SME
Classical Mechanics
Coupled oscillations
Use this for “Two 0.5 kg carts sit between walls on 200 N/m springs and are joined by a 20 N/m spring. Find both normal-mode frequencies, and how long the motion takes to pass to the other cart and back.”
Diagram
det(K − ω²M) = 0: ω₁² = 100, ω₂² = 121, so ω₁ = 10 and ω₂ = 11 rad/s
Mode shapes x₂/x₁ = 1 (slow) and −1 (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.
m (kg)
(
ω₁ (rad/s)
)
2
=
k (N/m)
m
(
ω₂ (rad/s)
)
2
=
k
+
2×
k′ (N/m)
f₁
Slow mode frequency
calculated · Hz
f₂
Fast mode frequency
calculated · Hz
Tₑₓ
Energy-exchange period
worked out (not typed)
Formulas
ω₁
=
√(
k
m
)
10
=
√(
100
1
)
ω₂
=
√(
k
+ 2 ×
k
′
m
)
11
=
√(
100 + 2 × 10.5
1
)
f₁
=
ω₁
2 × π
1.5915
=
10
2 × π
f₂
=
ω₂
2 × π
1.7507
=
11
2 × π
Tₑₓ
=
2
×
π
ω₂
−
ω₁
6.2832
=
2
×
π
11 − 10
Assumptions
•
Small motions along the line, no friction. Newton’s law for each mass gives mẍ = −Kx with K = [k + k′, −k′; −k′, k + k′].
•
det(K − mω²) = 0 gives the two normal modes: in step, shape (1, 1), mω₁² = k; opposite, shape (1, −1), mω₂² = k + 2k′.
•
Any motion is a mix of the two modes. Start one mass alone and the modes drift out of step and back: its motion passes to the other mass and returns once every Tₑₓ.
Step-by-step
WE KNOW
m = 1 kg, k = 100 N/m, k′ = 10.5 N/m
FIND
slow-mode ω (in step) (ω₁), fast-mode ω (opposite) (ω₂), slow mode frequency (f₁), fast mode frequency (f₂), energy-exchange period (Tₑₓ)
Step 1 · Find slow-mode ω (in step): ω₁
Use ω₁ = √(k ÷ m)
In
step,
x₁
=
x₂:
the
coupling
spring
keeps
its
length,
so
each
mass
feels
only
its
outer
spring
k
.
ω₁
=
√(
100
1
)
ω₁
=
10
rad/s
Step 2 · Find fast-mode ω (opposite): ω₂
Use ω₂ = √((k + 2 × k′) ÷ m)
Opposite,
x₂
=
−x₁:
the
coupling
spring
stretches
from
both
ends,
so
each
mass
feels
k
plus
2
k
′.
ω₂
=
√(
100 + 2 × 10.5
1
)
ω₂
=
√
100 + 21
ω₂
=
√
121
ω₂
=
11
rad/s
Step 3 · Find slow mode frequency (f₁)
Use f₁ = ω₁ ÷ (2 × π)
Cycles each second: one cycle is 2π radians.
f₁
=
10
2 × π
f₁
=
1.5915
Hz
Step 4 · Find fast mode frequency (f₂)
Use f₂ = ω₂ ÷ (2 × π)
Cycles each second: one cycle is 2π radians.
f₂
=
11
2 × π
f₂
=
1.7507
Hz
Step 5 · Find energy-exchange period (Tₑₓ)
Use Tₑₓ = 2 × π ÷ (ω₂ − ω₁)
The
two
modes
slip
one
full
turn
apart,
and
back
into
step,
once
every
2π
ω₂
−
ω₁
:
the
beat.
Tₑₓ
=
2
×
π
11 − 10
Tₑₓ = 6.2832 s
Check
Put the numbers back in. Both sides should match.
10
=
√(
100
1
)
✓
11
=
√(
100 + 2 × 10.5
1
)
✓
1.5915
=
10
2 × π
✓
1.7507
=
11
2 × π
✓
6.2832
=
2
×
π
11 − 10
✓
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The Coriolis acceleration and the Foucault pendulum
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Normal modes of a two-mass chain
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