Lecture 4 Single Degree of Freedom Model
Lecture 4 Single Degree of Freedom Model
Lecture 4 Single Degree of Freedom Model
𝒎𝒙 + 𝒌𝒙 = 𝟎 4-1
𝒙 𝒕 + 𝝎𝟐𝒏 𝒙 𝒕 = 𝟎 4-2
The free response depends only on the initial conditions of the system, that is, 𝑥 0 and 𝑥(0),
the initial displacement and initial velocity, respectively. Since the left-hand side of Equation
4-2 includes the displacement 𝑥(𝑡) and its second derivative 𝑥(𝑡), the solution, when
differentiated twice, is a function that reappears: 𝑥(𝑡) = −𝜔23 𝑥(𝑡).
Functions with this property are harmonic functions, such as sines and cosines. Therefore,
we assume a solution of the form
where 𝜔 needs to be determined and constants 𝐶< and 𝐶3 can be found by satisfying the initial
conditions. Double differentiating and substituting the assumed solution into the governing equation
results in
−𝝎𝟐 + 𝝎𝟐𝒏 = 𝟎
meaning 𝜔 equals the natural frequency of oscillation, and, therefore, 𝜔 = 𝜔2 = 𝑘 𝑚 for the
mass-spring system. (The negative value of the square root is discarded since there are no negative
frequencies.)
By replacing 𝜔 by 𝑘 𝑚 Equation 4-3 becomes
4-4
𝒌 𝒌
𝒙 𝒕 = 𝑪𝟏 𝒔𝒊𝒏 𝒕 + 𝑪𝟐 𝒄𝒐𝒔 𝒕
𝒎 𝒎
The constants 𝐶< and 𝐶3 can be evaluated from the initial displacement 𝑥
0 and initial velocity
𝑥(0) . From Equation 4-4, we write equations for 𝑥 0 and 𝑥(0) to find that 𝐶3 = 𝑥 0 and
𝐶< = 𝑥(0) 𝜔𝑛 .
The general response in terms of the initial conditions is then
𝑥(0) 4-5
𝒙 𝒕 = 𝒔𝒊𝒏𝜔2 𝒕 + 𝑥 0 𝒄𝒐𝒔𝜔2 𝒕
𝜔2
For the undamped system the oscillation continues for all time with the period 𝑇 = 2𝜋 𝜔2 =
2𝜋 𝑚 𝑘 . The response is a function of the initial conditions, with the frequency of oscillation
𝜔2 .
Note: the angular frequency is the intrinsic property of the system, that is why it is
sometimes called the natural frequency of the system,
Equation 4-5 can be expressed equivalently in terms of an amplitude, frequency, and phase of
vibration by writing
𝑥 𝑡 = 𝐴 𝑐𝑜𝑠 𝜔2 𝑡 − 𝜙 4-6
3 4-7
3 𝑥 0
𝐴= 𝑥 0 +
𝜔2
𝑥 0 4-8
𝜙 = 𝑡𝑎𝑛I<
𝑥 0 𝜔2
In this form we can see the effects of the initial conditions on the response amplitude and phase.
Note: The amplitude 𝐴 and phase 𝜙, can be found given the initial conditions 𝑥 0 and
𝑥(0)
The period of oscillations 𝑇, does not depend on the initial conditions and hence on the
amplitude, which is the property of linear systems.
Free Vibration of an Undamped Torsional System (simple pendulum)
The equation of motion in standard form for simple pendulum is expressed as
𝒈 4-9
𝜽+ 𝜽 = 𝟎
𝒍
The general solution of Equation 4-9 can be obtained as in case of Equation 4-3
where 𝜔2 is given by 𝑔 𝑙 and 𝐶< and 𝐶3 can be determined from initial condition
𝜃 0 = 𝜃Q 𝑎𝑛𝑑 𝜃 0 = 𝜃Q
𝐶3 = 𝜃 0
𝐶< = 𝜃(0) 𝜔𝑛 .
𝐴 = 2 𝑎𝑛𝑑 𝑡𝑎𝑛𝜙 = 1
𝑥 𝑡 = 2𝑠𝑖𝑛 𝑡 − 𝜋 4 m
𝐴 = 2 𝑎𝑛𝑑 𝑡𝑎𝑛𝜙 = 1
𝑥 𝑡 = 2𝑠𝑖𝑛 𝑡 − 5𝜋 4 m
Alternate Formulation
Instead of the harmonic form of Equation 4-3, an alternate approach in solving the equation of
motion is to assume the form 𝑥 𝑡 = 𝐶𝑒 XY , where 𝐶 and 𝜆 are to be determined. The exponential
form of the solution is often used in the solution of differential equations. In vibration studies it is
used because it leads to functions that are important for characterizing a system. An example is
the frequency response function, derived in the subsequent lecture.
Differentiating the assumed form and substituting into the governing equation leads to 𝜆3 + 𝜔23 =
0 or 𝜆<,3 = ±𝑖𝜔2 and thus
𝒙 𝒕 = 𝑫𝟏 𝒆𝒊𝝎𝒏 𝒕 + 𝑫𝟐 𝒆I𝒊𝝎𝒏 𝒕
𝒙 𝟎 = 𝑩𝟏 𝒂𝒏𝒅 𝒙 𝟎 = 𝑩𝟐 𝝎𝒏 4-12
Therefore, the general solution, in terms of the initial conditions and the natural frequency, matches
Equation 4-5.
Phase Plane
Another way to envision the motion of an unforced harmonic vibration is in the phase plane. Phase
plane analysis is a graphical method for studying second-order systems. An advantage of the phase
plane is that the trajectory paths indicate the nature of the oscillation.
Note: The word “phase” used in phase plane is distinct from the phase between input
and output of a dynamic system.
In a qualitative way, one can follow the trajectories and observe whether the system is stable,
oscillatory, periodic, or otherwise. In a qualitative method, the differential equation does not have
to be solved and yet it is possible to ascertain much about the behavior of the system.
The phase plane is created by replacing the second-order governing equation 𝑥 𝑡 + 𝜔23 𝑥 𝑡 = 0
by two equivalent first-order differential equations,
𝒅𝒙(𝒕)
= 𝒚(𝒕)
𝒅𝒕
𝒅𝒚(𝒕)
= −𝝎𝟐𝒏 𝒙(𝒕)
𝒅𝒕
Substituting the derivative of the first equation into the second equation recovers the original
second-order equation.
For a mass-spring system, the path of the mass in the plane can be found as follows:
𝒅𝒚 𝒅𝒕 𝝎𝟐𝒏 𝒙
=−
𝒅𝒙 𝒅𝒕 𝒚
𝒅𝒚 𝝎𝟐𝒏 𝒙
+ = 𝟎
𝒅𝒙 𝒚
𝒚𝟐 + 𝝎𝟐𝒏 𝒙𝟐 = 𝒄 4-13
where the last equation is obtained by integrating the third equation and 𝑐 is the constant of
integration.
Note: In the phase plane 𝑥, 𝑦 the velocity 𝑦 can be plotted as a function of position 𝑥
creating a plot of velocity versus displacement.
Since 𝑥 is the displacement and 𝑥 is the velocity,𝑥 3 is related to potential energy and 𝑦 3 is related
to kinetic energy. Thus, Equation 4-13. can be viewed as an equation for constant energy and 𝑐 is
related to the total system energy. Curves for different values of are shown in Fig. 4.2. The arrows
indicate the direction of motion, which can be determined by selecting a quadrant and evaluating
the rate of change of a parameter. For example, in the first quadrant 𝑦 = 𝑥 > 0 then 𝑦 > 0 and
𝑥 increases with time.
Fig.4.2 Phase plot for 𝒙 + 𝝎𝟐𝒏 𝒙 = 𝟎. The paths are concentric ellipses governed by 𝒚𝟐 + 𝝎𝟐𝒙 𝒙𝟐 = 𝒄
Example: Phase plane representation of a mass-spring system
The governing equation of the mass-spring system in Fig. 4.3 is linear second-order differential
equation
𝑚𝑥 + 𝑘𝑥 = 0 𝑖. 𝑒. 𝑥 + 𝑥 = 0
a b
Assume that the mass is initially at rest, at length 𝑥Q . Then the solution of this equation is
𝒙 𝒕 = 𝒙𝟎 𝒄𝒐𝒔(𝒕)
𝒙 𝒕 = −𝒙𝟎 𝒔𝒊𝒏(𝒕)
Eliminating time t from the above equations, we obtain the equation of the trajectories
𝒙𝟐 + 𝒙𝟐 = 𝒙𝟐𝟎
This represents a circle in the phase plane. Its plot is shown in Fig. 4.3b
The nature of the system response corresponding to various initial conditions is directly displayed
on the phase plane. In the above example, we can easily see that the system trajectories neither
converge to the origin nor diverge to infinity. They simply circle around the origin, indicating the
marginal nature of the system’s stability.
Interpretation of results
The solution of spring-mass system predicts that the oscillations continue forever (Fig. 4.1), which
is not true for the real systems. The reason for this is that we assumed that there is no damping.
Now consider the case when 𝑐 ≠ 0. Such a system is called as damped system.