02-08-2022

Solving the equations of motion

Recall: 𝑑2 𝜃 𝑔
+ 𝜃=0 EOM for a simple pendulum
𝑑𝑡 2 𝐿

𝑑2𝑥
𝑚 2 + 𝑘𝑥 = 0 EOM for a spring-mass system
𝑑𝑡

Note the similarity in the form of these equations.

These are second order “ordinary differential equations (ODE)” in time.

Discrete mechanical systems have governing equations in the form of

ODE.

Solving the equations of motion

Equation of motion (EOM) for a spring-mass system:
𝑑2𝑥
𝑚 2 + 𝑘𝑥 = 0
𝑑𝑡
The solution can be found by assuming 𝑥 𝑡 = 𝐶𝑒 𝑠𝑡
where C and s are constants (TBD).
Substituting this form into EOM gives:
𝐶 𝑚𝑠 2 + 𝑘 = 0

C cannot be 0 because that will be a trivial solution. Therefore, we have:

𝑚𝑠 2 + 𝑘 = 0 Characteristic Equation
for a spring-mass system

Roots are: 𝑘
𝑠 = ±𝑗𝜔𝑛 where 𝑗 = −1 and 𝜔𝑛 = 𝑚
Here, 𝝎𝒏 is the natural / resonance / eigen frequency of the system.

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Solving the equations of motion

Two solutions: Recall:
𝑠 = 𝑗𝜔𝑛 and 𝑠 = − 𝑗𝜔𝑛 𝑥 𝑡 = 𝐶𝑒 𝑠𝑡

What is the significance of complex (vs. real) roots?

The general solution to EOM can therefore be expressed as:

𝑥 𝑡 = 𝐶1 𝑒 𝑗𝜔𝑛 𝑡 + 𝐶2 𝑒 −𝑗𝜔𝑛 𝑡
OR
𝑥 𝑡 = 𝐴1 cos 𝜔𝑛 𝑡 + 𝐴2 sin 𝜔𝑛 𝑡 ……… (1)
Here A1 and A2 are new constants.
C1, C2 or A1, A2 can be determined from the initial conditions.

Solving the equations of motion

Initial Conditions:

Second order ODE will have two initial conditions. Here, they are:

𝑥 𝑡 = 0 = 𝑥0 …… Initial displacement
𝑥ሶ 𝑡 = 0 = 𝑥ሶ 0 …… Initial velocity

Substituting into 𝑥 𝑡 = 𝐴1 cos 𝜔𝑛 𝑡 + 𝐴2 sin 𝜔𝑛 𝑡

gives:
𝐴1 = 𝑥0
𝑥 𝑡 = 0 = 𝐴1 = 𝑥0
𝑥ሶ 0
𝑥ሶ 𝑡 = 0 = 𝜔𝑛 𝐴2 = 𝑥ሶ 0 𝐴2 =
𝜔𝑛

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Solving the equations of motion

The solution is therefore given by:

𝑥ሶ 0
𝑥 𝑡 = 𝑥0 cos 𝜔𝑛 𝑡 + sin 𝜔𝑛 𝑡
𝜔𝑛
Free vibrations of a spring-mass system

This solution is a harmonic function of time:

The motion is sinusoidal in time and demonstrates a single

resonant frequency.

So, the spring-mass system is also called a “harmonic oscillator”

Form 1: 𝑥 𝑡 = 𝐶1 𝑒 𝑗𝜔𝑛 𝑡 + 𝐶2 𝑒 −𝑗𝜔𝑛 𝑡

= 𝐶1 (cos 𝜔𝑛 𝑡 + 𝑗 sin 𝜔𝑛 𝑡) + 𝐶2 (cos 𝜔𝑛 𝑡 − 𝑗 sin 𝜔𝑛 𝑡)
= (𝐶1 +𝐶2 ) cos 𝜔𝑛 𝑡 + 𝑗(𝐶1 −𝐶2 ) sin 𝜔𝑛 𝑡

Form 2: 𝑥 𝑡 = 𝐴1 cos 𝜔𝑛 𝑡 + 𝐴2 sin 𝜔𝑛 𝑡

𝐴1 = (𝐶1 + 𝐶2 ), 𝐴2 = 𝑗(𝐶1 −𝐶2 )

Form 3: 𝑥 𝑡 = 𝐴 cos 𝜔𝑛 𝑡 − 𝜑 = 𝐴 cos 𝜔𝑛 𝑡 cos 𝜑 + sin 𝜔𝑛 𝑡 sin 𝜑

𝐴1 = 𝐴 cos 𝜑, 𝐴2 = 𝐴 sin 𝜑

𝐴= 𝐴12 + 𝐴22
𝐴2
𝜑 = tan−1
𝐴1

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Review: Basic terminology

x(t )

0 t

2


x(t): Vibration response
Cycle: One revolution, as shown in RED
τ: Time period = time to complete one cycle
ω : Radian or angular frequency = 2πf, where ‘f’ is the frequency
of oscillation; Frequency: Number of cycles per unit time

Oscillatory free response

The black mass is un-damped and the blue mass is underdamped.

(damper not shown in figure)

This animation is from:

https://www.acs.psu.edu/drussell/Demos/SHO/damp.html

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Free vibrations of a spring-mass-damper system

FREE BODY DIAGRAM

𝑘 𝑐 𝑘𝑥 𝑐𝑥ሶ
𝑐 : viscous
𝑚 𝑚 damping
coefficient
𝑥
𝑥

Recall: 𝑥 is measured from the static equilibrium position of mass 𝑚.

Using Newton’s second law of motion and the Free-Body-Diagram:
𝑚𝑥ሷ = −𝑘𝑥 − 𝑐𝑥ሶ
“Equation of motion”
Or, 𝑚𝑥ሷ + 𝑐 𝑥ሶ + 𝑘𝑥 = 0
for the system

Free vibrations of a spring-mass-damper system

The solution can be found by assuming 𝑥 𝑡 = 𝐶𝑒 𝑠𝑡 where

𝐶 and 𝑠 are constants (TBD).

Substituting this form into EOM gives:

𝐶 𝑚𝑠 2 + 𝑐𝑠 + 𝑘 = 0
𝐶 cannot be 0 because that will be a trivial solution.
Therefore, we have:
𝑚𝑠 2 + 𝑐𝑠 + 𝑘 = 0

Characteristic Equation for a spring-mass-damper system

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Free vibrations of a spring-mass-damper system

𝑚𝑠 2 + 𝑐𝑠 + 𝑘 = 0
The two roots of this equation are:

−𝑐 ± 𝑐 2 − 4𝑚𝑘 𝑐 𝑐 2 𝑘
𝑠1,2 = =− ± −
2𝑚 2𝑚 2𝑚 𝑚

Thus, the general solution to the EOM is

𝑥 𝑡 = 𝐶1 𝑒 𝑠1𝑡 + 𝐶2 𝑒 𝑠2𝑡
Here 𝐶1 and 𝐶2 are arbitrary constants TBD from the
initial conditions.
When will this solution be oscillatory?

Three cases for the roots

𝑐 𝑐 2 𝑘
𝑠1,2 =− ± −
2𝑚 2𝑚 𝑚
𝑐 2 𝑘
Case 1: Roots are real and distinct >
2𝑚 𝑚
𝑐 2 𝑘
Case 2: Roots are real and equal =
2𝑚 𝑚
𝑐 2 𝑘
Case 3: Roots are complex conjugates <
2𝑚 𝑚

When will this solution be oscillatory?

Only for Case 3

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Three cases for the roots

So, only for one case, the solution is oscillatory.

𝑐 2 𝑘
A change in the parameter relative to causes
2𝑚 𝑚
qualitative change in the response.

The value of 𝑐 for this qualitative change is given by

𝑐𝑐 = 4 𝑚𝑘 = 2𝑚𝜔𝑛

Critical Damping and the Damping ratio

The critical damping is defined as the value of 𝑐
for which the radical in 𝑠 becomes zero:

𝑐𝑐 = 4 𝑚𝑘 = 2𝑚𝜔𝑛

The damping ratio for any damped system is then defined as:
𝑐 𝑐
𝜁= =
𝑐𝑐 2𝑚𝜔𝑛

For 𝑐 = 𝑐𝑐, 𝜁 = 1
For 𝑐 < 𝑐𝑐, 𝜁 < 1
For 𝑐 > 𝑐𝑐, 𝜁 > 1

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Critical Damping and the Damping ratio

𝑐 𝑐 2 𝑘 𝑐
𝑠1,2 =− ± − and 𝜁=
2𝑚 2𝑚 𝑚 2𝑚𝜔𝑛

Thus, the roots can be written as:

𝑠1,2 = −𝜁 ± 𝜁 2 − 1 𝜔𝑛

Two unknowns TBD

The solution can be written as: from two initial conditions

−𝜁+ 𝜁 2 −1 𝜔𝑛 𝑡 −𝜁− 𝜁 2 −1 𝜔𝑛 𝑡
𝑥 𝑡 = 𝐶1 𝑒 + 𝐶2 𝑒

Case 1 / 3: Underdamped system

𝑠1 = −𝜁 + 𝑗 1 − 𝜁 2 𝜔𝑛 , 𝑠2 = −𝜁 − 𝑗 1 − 𝜁 2 𝜔𝑛
Case 1: 𝜁 < 1 or 𝑐 < 𝑐𝑐

𝑥 𝑡 = 𝑒 −𝜁𝜔𝑛 𝑡 𝐶1′ cos( 1 − 𝜁 2 𝜔𝑛 𝑡) + 𝐶2′ sin( 1 − 𝜁 2 𝜔𝑛 𝑡)

For initial conditions 𝑥(𝑡 = 0) = 𝑥0, and 𝑥(𝑡

ሶ = 0) = 𝑥ሶ 0,
𝑥ሶ +𝜁𝜔 𝑥
𝐶1′ = 𝑥0 and 𝐶2′ = 0 2𝑛 0
1−𝜁 𝜔𝑛

Solution for an underdamped system

𝑥ሶ 0 + 𝜁𝜔𝑛 𝑥0
𝑥(𝑡) = 𝑒 −𝜁𝜔𝑛𝑡 𝑥0cos( 1 − 𝜁 2 𝜔𝑛 𝑡) + sin( 1 − 𝜁 2 𝜔𝑛 𝑡)
1 − 𝜁 2 𝜔𝑛

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Case 1 / 3: Underdamped system

𝑥ሶ 0 + 𝜁𝜔𝑛 𝑥0
𝑥(𝑡) = 𝑒 −𝜁𝜔𝑛𝑡 𝑥0cos( 1 − 𝜁 2 𝜔𝑛 𝑡) + sin( 1 − 𝜁 2 𝜔𝑛 𝑡)
1 − 𝜁 2 𝜔𝑛
Can be simplified as:
𝑥(𝑡) = 𝑋𝑒 −𝜁𝜔𝑛𝑡 cos( 1 − 𝜁 2 𝜔𝑛 𝑡 − 𝜑)

Exponential decay 𝜔𝑑 = 1 − 𝜁 2 𝜔𝑛 is the frequency of

with time damped vibration. Note that 𝜔𝑑 < 𝜔𝑛
𝑥(𝑡) slope = 𝑥ሶ0
2𝜋
𝜏𝑑 =
𝜔𝑑
𝑋 𝑋𝑒 −𝜁𝜔𝑛𝑡
𝜔𝑑 𝑡
𝜑

Case 1 / 3: Underdamped system

Estimating damping from data: Logarithmic decrement
𝑥(𝑡) 2𝜋
𝜏𝑑 =
𝜔𝑑
𝑋𝑒 −𝜁𝜔𝑛 𝑡
𝑥1
𝑥2
𝑡2
𝜔𝑑 𝑡
𝑡1
Determine 𝑥1/𝑥2 from measured
data.
𝑥1 𝑋𝑒 −𝜁𝜔𝑛𝑡1 cos(𝜔𝑑 𝑡1 − 𝜑) Determine log decrement 𝛿.
= 𝛿
𝑥2 𝑋𝑒 −𝜁𝜔𝑛𝑡2 cos(𝜔𝑑 𝑡2 − 𝜑) Estimate damping ratio as 𝜁 ≈
2𝜋
2𝜋
We choose 𝑡2 = 𝑡1 + 𝜔 . So, cos 𝜔𝑑 𝑡2 − 𝜑 = cos(𝜔𝑑 𝑡1 − 𝜑)
𝑑
𝑥1 𝑒 −𝜁𝜔𝑛 𝑡1
= = 𝑒 𝜁𝜔𝑛 (𝑡2 −𝑡1 ) = 𝑒 2𝜋𝜁𝜔𝑛 /𝜔𝑑
𝑥2 𝑒 −𝜁𝜔𝑛 𝑡2
𝑥1 2𝜋𝜁𝜔𝑛 2𝜋𝜁
Define Log Decrement: 𝛿 = ln = =
𝜔
𝑥2 𝜔𝑑 1 − 𝜁2
For small 𝜁 ≪ 1, 𝜔𝑛 ≈ 1. Therefore, 𝛿 ≈ 2𝜋𝜁 for 𝜁 ≪ 1
𝑑

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Case 2 / 3: Critically damped system

Case 2: 𝜁 = 1 or 𝑐 = 𝑐𝑐

In the solution we derived for the underdamped system,

𝑥ሶ 0 + 𝜁𝜔𝑛 𝑥0
𝑥(𝑡) = 𝑒 −𝜁𝜔𝑛 𝑡 𝑥0cos(𝜔𝑑 𝑡) + sin(𝜔𝑑 𝑡)
𝜔𝑑
If we take the limit 𝜁 → 1, then 𝜔𝑑 → 0
Therefore, cos(𝜔𝑑 𝑡) → 1
sin(𝜔𝑑 𝑡) → 𝜔𝑑 𝑡
Using these limits in the 𝑥(𝑡) expression gives:
𝑥ሶ 0 + 𝜁𝜔𝑛 𝑥0
𝑥(𝑡) = 𝑒 −𝜁𝜔𝑛 𝑡 𝑥0 + 𝜔𝑑 𝑡
𝜔𝑑

Case 2 / 3: Critically damped system

𝑥ሶ 0 + 𝜁𝜔𝑛 𝑥0
𝑥(𝑡) = 𝑒 −𝜁𝜔𝑛 𝑡 𝑥0 + 𝜔𝑑 𝑡
𝜔𝑑
Now substituting 𝜁 = 1:

𝑥(𝑡) = 𝑒 −𝜔𝑛𝑡 𝑥0 + 𝑥ሶ 0 + 𝜔𝑛 𝑥0 𝑡 Significance of critical

damping:
Exponentially Linearly  Smallest value of damping
decreasing in time increasing in time for which the system will
have no oscillations.
Solution for critically-damped system
 System returns to rest
fastest (even faster than
overdamped case).

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Case 3 / 3: overdamped system

Case 3: 𝜁 > 1 or 𝑐 > 𝑐𝑐

As 𝜁 2 − 1 > 0 the roots are real and distinct and both are negative:

𝑠1 = −𝜁 + 𝜁 2 − 1 𝜔𝑛 < 0
𝑠2 = −𝜁 − 𝜁 2 − 1 𝜔𝑛 < 0
with 𝑠2 < 𝑠1

The solution is given by:

−𝜁+ 𝜁 2 −1 𝜔𝑛 𝑡 −𝜁− 𝜁 2 −1 𝜔𝑛 𝑡
𝑥 𝑡 = 𝐶1 𝑒 + 𝐶2 𝑒

Case 3 / 3: overdamped system

The solution is given by:

−𝜁+ 𝜁 2 −1 𝜔𝑛 𝑡 −𝜁− 𝜁 2 −1 𝜔𝑛 𝑡
𝑥 𝑡 = 𝐶1 𝑒 + 𝐶2 𝑒

As before, the constants 𝐶1 and 𝐶2 are determined from the initial

conditions 𝑥0 and 𝑥ሶ 0 :
𝑥0 𝜔𝑛 𝜁 + 𝜁 2 − 1 + 𝑥ሶ 0
𝐶1 = and
2𝜔𝑛 𝜁 2 − 1

−𝑥0 𝜔𝑛 𝜁 − 𝜁 2 − 1 − 𝑥ሶ 0
𝐶2 =
2𝜔𝑛 𝜁 2 − 1
Solution for overdamped system

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Comparison of motions with different types of damping

Critically
𝑥(𝑡) Undamped
damped
overdamped

Underdamped

Summary for free response

 Basics and the simple pendulum
 Free vibrations of undamped SDOF system
• Deriving the equations of motion
• Characteristic or frequency equation
• Natural or resonance frequency
• Solving the equations of motion
 Free vibrations of a damped SDOF system
• Equations of motion
• Critical damping and the damping ratio
• Underdamped, critically-damped and
overdamped cases

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An Example

Example
Log. Decrement
𝑥1 𝑚 = 200 kg
𝛿 = ln = 2𝜋𝜁𝜔𝑛 /𝜔𝑑
𝑥2 𝜏𝑑 = 2 s
𝑥1 2𝜋𝜁 𝑥1.5 = 𝑥1 /4
𝛿 = ln = ln(16) = 2.77 =
𝑥2 1 − 𝜁2
𝜁 = 0.4

Damped period of vibration

2𝜋 2𝜋
𝜏𝑑 = = =2
𝜔𝑑 𝜔𝑛 1 − 𝜁 2

This gives 𝜔𝑑 = 𝜋 rad/s and 𝜔𝑛 = 3.43 rad/s Note the units

Value for critical damping 𝑐𝑐 = 2𝑚𝜔𝑛 = 2 ∗ 200 ∗ 3.43 = 1373.5 𝑁𝑠/𝑚

Damping constant 𝑐 = 𝜁𝑐𝑐 = 0.4 ∗ 1373.5 = 554.5 𝑁𝑠/𝑚

Stiffness: 𝑘 = 𝑚𝜔𝑛 2 = 200 ∗ 3.432 = 2358.3 𝑁𝑠/𝑚

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Example
At what time will the mass have maximum displacement?

𝑥(𝑡) = 𝑋𝑒 −𝜁𝜔𝑛 𝑡 sin(𝜔𝑑 𝑡) because 𝜑 = 90° based on the plot.

𝑑𝑥
Say, the maximum displacement occurs at 𝑡 = 𝑡1 . At that time, =0
𝑑𝑡
(also, 𝑥ሷ < 0)
𝑥ሶ 𝑡 = 𝑋𝑒 −𝜁𝜔𝑛 𝑡 − 𝜁𝜔𝑛 sin 𝜔𝑑 𝑡 + 𝜔𝑑 cos 𝜔𝑑 𝑡

𝑥ሶ 𝑡1 = 0 = 𝑋𝑒 −𝜁𝜔𝑛 𝑡1 −𝜁𝜔𝑛 sin 𝜔𝑑 𝑡1 + 𝜔𝑑 cos 𝜔𝑑 𝑡1

−𝜁𝜔𝑛 sin 𝜔𝑑 𝑡1 + 𝜔𝑑 cos 𝜔𝑑 𝑡1 = 0

1 − 𝜁2
𝜔𝑑 1 − 𝜁2 𝜔𝑑 𝑡1
tan 𝜔𝑑 𝑡1 = =
𝜁𝜔𝑛 𝜁
𝜁

OR sin 𝜔𝑑 𝑡1 = 1 − 𝜁2

Example
sin 𝜔𝑑 𝑡1 = sin 𝜋𝑡1 = 1 − 𝜁 2 = 0.915 𝜔𝑑 = 𝜋 from our calculations
earlier for this problem.
sin−1 (0.915)
This gives 𝑡1 = = 0.368 sec Note that 𝑡1 is NOT
𝜋
equal to 0.5s or 𝜏𝑑 /4

We are given that the maximum displacement 𝑥(𝑡 = 𝑡1 ) = 250 mm. Therefore,

0.25 = 𝑋𝑒 −0.4∗3.43∗0.368 1 − 0.42

This gives 𝑋 = 0.455 m.

The velocity can be determined from displacement:

𝑥(𝑡) = 𝑋𝑒 −𝜁𝜔𝑛 𝑡 sin 𝜔𝑑 𝑡 because 𝜑 = 90° based on the plot.

𝑥(𝑡)
ሶ = 𝑋𝑒 −𝜁𝜔𝑛 𝑡 (−𝜁𝜔𝑛 sin 𝜔𝑑 𝑡 + 𝜔𝑑 cos 𝜔𝑑 𝑡)

For 𝑡 = 0, 𝑥ሶ 𝑡 = 0 = 𝑥ሶ 0 = 𝑋𝜔𝑑 = 0.455 ∗ 𝜋 = 1.43 m/s

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Forced response of single degree of freedom systems

 Transient: 𝑥(𝑡) = 𝑥ℎ(𝑡) + 𝑥𝑝(𝑡)

 Sensitive to initial conditions Total transient steady-state
 Vibrates at 𝜔𝑑
 Dies out with time Transient or free response

 Steady state:
 If harmonic force, response Steady-state forced response
is at the same frequency as
forcing
 Response can last for ever Total or general response
(as long as force is applied)

Forced response of single degree of freedom systems

Forcing F (t)
Type of forcing Examples
𝑘 𝑐
Sinusoidal or Steady-state
Harmonic machinery
𝑚 Periodic Dirt roads
𝑥
Impact or Impulse Pot-holes, explosions,
𝐹(𝑡) shock
Random wind

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Equation of motion
FREE BODY DIAGRAM
𝑘𝑥 𝑐𝑥ሶ
𝑘 𝑐

𝑚
𝑚
𝑥 𝑥
𝐹(𝑡)
𝐹(𝑡)
From Newton’s second law of motion (see FBD)
𝑚𝑥ሷ = 𝐹(𝑡) − 𝑘𝑥 − 𝑐𝑥ሶ
This equation can be rearranged to give the EOM:
𝑚𝑥ሷ + 𝑐𝑥ሶ + 𝑘𝑥 = 𝐹(𝑡) Non-homogenous equation
Recall …….𝑚𝑥ሷ + 𝑐𝑥ሶ + 𝑘𝑥 = 0 Homogenous equation
In addition to forcing, the system could have initial conditions. The total
solution is a sum of “particular solution” and “homogeneous solution”.

Nomenclature
Forced Free

Particular or Non- Homogeneous Mathematical

homogeneous
Steady-state Transient Physical

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Harmonic forcing of an undamped system

Harmonic force is a force at a single frequency ‘𝜔’ given by:
𝐹 𝑡 = 𝐹0 cos 𝜔𝑡 Forcing frequency

NOTE: In general, 𝜔 is not equal to 𝜔𝑛 . (That can be one of the values)

For simplicity, first consider an undamped system. The EOM is:

𝑚𝑥ሷ + 𝑘𝑥 = 𝐹0 cos 𝜔𝑡 ….. (1)

The homogenous solution for this equation is given by

𝑥ℎ 𝑡 = 𝐶1 cos 𝜔𝑛 𝑡 + 𝐶2 sin 𝜔𝑛 𝑡 ….. (2)

where 𝑘 is the natural frequency of the system.

𝜔𝑛 =
𝑚

Harmonic forcing of an undamped system

𝑚𝑥ሷ + 𝑘𝑥 = 𝐹0 cos 𝜔𝑡 ….. (1)

Because the exciting force is harmonic, the particular solution

(or steady-state) is also harmonic:

𝑥𝑝 𝑡 = 𝑋 cos 𝜔𝑡 ….. (3)

Substituting (3) into (1), we get:

𝐹0 𝜂𝑠𝑡
𝑋= = 2
….. (4)
𝑘 − 𝑚𝜔 2 𝜔
1− 𝜔
𝑛

𝜂𝑠𝑡 = 𝐹0 /𝑘 : Static displacement

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Harmonic forcing of an undamped system

In-phase
with force

𝑋
𝜂𝑠𝑡 Out of phase
with force

Amplitude 𝑋 can be expressed as:

𝑋 1 This expression shows how the
= 2 dynamic displacement is different
𝜂𝑠𝑡 𝜔
1− 𝜔 from static displacement.
𝑛

Harmonic forcing of an undamped system

𝜔 < 𝜔𝑛 𝜔 > 𝜔𝑛
𝐹 𝑡 = 𝐹0 cos 𝜔𝑡
𝐹 𝑡 = 𝐹0 cos 𝜔𝑡
𝐹0
𝐹0

𝑥𝑝 𝑡 = 𝑋 cos 𝜔𝑡
𝑥𝑝 𝑡 = 𝑋 cos 𝜔𝑡
𝑋

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Harmonic forcing of an undamped system

𝜔 < 𝜔𝑛 𝜔 > 𝜔𝑛
𝐹 𝑡 = 𝐹0 cos 𝜔𝑡
𝐹 𝑡 = 𝐹0 cos 𝜔𝑡
𝐹0
𝐹0

𝑥𝑝 𝑡 = 𝑋 cos 𝜔𝑡
𝑥𝑝 𝑡 = 𝑋 cos 𝜔𝑡
𝑋

𝑘𝑥 ≈ 𝐹0 cos 𝜔𝑡 𝑚𝑥ሷ ≈ 𝐹0 cos 𝜔𝑡

Stiffness-dominated mass-dominated

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