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Mechanical Vibrations · Problem type

Natural frequency from static deflection

Use this for “A 50 kg machine sags 2 mm on its mounts. Find the mount stiffness, ωₙ and fₙ.”

Diagram

δst = 2 mm50 kgkδst = 490.5 Nmg = 490.5 N(0.002, 490.5)U = 0.4905 JF (N)δₛₜ (m)slope k = 245,300 N/m
F = mg = 50 × 9.81 = 490.5 N
k = F/δₛₜ = 490.5/0.002
k = 245,300 N/m
U
=
1
2
kδₛₜ²
=
1
2
×
245,300
×
0.002
2
U = 0.4905 J
U is the triangle under the line.
This is an example. Type your own values to start a new problem.
k (N/m)
×
δₛₜ (mm)
=
m (kg)
×
g
ωₙ (rad/s)
=
g
÷
δₛₜ (mm)
The picture’s line has slope k = F ÷ δₛₜ, with F = mg: that is the first line.
fₙ
Natural frequency in hertz
calculated · Hz
T
Period
calculated

Formulas

k
×
δₛₜ
=
m
×
9.81
245,250 × 0.002 = 50 × 9.81
ωₙ
=
√(
9.81
δₛₜ
)
70.0357
=
√(
9.81
0.002
)
fₙ
=
ωₙ
2π
11.1465
=
70.0357
2π
T
=
1
fₙ
0.08971
=
1
11.1465
Formulas are worked in m. The step-by-step shows the conversions.

Assumptions

•
The spring is linear and light; at rest it holds the weight: kδₛₜ = mg, with g = 9.81 m/s².
•
So k ÷ m = g ÷ δₛₜ, and ωₙ = √(g ÷ δₛₜ) for any mass.
•
No damping; the mass swings up and down about the sagged rest position.

Step-by-step

WE KNOW
m = 50 kg, δₛₜ = 2 mm
FIND
spring constant (k), natural frequency (ωₙ), natural frequency in hertz (fₙ), period (T)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m).
δₛₜ = 2 mm = 0.002 m (1 m = 1000 mm)
Step 2 · Find spring constant (k)
Use k × δₛₜ = m × 9.81
At rest the spring force kδₛₜ holds the weight mg: divide the weight by the sag.
k
=
m
×
9.81
δₛₜ
k
=
50
×
9.81
0.002
k
=
490.5
0.002
k
=
245,250
N/
m
Step 3 · Find natural frequency (ωₙ)
Use ωₙ = √(9.81 ÷ δₛₜ)
ωₙ
=
√(
k
m
),
and
kδₛₜ
=
mg
makes
k
m
=
g
δₛₜ
.
ωₙ
=
√(
9.81
0.002
)
ωₙ
=
√
4
,905
ωₙ = 70.0357 rad/s
Step 4 · Find natural frequency in hertz (fₙ)
Use fₙ = ωₙ ÷ 2π
One cycle is 2π radians, so divide the radians each second by 2π.
fₙ
=
70.0357
2π
fₙ = 11.1465 Hz
Step 5 · Find period (T)
Use T = 1 ÷ fₙ
The time for one cycle is one over the cycles each second.
T
=
1
11.1465
T
=
0.08971
s
Check
Put the numbers back in. Both sides should match.
245,250 × 0.002 = 50 × 9.81 ✓
70.0357
=
√(
9.81
0.002
)
✓
11.1465
=
70.0357
2π
✓
0.08971
=
1
11.1465
✓
Topic check
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