 2 · Find fast-mode ω (ω₂)
Use ω₂ = √((m₂ × (k₁ + k₂) + m₁ × k₂ + √((m₂ × (k₁ + k₂) + m₁ × k₂)² − 4 × m₁ × m₂ × k₁ × k₂)) ÷ (2 × m₁ × m₂))
The larger root of the same quadratic is the fast mode.
ω₂
=
√((400
+
√
400
2
− 4 × 1 × 20,000
)
÷
(2
×
1))
ω₂
=
18.4776
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
blocks
move
together.
r₁
=
200 + 100 − 1 ×
7.6537
2
100
r₁
=
200 + 100 − 58.58
100
r₁
=
241.42
100
r₁
=
2.4142
Step 4 · Find fast mode shape x₂ ÷ x₁ (r₂)
Use r₂ = k₂ ÷ (k₂ − m₂ × ω₂²)
Use
the
second
row
of
(K
−
ω
2
M)x
=
0;
it
gives
the
same
ratio
and
avoids
a
near-cancellation.
Negative,
so
the
blocks
move
opposite.
1
×
18.4776
2
≥
2
×
100:
mass
2
moves
less,
so
the
second
row
r₂
=
100
100 − 1 ×
18.4776
2
r₂
=
100
100 − 341.42
r₂
=
100
−241.42
r₂
=
−0.4142
Check
Put the numbers back in. Both sides should match.
7.6537 = √((1 × (200 + 100) + 1 × 100 − √((1 × (200 + 100) + 1 × 100)² − 4 × 1 × 1 × 200 × 100)) ÷ (2 × 1 × 1)) ✓
18.4776 = √((1 × (200 + 100) + 1 × 100 + √((1 × (200 + 100) + 1 × 100)² − 4 × 1 × 1 × 200 × 100)) ÷ (2 × 1 × 1)) ✓
2.4142
=
200 + 100 − 1 ×
7.6537
2
100
✓
−0.4142
=
100
100 − 1 ×
18.4776
2
✓
Topic check
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