Upload
others
View
12
Download
0
Embed Size (px)
Citation preview
After studying the case of the free vibration, let us turn our attention to forced vibration. We will just study the mass-spring system, which is governed by the equation
๐๐ฆโฒโฒ ๐ก + ๐พ๐ฆโฒ ๐ก + ๐๐ฆ ๐ก = ๐น(๐ก) A simple case of forced vibration is a vertical mass-spring system that is subject to gravity.
The vibrating spring under the influence of gravity
The gravitational force acting on the mass is ๐น ๐ก = ๐๐. Therefore, the spring equation becomes ๐๐ฆโฒโฒ ๐ก + ๐พ๐ฆโฒ ๐ก + ๐๐ฆ ๐ก = ๐๐. We previously learned how to find the general solution of the homogeneous equation. We will now find a particular solution of the nonhomogeneous equation using the method of undetermined coefficients. Since the right side is constant, we will find a constant solution ๐ฆ ๐ก = ๐ถ. By substituting that into the equation, we get
๐๐ถ = ๐๐ or
๐ถ =๐๐๐
Hence, if z ๐ก is the general solution for the homogeneous case, then the general solution for this nonhomogeneous case is
๐ฆ ๐ก = ๐ง ๐ก +๐๐๐
Thus, the effect of gravity is very simple: it adds a constant amount of elongation.
Image Source: Wikimedia Commons, licensed for unlimited public use.
Periodic forcing functions
We will now study a more difficult case, that of a periodic forcing function. This case is of great physical interest because mechanical and electric oscillators are often subject to periodic external forces. A circuit for example may be attached to an AC power source, or be exposed to radio waves. A mass-spring system like the shock absorbers in your car can be subject to periodic external forces as you drive on certain roads.
A general analysis of forced vibration with a periodic forcing term
Let us assume that the ๐น(๐ก) in
๐๐ฆโฒโฒ ๐ก + ๐พ๐ฆโฒ ๐ก + ๐๐ฆ ๐ก = ๐น(๐ก) is periodic with frequency ๐ > 0 and of the form
๐น ๐ก = ๐น0 cos๐๐ก with a constant ๐น0 > 0. We will try to find a particular solution y ๐ก by using the method of undetermined coefficients:
y ๐ก = ๐ด cos๐๐ก + ๐ต sin๐๐ก
By substituting the assumed form of the solution into the differential equation, we get
๐๐2 โ๐ด cos๐๐ก โ ๐ต sin๐๐ก + ๐พ๐ โ๐ด sin๐๐ก + ๐ต cos๐๐ก+ ๐ ๐ด cos๐๐ก + ๐ต sin๐๐ก = ๐น0 cos๐๐ก
We simplify by combining cos and sin terms: cos๐๐ก โm๐2A + ๐พ๐๐ต + ๐๐ด + sin๐๐ก โ๐๐2๐ต โ ๐พ๐๐ด + ๐๐ต = ๐น0 cos๐๐ก which leads to cos๐๐ก ๐พ๐๐ต + (๐ โ๐๐2)๐ด + sin๐๐ก โ๐พ๐๐ด + (๐ โ๐๐2)๐ต = ๐น0 cos๐๐ก In order for the two sides to be equal, we require
๐ โ๐๐2 ๐ด + ๐พ๐๐ต = ๐น0
โ๐พ๐๐ด + ๐ โ๐๐2 ๐ต = 0
Our 2x2 system ๐ โ๐๐2 ๐ด + ๐พ๐๐ต = ๐น0
โ๐พ๐๐ด + ๐ โ๐๐2 ๐ต = 0
is solvable precisely if its determinant is nonzero. The determinant is
๐ท = ๐ โ๐๐2 2 + ๐พ2๐2 = ๐2 ๐02 โ ๐2 2 + ๐พ2๐2 But a sum of squares of real numbers can only be zero when the individual terms are zero- thus our 2x2 system fails to be solvable only when ๐พ = 0 and ๐ = ๐๐2. Indeed, when ๐ = ๐๐2 and ๐พ = 0, then the first equation of our system reduces to 0 = F0, which is a contradiction. Observe that ๐พ = 0 means no damping, and ๐ = ๐๐2 is equivalent to ๐ = ๐
๐, which we recognize as the natural frequency ๐0 of the
undamped case. We have thus showed:
A particular solution
y ๐ก = ๐ด cos๐๐ก + ๐ต sin๐๐ก
of the differential equation
๐๐ฆโฒโฒ ๐ก + ๐พ๐ฆโฒ ๐ก + ๐๐ฆ ๐ก = ๐น0 cos๐๐ก
(all constants positive)
is guaranteed to exist, except in the undamped case (๐พ = 0) where the external frequency ๐ is equal to the natural frequency ๐0 of the mass-spring system.
Discussion of the undamped resonance case (D=0)
It should not come as a surprise that the undamped case where ๐ = ๐0 is special. In this case, we have the physical phenomenon of resonance โ the boundless amplification of oscillations when a system is being excited at its own natural frequency. We would not expect a constant amplitude particular solution in this case, but rather one whose oscillations get arbitrarily large. The theory for solving the nonhomogeneous case that we have already learned confirms this. If ๐พ = 0 and ๐ = ๐0, the differential equation becomes
๐๐ฆโฒโฒ ๐ก + ๐๐ฆ ๐ก = ๐น0 cos๐0๐ก Since y ๐ก = ๐ด cos๐0๐ก + ๐ต sin๐0๐ก is the general solution of the homogeneous equation, no solution of the nonhomogeneous equation of that form can exist. However, a solution of the form
y ๐ก = ๐ด๐ก cos๐0๐ก + ๐ต๐ก sin๐0๐ก exists and we will now find it. Computing derivatives, we get yโฒ ๐ก = ๐ด cos๐0๐ก โ ๐0๐ด๐ก sin๐0๐ก +๐ต sin๐0๐ก + ๐0๐ต๐ก cos๐0๐ก yโฒโฒ ๐ก = โ๐ด๐0 sin๐0๐ก โ ๐0๐ด sin๐0๐ก โ ๐02๐ด๐ก cos๐0๐ก + ๐ต๐0 cos๐0๐ก + ๐0๐ต cos๐0๐ก โ ๐02๐ต๐ก sin๐0๐ก We can simply this as follows:
yโฒโฒ ๐ก = โ๐02๐ก ๐ด cos๐0๐ก + ๐ต sin๐0๐ก + 2๐0๐ต cos๐0๐ก โ 2๐0๐ด sin๐0๐ก
We now plug yโฒโฒ ๐ก and ๐ฆ ๐ก into ๐๐ฆโฒโฒ + ๐๐ฆ = ๐น0 cos๐0๐ก and get ๐(โ๐0
2๐ก ๐ด cos๐0๐ก + ๐ต sin๐0๐ก + 2๐0๐ต cos๐0๐ก โ 2๐0๐ด sin๐0๐ก) + ๐(๐ด๐ก cos๐0๐ก + ๐ต๐ก sin๐0๐ก)= ๐น0 cos๐0๐ก
This looks much worse than it is. By combining terms on the left side, we get
(A๐ก cos๐0๐ก + ๐ต๐ก sin๐0๐ก) (๐ โ ๐๐02) + 2๐0๐๐ต cos๐0๐ก โ 2๐0๐ด sin๐0๐ก = ๐น0 cos๐0๐ก
Since ๐ โ ๐๐0
2 = 0, this simplifies to
2๐0๐๐ต cos๐0๐ก โ 2๐0๐ด sin๐0๐ก = ๐น0 cos๐0๐ก For this to be true, we need ๐ด = 0 and ๐ต = ๐น0
2๐0๐ . Thus, we have shown:
y ๐ก = ๐น02๐0๐
๐ก sin๐0๐ก is a particular solution of ๐๐ฆโฒโฒ ๐ก + ๐๐ฆ ๐ก = ๐น0 cos๐0๐ก . The general solution therefore is
y ๐ก = ๐1 cos๐0๐ก + ๐2 sin๐0๐ก +๐น0
2๐0๐๐ก sin๐0๐ก
The presence of the โtโ factor in the third term causes the oscillations to grow in magnitude beyond all bounds as t gets larger and larger.
Having successfully explored the special case where ๐ท = ๐ โ ๐๐2 2 + ๐พ2๐2 = 0, we now turn our attention to the case where ๐ท > 0. Recall that in that case, a particular solution of the form y ๐ก = ๐ด cos๐๐ก + ๐ต sin๐๐ก exists for the equation
๐๐ฆโฒโฒ ๐ก + ๐พ๐ฆโฒ ๐ก + ๐๐ฆ ๐ก = ๐น0 cos๐๐ก and the coefficients A,B must satisfy
๐ โ ๐๐2 ๐ด + ๐พ๐๐ต = ๐น0
โ๐พ๐๐ด + ๐ โ ๐๐2 ๐ต = 0 Since the determinant ๐ท is now assumed to be nonzero, we can use Cramerโs rule to solve for A and B:
๐ด =๐น0 ๐ โ ๐๐2
๐ท =๐น0๐ ๐02 โ ๐2
๐ท
B =๐น0๐พ๐๐ท
We have thus found a particular solution of
๐๐ฆโฒโฒ ๐ก + ๐พ๐ฆโฒ ๐ก + ๐๐ฆ ๐ก = ๐น0 cos๐๐ก for the case ๐ท > 0:
y ๐ก =๐น0๐ ๐02 โ ๐2
๐ท cos๐๐ก +๐น0๐พ๐๐ท
sin๐๐ก Recall from a previous presentation that ๐1 cos๐๐ก + ๐2 sin ๐๐ก can be rewritten as a single, phase-shifted sine (or cosine) wave with amplitude
A = ๐12 + ๐22 It follows that the amplitude of our particular solution is
A =๐น0๐ท ๐2 ๐02 โ ๐2 2 + ๐พ2๐2 =
๐น0๐ท
Having found a particular solution, we turn our attention to the general solution. If ๐ฆโ ๐ก is the general solution of the homogeneous equation, then the general solution of the nonhomogeneous case is
y ๐ก = ๐ฆโ ๐ก +๐น0๐ ๐02 โ ๐2
๐ท cos๐๐ก +๐น0๐พ๐๐ท sin๐๐ก
Discussion of the case D>0 Recall that the determinant is ๐ท = ๐2 ๐02 โ ๐2 2 + ๐พ2๐2. We will discuss the damped and the undamped situations separately. First let us discuss the damped case ๐พ > 0. In that case, ๐ท > 0 even when ฯ = ๐0. The general solution we found is
y ๐ก = ๐ฆโ ๐ก +๐น0๐ ๐02 โ ๐2
๐ทcos๐๐ก +
๐น0๐พ๐๐ท
sin๐๐กparticular solution
Let us simplify the notation by calling the particular solution ๐ฆ๐ ๐ก :
y ๐ก = ๐ฆโ ๐ก + ๐ฆ๐ ๐ก Recall from a previous presentation that in the damped case, all solutions of the homogeneous equation decay exponentially. This means that over time, as ๐ก โ โ, the term ๐ฆโ ๐ก goes to zero, and only the particular solution ๐ฆ๐ ๐ก remains.
This is why ๐ฆโ ๐ก is called the transient solution, whereas ๐ฆ๐ ๐ก is called the steady state solution or forced response. Over time, ALL solutions, regardless of the initial conditions, โsettle intoโ (become approximately equal to ) the steady state solution. This is important because the transient solution ๐ฆโ ๐ก contains the information about the initial conditions whereas the steady state solution ๐ฆ๐ ๐ก does not. The damped, forced system โforgetsโ its initial condition over time. A physical interpretation of this phenomenon is that damping dissipates the energy embodied in the initial position and velocity, and that the long-term behavior of the system is solely determined by the external periodic force.
The damped case with D>0
Steady state amplitude as a function of excitation frequency and damping
Let us examine how the amplitude of the steady state solution depends on the frequency of the external force and on damping. Our physical intuition would predict that the amplitude will be largest if the system is excited at its natural frequency. We will see that this is approximately correct. Let us recall the amplitude of the steady state solution:
A =๐น0๐ท
=๐น0
๐2 ๐02 โ ๐2 2 + ๐พ2๐2
If ๐พ is very small to begin with, then ๐ท would be small (not necessarily precisely the smallest) for ฯ = ๐0, and ๐ด would correspondingly be near its maximum:
๐ด =๐น0
๐พ2๐02=
๐น0๐พ๐0
We therefore expect this to be an approximation to the maximum amplitude in the lightly damped case.
A precise determination of the external frequency ๐max that maximizes the amplitude of the steady state solution requires us to find the derivative of A with respect to ๐, and where it is zero. We have
๐A๐๐ = โ
๐น0(2๐2 ๐02 โ ๐2 โ2๐ + 2๐พ2๐)
2 ๐2 ๐02 โ ๐2 2 + ๐พ2๐232
Setting the second numerator factor to zero and simplifying, we get the equation
๐(๐พ2 โ 2 ๐2 ๐02 โ ๐2 ) = 0
Since we assumed ๐ > 0, we have a critical value only where ๐2 = ๐02 โ๐พ2
2๐2 , and
that only exists if ๐02 โฅ๐พ2
2๐2, which can be rewritten as ๐พ โค 2๐๐. Since (๐พ2 โ 2 ๐2 ๐02 โ ๐2 ) increases with ๐, ๐A
๐๐ decreases with ๐, and must
therefore pass from positive to negative values at the critical value we just found. Since that critical value is also the only critical value on the open interval (0,โ) on which ๐ is defined, the amplitude must have an absolute maximum there. To get that absolute maximum, we substitute the critical value for ๐ into ๐ด (calculation omitted).
We therefore have the following result:
The steady state solution (aka forced response) of the damped system
๐๐ฆโฒโฒ + ๐พ๐ฆโฒ + ๐๐ฆ = ๐น0 cos๐๐ก
with ๐พ > 0 has its greatest amplitude at the external excitation frequency
๐max = ๐02 โ๐พ2
2๐2
assuming that the system is not damped too much (๐พ โค 2๐๐).
This frequency is always smaller than the natural frequency of the undamped system (๐max < ๐0) and in the case of light damping, ๐max โ ๐0. The
corresponding maximum amplitude is
๐ดmax =๐น0
๐พ๐0 (1 โ ๐พ24๐๐)
โ ๐น0๐พ๐0
for small ๐พ
As ๐พ โ 0, ๐ดmax goes to infinity.
The undamped case with ๐ท > 0 Finally, let us briefly discuss our general solution
y ๐ก = ๐ฆโ ๐ก + ๐ฆ๐ ๐ก with ๐ฆ๐ ๐ก = ๐น0๐ ๐0
2โ๐2
๐ทcos๐๐ก + ๐น0๐พ๐
๐ทsin๐๐ก
for the undamped case with ๐ท > 0 . This is the case with ๐พ = 0 and ๐0 โ ๐. Since ๐พ = 0, the determinant simplifies to ๐ท = ๐2 ๐02 โ ๐2 2, and the particular solution becomes
๐ฆ๐ ๐ก =๐น0
๐ ๐02 โ ๐2 cos๐๐ก. Observe that for frequencies that are close to the natural frequency, the amplitude of ๐ฆ๐ ๐ก becomes large. The general solution is y ๐ก = ๐1 cos๐0๐ก + ๐2 sin๐0๐ก + ๐น0
๐ ๐02โ๐2 cos๐๐ก.
Summary of general solutions The general solution of the forced vibration system ๐๐ฆโฒโฒ + ๐พ๐ฆโฒ + ๐๐ฆ = ๐น0 cos๐๐ก is..
If..
y ๐ก = ๐1 cos๐0๐ก + ๐2 sin๐0๐ก +๐น0
2๐0๐๐ก sin๐0๐ก
๐พ = 0, ๐ = ๐0
y ๐ก = ๐1 cos๐0๐ก + ๐2 sin๐0๐ก +๐น0
๐ ๐02 โ ๐2 cos๐๐ก ๐พ = 0, ๐0 โ ๐
y ๐ก = ๐ฆโ ๐ก +๐น0๐ ๐02 โ ๐2
๐ท cos๐๐ก +๐น0๐พ๐๐ท sin๐๐ก
where D = ๐2 ๐02 โ ๐2 2 + ๐พ2๐2 and ๐ฆโ ๐ก is the general solution of the homogeneous case: ๐ฆโ ๐ก = ๐1๐๐1๐ก + ๐2๐๐2๐ก if the system is overdamped
๐ฆโ ๐ก = (๐1+๐2๐ก)๐โ ๐พ2๐๐ก if the system is critically damped
๐ฆโ ๐ก = (๐1cos๐๐ก + ๐2 sin๐๐ก)๐โ๐พ2๐๐ก if the system is underdamped.
๐พ > 0 ๐พ2 > 4๐๐ ๐พ2 = 4๐๐ ๐พ2 < 4๐๐