In this slides you will gain the knowledge of different types of methods for vibrating systems and his derivations.

1. GANDHINAGAR INSTITUTE OF
TECHNOLOGY
MECHANICAL DEPARTMENT
DYNAMICS OF MACHINE

2. CONTENT :-
• Introduction
• Determination of Natural frequency
• Types of Method
• Equilibrium Method
• Energy Method
• Rayleigh’s Method

3. INTRODUCTION :-
• If the external forces is removed after giving an initial displacement to the system,
then the system vibrates on its own due to internal elastic forces. Such vibrations
are known as free vibration.
• There is no external artificial resistance to the vibrations then such vibrations are
known as undamped free vibration.
• In most of the free vibration there is always certain amount of damping associated
with the system.
• However damping is very small, for all practical purpose it can be neglected and
the vibrations considered as undamped vibration.

4. DETERMINATION OF NATURAL
FREQUENCY :-
• The natural frequency of any body or a system is depends upon the geometrical
parameters and mass properties of the body.
• There are various methods to obtained the equation of vibrating system,
1) Equilibrium method
2) Energy method
3) Rayleigh’s method

5. EQUILIBRIUM METHOD :-
• According to D’Alembert’s Principal, a body or a system which is not in static
equilibrium due to acceleration it possess, can be brought to static equilibrium by
introducing the inertia force on it.
• This Principal is used for developing the equation of the motion for vibrating the
equation of the motion for vibrating system which if further used to find the
natural frequency of the vibrating system.

6. Consider a spring-mass system as shown in figure;

7. A Spring has a negligible mass. The forces acting on the masses are :-
(i) Inertia Force, 𝑚 𝑥
(ii) Spring Force, 𝑘 𝑥 + 𝛿
(iii) Gravitational Force, 𝑚𝑔
According to D’Alembert’s Principal,
𝛴[Inertia Force + External Force] = 0
𝑚 𝑥 + 𝑘 𝑥 + 𝛿 - 𝑚𝑔 = 0
𝑥 +
𝑘
𝑚
𝑥 = 0

8. Comparing the above equation with the fundamental equation of the simple
harmonic equation.
We get,
𝑥 + 𝑤 𝑛
2 𝑥 = 0
The Natural Frequency is,
𝜔 𝑛 =
𝑘
𝑚
, 𝑓𝑛 =
𝜔 𝑛
2𝜋
The Time Period equation is,
𝑡 𝑝 = 2𝜋
𝑚
𝑘

9. ENERGY METHOD :-
• According to law of conservation of energy, the energy can neither be created nor
be destroyed but it can be transfer from the one from of energy to another form
of energy.
• In free damped vibration, no energy is transferred to the system or from the
system, therefore total mechanical energy remains constant.
• The potential energy due to
1. Gravitational potential energy
2. Strain energy
At equilibrium position the kinetic energy is maximum and the potential energy is
zero and vice versa.

10. According to law of energy conservation,
Total energy = Constant
KE + PE = Constant
Differentiating equation,
ⅆ
ⅆ𝑡
𝑘𝐸 + 𝑃𝐸 = 0
Kinetic Energy (KE) =
1
2
𝑚𝑥2
Potential Energy (PE) =
1
2
𝑘𝑥2
Substituting equations we get,
𝑥 +
𝑘
𝑚
𝑥 = 0

11. Comparing it with fundamental equation of Simple Harmonic Equation, we get,
𝜔 𝑛 =
𝑘
𝑚
, 𝑓𝑛 =
𝜔 𝑛
2𝜋

12. RAYLEIGH’S METHOD
• This is the extension of the energy method, which is developed by the Lord
Rayleigh.
Total energy = constant
𝐾𝐸 1 + 𝑃𝐸 1 = 𝐾𝐸 2 + 𝑃𝐸 2
• The subscripts 1 & 2 denotes the two different positions. Let subscripts 1 will
denotes the mean position where the potential energy is zero. And subscripts 2 will
denotes the extreme position where kinetic energy is zero.
The above equation will be,
𝐾𝐸 1 = 𝑃𝐸 2

13. • But at mean position the kinetic energy is maximum and at extreme position the
potential energy is maximum.
𝐾𝐸 ma𝑥 = 𝑃𝐸 max
• According to Lord Rayleigh’s the maximum kinetic energy which is at the mean
position is equal to maximum potential energy which is the extreme position.
• Let’s Body is moving with simple harmonic motion, therefore the displacement of
the body is given by,
𝑥 = 𝑋 sin 𝑤 𝑛 𝑡
Differentiating above equation,
𝑥 =
𝑑𝑥
ⅆ𝑡
𝑥 =𝑤 𝑛 𝑋 cos 𝑤 𝑛 𝑡
𝑥 𝑚𝑎𝑥 = 𝑤 𝑛 𝑋 (t = 0, at mean position)

14. Maximum kinetic energy at mean position,
𝐾𝐸 ma𝑥 =
1
2
𝑚( 𝑥 𝑚𝑎𝑥)2
𝐾𝐸 ma𝑥 =
1
2
𝑚𝑤 𝑛
2
𝑥2
Maximum potential energy at extreme position,
PE =
1
2
𝑘𝑥2
Comparing both equations,
1
2
𝑚𝑤 𝑛
2 𝑥2 =
1
2
𝑘𝑥2
The Natural Frequency is,
𝜔 𝑛 =
𝑘
𝑚
, 𝑓𝑛 =
𝜔 𝑛
2𝜋
The Time Period equation is,
𝑡 𝑝 = 2𝜋
𝑚
𝑘

15. THANK YOU…