SME
Mechanical Vibrations · Problem type
Two masses on one spring, free at both ends
Use this for “A 5 kg glider and a 2 kg glider on an air track are joined by a 400 N/m spring. Find the frequency they vibrate at and the mode shape.”
Diagram
det(K
−
ω
2
M)
=
0:
ω₁
=
0,
the
pair
sliding
as
one,
and
ω₂
=
√(
k(m₁ + m₂)
m₁m₂
)
=
22.4
rad/s
x₂/x₁
=
−
m₁
m₂
=
−0.667:
the
center
of
mass
stays
put
This is an example. Type your own values to start a new problem.
(
ω (rad/s)
)
2
=
k (N/m)
×
(
m₁ (kg)
+
m₂ (kg)
)
÷
(
m₁ (kg)
×
m₂ (kg)
)
r
=
−
m₁ (kg)
÷
m₂ (kg)
The first line is ω = √(k(m₁ + m₂) ÷ (m₁m₂)) squared; the second is the mode shape, r = x₂ ÷ x₁.
f
Frequency in hertz
calculated · Hz
Formulas
ω
=
√(
k
×
m₁
+
m₂
m₁
×
m₂
)
22.3607
=
√(600
×
2 + 3
2 × 3
)
f
=
ω
2π
3.5588
=
22.3607
2π
r
=
−
m₁
m₂
−0.6667
=
−
2
3
Assumptions
•
Two masses joined by one spring k, with no walls and no friction holding them.
•
One root of det(K − ω²M) = 0 is ω = 0: the pair slides as one and the spring never stretches.
•
In the other mode the center of mass stays put, m₁x₁ + m₂x₂ = 0, so r = x₂ ÷ x₁ = −m₁ ÷ m₂.
Step-by-step
WE KNOW
m₁ = 2 kg, m₂ = 3 kg, k = 600 N/m
FIND
natural frequency (ω), frequency in hertz (f), mode shape x₂ ÷ x₁ (r)
Step 1 · Find natural frequency (ω)
Use ω = √(k × (m₁ + m₂) ÷ (m₁ × m₂))
The
nonzero
root
of
det(K
−
ω
2
M)
=
0:
the
spring
against
the
reduced
mass
m₁
m₂
m₁
+
m₂
.
ω
=
√(600
×
2 + 3
2 × 3
)
ω
=
√(600
×
5
6
)
ω
=
√(
3,000
6
)
ω
=
√
500
ω
=
22.3607
rad/s
Step 2 · Find frequency in hertz (f)
Use f = ω ÷ 2π
Cycles each second: one cycle is 2π radians.
f
=
22.3607
2π
f
=
3.5588
Hz
Step 3 · Find mode shape x₂ ÷ x₁ (r)
Use r = −m₁ ÷ m₂
The center of mass stays put, m₁x₁ + m₂x₂ = 0: the lighter mass swings farther, the other way.
r
=
−
2
3
r
=
−0.6667
Check
Put the numbers back in. Both sides should match.
22.3607
=
√(600
×
2 + 3
2 × 3
)
✓
3.5588
=
22.3607
2π
✓
−0.6667
=
−
2
3
✓
Topic check
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