Summary of Free Undamped Vibrations
1. Key Concept: Free Undamped Vibrations
What is Free Vibration?


Free vibration occurs when a system oscillates naturally after being
displaced from its equilibrium position without any external forces acting on
it (e.g., no driving forces, no damping).
The system continues oscillating indefinitely as long as there’s no energy
loss, meaning no damping is involved.
In the mass-spring system, the restoring force (spring) causes the mass to
oscillate. The motion is simple harmonic (i.e., sinusoidal) and determined by the
system's natural frequency.
Key Points to Remember:


The system is assumed to be undamped (no energy loss).
The oscillations are determined solely by the system’s stiffness and mass.
2. The Equation of Motion for Free Undamped Vibrations
For a mass-spring system, the general equation of motion is:
¨
m x + k x=0
Where:

m = mass (kg)

x = acceleration (second derivative of displacement x (t ) , m/s )
k = spring stiffness (N/m)


¨
2
x (t ) = displacement (m)
This is a second-order linear differential equation that governs the vibration of
the system.
Key Concept:

This equation describes the motion of an object where the displacement is
proportional to the restoring force from the spring, and there are no
external forces or energy losses.
3. The General Solution to the Equation of Motion
¨
The solution to the equation m x + k x=0 is:
x (t )=C sin ( ω n t+ ψ )
Where:


C = amplitude of the motion (depends on initial displacement)
ω n = natural frequency (rad/s), given by:
ω n=

√
k
m
ψ = phase angle (depends on initial velocity and displacement)
4. Period and Frequency of the Vibration

Period T is the time taken for the system to complete one full oscillation:
T=

2π
ωn
Frequency f n is the number of oscillations per second:
1 ωn
f n= =
T 2π
This tells you how fast the system oscillates. The natural frequency ω n is inversely
related to the period T .
5. Initial Conditions
To fully solve the motion, we need initial conditions, which are usually given:

Displacement at t=0 : x ( 0 )=x 0

Velocity at t=0 : x ( 0 )=v 0
˙
From these initial conditions, we can find C and ψ :
√ ()

v0 2
C= x +
ωn

ψ=tan
2
0
−1
( )
x 0 ωn
v0
These formulas help you solve the system’s exact displacement at any point in
time.
6. Spring Stiffness
The key factor governing the motion of the system is the spring stiffness k . Here,
we’ll review different types of spring stiffness and their formulas, which are
crucial for the exam.
A. Linear Spring Stiffness
For most mass-spring systems, the spring stiffness is defined as:
k=
F
Δx
Where:


F = Force applied (N)
Δ x = Displacement from equilibrium (m)
This is a simple linear spring, where the stiffness k is constant.
B. Helical Spring Stiffness (For torsion)
For a helical spring (one that twists under load), the stiffness k is given by:
4
k=
Gd
3
64 n R
Where:




G = shear modulus (N/m²) (material property)
d = diameter of the wire (m)
n = number of turns in the spring
R = radius of the spring (m)
This equation describes how much torsional resistance a helical spring offers
when twisted.
C. Beam Axial Stiffness
For a beam under axial load (compression or extension), the stiffness is given by:
k=
EA
l
Where:



E = Young’s modulus (N/m²) (material property)
A = cross-sectional area of the beam (m²)
l = length of the beam (m)
This formula describes the axial stiffness of a beam — how much the beam
resists deformation along its length.
D. Torsional Stiffness (for rotating shafts)
For a shaft subject to torsion (twisting), the stiffness k is:
k=
G Jp
l
Where:

G = shear modulus (N/m²)
J p = polar moment of inertia of the beam's cross-section (m⁴)

l = length of the shaft (m)

This is used when dealing with rotational vibration problems, where you need to
account for twisting of the shaft.
E. Beam Transverse Stiffness (Bending)
When a beam is subjected to bending (perpendicular force), its stiffness is:
k=
3EI
3
l
Where:



E = Young’s modulus (N/m²)
I = second moment of area (or area moment of inertia) of the beam’s
cross-section (m⁴)
l = length of the beam (m)
This formula gives us the resistance of a beam to bending under an applied
force.
7. Springs in Series and Parallel
When multiple springs are involved, they can be combined in series or parallel.
A. Springs in Series
For springs in series (connected end to end), the equivalent stiffness is:
1
1 1
= +
k eq k 1 k 2
This means the effective stiffness is lower than that of any individual spring.
B. Springs in Parallel
For springs in parallel (connected side by side), the equivalent stiffness is:
k eq=k 1+ k 2
This means the effective stiffness is the sum of the individual spring constants.
8. Energy Considerations
In free undamped vibrations, the total mechanical energy (kinetic + potential
energy) remains constant because there is no energy loss due to damping. The
energy simply oscillates between kinetic energy (due to velocity) and potential
energy (due to the spring's compression/extension).
9. Summary of Key Formulas to Learn for the Exam
Formulas Given in the Exam Formula Sheet:
 Natural frequency ω n=√ k /m
2π
 Period T = ω
n
ωn
 Frequency f n=
2π
Formulas You Need to Derive/Understand (Not on the Sheet):
 General solution for displacement x (t )=C sin ( ω n t+ ψ )

Initial condition solutions:
√ ()
( )
v0 2
−1 x 0 ωn
C= x +
, ψ=tan
ωn
v0
2
0

Combination of Springs:
1
1 1
o Series: k = k + k
eq
1
2
o Parallel: k eq=k 1+ k 2
4

Gd
Helical spring stiffness: k =
3
64 n R

Beam axial stiffness: k =

Torsional stiffness: k =

Beam bending stiffness: k =
EA
l
G Jp
l
3EI
3
l
10. Conclusion


Free undamped vibration is all about mass-spring systems and
understanding how they oscillate at a natural frequency ω n=√ k /m .
Springs in series and parallel are crucial for problems involving multiple
springs.


Make sure to memorize the key spring stiffness formulas (axial, torsional,
beam) and be able to recognize when to use each one in the context of a
problem.
The initial condition solutions and energy conservation are vital when
you need to fully describe the motion of the system.
Summary of Free Damped Vibrations for the Exam
Key Concepts in Free Damped Vibrations
In free damped vibrations, damping forces are introduced, which causes energy
loss and results in a decreasing amplitude over time. This is contrasted with free
undamped vibrations, where the oscillation continues indefinitely with constant
amplitude.
Damping Force:

The damping force is proportional to the velocity and acts in the opposite
direction:
˙
f c =− c x ( t )
Where:
o c = damping coefficient (N·s/m)
˙
o x (t ) = velocity (m/s)
Equation of Motion for Damped Free Vibration
The general equation of motion for a damped system is:
¨
˙
m x ( t )+ c x ( t ) + k x ( t )=0
Where:

m = mass (kg)

x (t ) = acceleration (m/s²)
c = damping coefficient (N·s/m)


¨
k = spring stiffness (N/m)

x (t ) = displacement (m)

x (t ) = velocity (m/s)
˙
This is a second-order linear differential equation that governs the motion of
the system.
Damping Ratio ζ and Types of Damping
The behavior of the system is determined by the damping ratio ζ :
ζ=
c
2 √k m
Where:



c = damping coefficient (N·s/m)
k = spring stiffness (N/m)
m = mass (kg)
Types of Damping:
1. Underdamped (ζ < 1): The system oscillates with exponentially decaying
amplitude.
2. Critically Damped (ζ =1): The system returns to equilibrium without
oscillating in the shortest possible time.
3. Overdamped (ζ > 1): The system decays without oscillation but at a slower
rate.
Solution for Underdamped Vibration
For underdamped systems, the displacement is given by:
x (t )= A 1 e
− ζ ωn t
cos ( ω d t ) + A 2 e
− ζ ωn t
sin ( ωd t )
Where:

A1, A2 = constants determined by initial conditions

ω n=√ k /m = natural frequency (rad/s)

ω d=ωn √ 1 − ζ 2 = damped natural frequency (rad/s)
The motion decays exponentially with the factor e −ζ ω t , while oscillating at the
damped frequency ω d.
n
Damped Natural Frequency and Period
The damped natural frequency is defined as:
ω d=ωn √ 1 − ζ 2
And the damped period is:
T d=
2π
ωd
The system oscillates at a lower frequency than the undamped system due to
damping.
Logarithmic Decrement Method (for Lightly Damped Systems)
For lightly underdamped systems, you can estimate the damping ratio ζ using the
logarithmic decrement method:
1. Measure the amplitude ratio between successive peaks (at t 0 and t n):
( )
x (t 0)
1
δ= ln
n
x (t n)
2. The damping ratio ζ can then be estimated as:
ζ=
δ
√ 4 π 2+ δ 2
For small damping (ζ ≈ δ /2 π ).
This formula allows for experimental determination of damping from the decay
of oscillations. It's important to understand how to use this method for
measuring damping in practical systems, especially in light damping cases.
Important Equations for the Exam
Given in the Exam Formula Sheet:
1. Equation of Motion for Damped Free Vibration:
¨
˙
m x ( t )+ c x ( t ) + k x ( t )=0
2. Damping Ratio:
ζ=
3. Damped Natural Frequency:
c
2 √k m
ω d=ωn √ 1 − ζ 2
4. Solution for Underdamped Vibration:
x (t )= A 1 e
− ζ ωn t
cos ( ω d t ) + A 2 e
− ζ ωn t
sin ( ωd t )
Not on the Formula Sheet (but should be memorized):
1. Logarithmic Decrement and Damping Ratio:
( )
x (t 0)
1
δ
δ= ln
,ζ=
n
2π
x (t n)
ζ=
δ
√ 4 π 2+ δ 2
Where δ is the logarithmic decrement.
Relevant Concepts for the Exam
1. Sketching Responses for Different Damping Types: You may be asked to
sketch or interpret the responses of the system for underdamped,
critically damped, and overdamped motions. The general shape of the
graph depends on the damping ratio ζ :
o Underdamped: Oscillations with exponentially decaying amplitude.
o Critically damped: Fastest return to equilibrium without oscillation.
o Overdamped: Slow return to equilibrium without oscillation.
2. Estimating Damping Using the Logarithmic Decrement: For lightly
underdamped systems, estimate damping using the logarithmic
decrement method. This is important for experimental analysis and might
be directly tested in your exam.
Conclusion
In preparation for the exam on free damped vibrations:


Memorize the key equations: Understand how to apply them to solve for
damping ratio ζ , damped natural frequency ω d, and displacement.
Be able to sketch and describe responses for underdamped, critically
damped, and overdamped systems.

Practice using the logarithmic decrement method for estimating the
damping ratio in practical applications.
1. Forced Vibrations
In forced vibrations, a system is subject to an external force, causing it to
oscillate. Unlike free vibrations, forced vibrations persist as long as the external
force is applied.
Key Equations for Forced Vibrations (Given in the Exam Formula Sheet)

Undamped Forced Vibration:
¨
m x + ω2n x=
F 0 sin ( ω t )
m
Where:
o ω n is the natural frequency,
o F 0 is the amplitude of the external force,
o ω is the driving frequency,
o x (t ) is the displacement in the system.
The solution for the displacement x (t ) is:
x p (t )= X sin ( ω t )
Where:
o X=

F 0 /k
is the amplitude.
2
1 − ( ω/ω n )
Damped Forced Vibration:
¨
˙
m x +2 ζ ω n x +ω 2n x=
F 0 sin ( ω t )
m
Where:
o ζ is the damping ratio.
o X=
F0 /k
2
( 1− ω2 /ω 2n ) + ( 2 ζ ω /ωn )2
o The phase angle ϕ is:
1/ 2
is the displacement amplitude.
ϕ=tan
−1
( (
2 ζ ω /ωn
2
1 − ω/ω n )
)
The Magnification Factor M is:
M=
1
√( 1 − ( ω /ω ) ) +( 2 ζ ω /ω )
2 2
2
n
n
This factor shows the amplification of the system's response as it nears
resonance.

Rotating Machine Unbalance:
m0 e ω 2r sin ( ωr t )
m x +2 ζ ω n x +ω x=
m
¨
˙
2
n
Where:
o m0 is the unbalanced mass,
o e is the eccentricity,
o ω r is the rotational frequency.
The solution for displacement and phase angle is similar to the general
damped forced vibration equations, but adjusted for the rotating unbalance.

Seismometers and Accelerometers: These devices measure forced
vibrations in buildings or other structures. The equation is:
¨
˙
2
m x +2 ζ ω n x +ω n x=
2
bω
sin ( ω t )
m
Where:
o b is a constant related to the measuring device.
Concepts to Remember for Forced Vibrations:
1. Resonance: Occurs when the driving frequency ω matches the natural
frequency ω n, causing the system's amplitude to grow large. The
Magnification Factor M describes this behavior.
2. Damping: Helps control resonance and reduce vibration amplitude. The
damping ratio ζ plays a critical role in system design.
2. Rigid Body Vibrations
In rigid body vibrations, we study the motion of entire bodies that rotate and
oscillate in space. These vibrations are critical in engineering when analyzing
rotating machinery, beams, or shafts.
Key Concepts for Rigid Body Vibrations:


Rotation: Unlike particle motion, rigid bodies can rotate, and we focus on
angular displacement.
Moment of Inertia: In rotational dynamics, the moment of inertia I 0 is
analogous to mass in linear motion. It resists changes in rotational velocity.
Key Equation for Rigid Body Vibrations (involving rotational motion):
The equation of motion for a rigid body undergoing rotational vibration is:
¨
˙
I 0 θ + c θ + k θ=F 0 cos ( ω t )
Where:

θ is the angular displacement,
I 0 is the moment of inertia of the body,

c is the damping coefficient,



k is the spring constant,
ω is the driving frequency.
Moment of Inertia (for a bar):
1 2
I 0= m l
9
Where:


m is the mass of the bar,
l is the length of the bar.
This formula is essential for calculating the system's resistance to rotational
motion.
Concepts to Remember for Rigid Body Vibrations:
1. Rotational Motion: The dynamics of rigid body vibrations are governed by
rotational motion, which involves angular displacement, velocity, and
acceleration.
2. Moment of Inertia: The moment of inertia plays a central role in how the
body resists angular acceleration. Larger I 0 means the body resists rotation
more.
Summary of Formulas to Memorize:
Forced Vibrations:
1. Undamped Forced Vibration:
¨
m x + ω2n x=
F 0 sin ( ω t )
m
2. Damped Forced Vibration:
¨
˙
m x +2 ζ ω n x +ω 2n x=
ϕ=tan
M=
−1
F 0 sin ( ω t )
m
( (
2 ζ ω /ωn
2
1 − ω/ω n )
)
1
√( 1 − ( ω /ω ) ) +( 2 ζ ω /ω )
2 2
2
n
n
3. Rotating Machine Unbalance:
2
m0 e ω r sin ( ωr t )
m x +2 ζ ω n x +ω x=
m
¨
˙
2
n
4. Seismometers & Accelerometers:
¨
˙
2
m x +2 ζ ω n x +ω n x=
2
bω
sin ( ω t )
m
Rigid Body Vibrations:
1. Equation of Motion:
¨
˙
I 0 θ + c θ + k θ=F 0 cos ( ω t )
2. Moment of Inertia:
1 2
I 0= m l
9
Key Exam Preparation Tips:




Resonance and Damping: Understand how resonance occurs when the
driving frequency matches the system's natural frequency. The damping
ratio ζ is crucial for reducing resonance and controlling vibration
amplitudes.
Practice Solving Forced Vibration Problems: Focus on problems involving
undamped and damped forced vibrations, systems with unbalanced
rotating masses, and vibrating sensors.
Rigid Body Vibration Analysis: Understand how to calculate angular
displacement and use the moment of inertia in rotational systems. Be able
to apply the equation of motion for rotating bodies.
Review Practical Applications: Know how these concepts apply in realworld systems such as rotating machinery, seismometers, and
accelerometers.
Conclusion:
This summary covers the critical formulas and concepts for forced vibrations and
rigid body vibrations. Make sure you understand the theoretical concepts,
memorize the key formulas, and practice solving problems to apply these concepts
effectively.