forced vibrations - arizona state universityboerner/mat275/3.8.pdfπ‘ does not. the damped, forced...
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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
ππ¦β²β² π‘ + πΎπ¦β² π‘ + ππ¦ π‘ = πΉ(π‘) 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 πΉ π‘ = ππ. Therefore, the spring equation becomes ππ¦β²β² π‘ + πΎπ¦β² π‘ + ππ¦ π‘ = ππ. 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 π¦ π‘ = πΆ. By substituting that into the equation, we get
ππΆ = ππ or
πΆ =πππ
Hence, if z π‘ is the general solution for the homogeneous case, then the general solution for this nonhomogeneous case is
π¦ π‘ = π§ π‘ +πππ
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 πΉ(π‘) in
ππ¦β²β² π‘ + πΎπ¦β² π‘ + ππ¦ π‘ = πΉ(π‘) is periodic with frequency π > 0 and of the form
πΉ π‘ = πΉ0 cosππ‘ with a constant πΉ0 > 0. We will try to find a particular solution y π‘ by using the method of undetermined coefficients:
y π‘ = π΄ cosππ‘ + π΅ sinππ‘
By substituting the assumed form of the solution into the differential equation, we get
ππ2 βπ΄ cosππ‘ β π΅ sinππ‘ + πΎπ βπ΄ sinππ‘ + π΅ cosππ‘+ π π΄ cosππ‘ + π΅ sinππ‘ = πΉ0 cosππ‘
We simplify by combining cos and sin terms: cosππ‘ βmπ2A + πΎππ΅ + ππ΄ + sinππ‘ βππ2π΅ β πΎππ΄ + ππ΅ = πΉ0 cosππ‘ which leads to cosππ‘ πΎππ΅ + (π βππ2)π΄ + sinππ‘ βπΎππ΄ + (π βππ2)π΅ = πΉ0 cosππ‘ In order for the two sides to be equal, we require
π βππ2 π΄ + πΎππ΅ = πΉ0
βπΎππ΄ + π βππ2 π΅ = 0
Our 2x2 system π βππ2 π΄ + πΎππ΅ = πΉ0
βπΎππ΄ + π βππ2 π΅ = 0
is solvable precisely if its determinant is nonzero. The determinant is
π· = π βππ2 2 + πΎ2π2 = π2 π02 β π2 2 + πΎ2π2 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 πΎ = 0 and π = ππ2. Indeed, when π = ππ2 and πΎ = 0, then the first equation of our system reduces to 0 = F0, which is a contradiction. Observe that πΎ = 0 means no damping, and π = ππ2 is equivalent to π = π
π, which we recognize as the natural frequency π0 of the
undamped case. We have thus showed:
A particular solution
y π‘ = π΄ cosππ‘ + π΅ sinππ‘
of the differential equation
ππ¦β²β² π‘ + πΎπ¦β² π‘ + ππ¦ π‘ = πΉ0 cosππ‘
(all constants positive)
is guaranteed to exist, except in the undamped case (πΎ = 0) where the external frequency π is equal to the natural frequency π0 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 π = π0 is special. In this case, we have the physical phenomenon of resonance β 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 πΎ = 0 and π = π0, the differential equation becomes
ππ¦β²β² π‘ + ππ¦ π‘ = πΉ0 cosπ0π‘ Since y π‘ = π΄ cosπ0π‘ + π΅ sinπ0π‘ 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 π‘ = π΄π‘ cosπ0π‘ + π΅π‘ sinπ0π‘ exists and we will now find it. Computing derivatives, we get yβ² π‘ = π΄ cosπ0π‘ β π0π΄π‘ sinπ0π‘ +π΅ sinπ0π‘ + π0π΅π‘ cosπ0π‘ yβ²β² π‘ = βπ΄π0 sinπ0π‘ β π0π΄ sinπ0π‘ β π02π΄π‘ cosπ0π‘ + π΅π0 cosπ0π‘ + π0π΅ cosπ0π‘ β π02π΅π‘ sinπ0π‘ We can simply this as follows:
yβ²β² π‘ = βπ02π‘ π΄ cosπ0π‘ + π΅ sinπ0π‘ + 2π0π΅ cosπ0π‘ β 2π0π΄ sinπ0π‘
We now plug yβ²β² π‘ and π¦ π‘ into ππ¦β²β² + ππ¦ = πΉ0 cosπ0π‘ and get π(βπ0
2π‘ π΄ cosπ0π‘ + π΅ sinπ0π‘ + 2π0π΅ cosπ0π‘ β 2π0π΄ sinπ0π‘) + π(π΄π‘ cosπ0π‘ + π΅π‘ sinπ0π‘)= πΉ0 cosπ0π‘
This looks much worse than it is. By combining terms on the left side, we get
(Aπ‘ cosπ0π‘ + π΅π‘ sinπ0π‘) (π β ππ02) + 2π0ππ΅ cosπ0π‘ β 2π0π΄ sinπ0π‘ = πΉ0 cosπ0π‘
Since π β ππ0
2 = 0, this simplifies to
2π0ππ΅ cosπ0π‘ β 2π0π΄ sinπ0π‘ = πΉ0 cosπ0π‘ For this to be true, we need π΄ = 0 and π΅ = πΉ0
2π0π . Thus, we have shown:
y π‘ = πΉ02π0π
π‘ sinπ0π‘ is a particular solution of ππ¦β²β² π‘ + ππ¦ π‘ = πΉ0 cosπ0π‘ . The general solution therefore is
y π‘ = π1 cosπ0π‘ + π2 sinπ0π‘ +πΉ0
2π0ππ‘ sinπ0π‘
The presence of the βtβ 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 π· = π β ππ2 2 + πΎ2π2 = 0, we now turn our attention to the case where π· > 0. Recall that in that case, a particular solution of the form y π‘ = π΄ cosππ‘ + π΅ sinππ‘ exists for the equation
ππ¦β²β² π‘ + πΎπ¦β² π‘ + ππ¦ π‘ = πΉ0 cosππ‘ and the coefficients A,B must satisfy
π β ππ2 π΄ + πΎππ΅ = πΉ0
βπΎππ΄ + π β ππ2 π΅ = 0 Since the determinant π· is now assumed to be nonzero, we can use Cramerβs rule to solve for A and B:
π΄ =πΉ0 π β ππ2
π· =πΉ0π π02 β π2
π·
B =πΉ0πΎππ·
We have thus found a particular solution of
ππ¦β²β² π‘ + πΎπ¦β² π‘ + ππ¦ π‘ = πΉ0 cosππ‘ for the case π· > 0:
y π‘ =πΉ0π π02 β π2
π· cosππ‘ +πΉ0πΎππ·
sinππ‘ Recall from a previous presentation that π1 cosππ‘ + π2 sin ππ‘ can be rewritten as a single, phase-shifted sine (or cosine) wave with amplitude
A = π12 + π22 It follows that the amplitude of our particular solution is
A =πΉ0π· π2 π02 β π2 2 + πΎ2π2 =
πΉ0π·
Having found a particular solution, we turn our attention to the general solution. If π¦β π‘ is the general solution of the homogeneous equation, then the general solution of the nonhomogeneous case is
y π‘ = π¦β π‘ +πΉ0π π02 β π2
π· cosππ‘ +πΉ0πΎππ· sinππ‘
Discussion of the case D>0 Recall that the determinant is π· = π2 π02 β π2 2 + πΎ2π2. We will discuss the damped and the undamped situations separately. First let us discuss the damped case πΎ > 0. In that case, π· > 0 even when Ο = π0. The general solution we found is
y π‘ = π¦β π‘ +πΉ0π π02 β π2
π·cosππ‘ +
πΉ0πΎππ·
sinππ‘particular solution
Let us simplify the notation by calling the particular solution π¦π π‘ :
y π‘ = π¦β π‘ + π¦π π‘ Recall from a previous presentation that in the damped case, all solutions of the homogeneous equation decay exponentially. This means that over time, as π‘ β β, the term π¦β π‘ goes to zero, and only the particular solution π¦π π‘ remains.
This is why π¦β π‘ is called the transient solution, whereas π¦π π‘ is called the steady state solution or forced response. Over time, ALL solutions, regardless of the initial conditions, βsettle intoβ (become approximately equal to ) the steady state solution. This is important because the transient solution π¦β π‘ contains the information about the initial conditions whereas the steady state solution π¦π π‘ does not. The damped, forced system βforgetsβ 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 =πΉ0π·
=πΉ0
π2 π02 β π2 2 + πΎ2π2
If πΎ is very small to begin with, then π· would be small (not necessarily precisely the smallest) for Ο = π0, and π΄ would correspondingly be near its maximum:
π΄ =πΉ0
πΎ2π02=
πΉ0πΎπ0
We therefore expect this to be an approximation to the maximum amplitude in the lightly damped case.
A precise determination of the external frequency πmax that maximizes the amplitude of the steady state solution requires us to find the derivative of A with respect to π, and where it is zero. We have
πAππ = β
πΉ0(2π2 π02 β π2 β2π + 2πΎ2π)
2 π2 π02 β π2 2 + πΎ2π232
Setting the second numerator factor to zero and simplifying, we get the equation
π(πΎ2 β 2 π2 π02 β π2 ) = 0
Since we assumed π > 0, we have a critical value only where π2 = π02 βπΎ2
2π2 , and
that only exists if π02 β₯πΎ2
2π2, which can be rewritten as πΎ β€ 2ππ. Since (πΎ2 β 2 π2 π02 β π2 ) increases with π, πA
ππ decreases with π, 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,β) on which π is defined, the amplitude must have an absolute maximum there. To get that absolute maximum, we substitute the critical value for π into π΄ (calculation omitted).
We therefore have the following result:
The steady state solution (aka forced response) of the damped system
ππ¦β²β² + πΎπ¦β² + ππ¦ = πΉ0 cosππ‘
with πΎ > 0 has its greatest amplitude at the external excitation frequency
πmax = π02 βπΎ2
2π2
assuming that the system is not damped too much (πΎ β€ 2ππ).
This frequency is always smaller than the natural frequency of the undamped system (πmax < π0) and in the case of light damping, πmax β π0. The
corresponding maximum amplitude is
π΄max =πΉ0
πΎπ0 (1 β πΎ24ππ)
β πΉ0πΎπ0
for small πΎ
As πΎ β 0, π΄max goes to infinity.
The undamped case with π· > 0 Finally, let us briefly discuss our general solution
y π‘ = π¦β π‘ + π¦π π‘ with π¦π π‘ = πΉ0π π0
2βπ2
π·cosππ‘ + πΉ0πΎπ
π·sinππ‘
for the undamped case with π· > 0 . This is the case with πΎ = 0 and π0 β π. Since πΎ = 0, the determinant simplifies to π· = π2 π02 β π2 2, and the particular solution becomes
π¦π π‘ =πΉ0
π π02 β π2 cosππ‘. Observe that for frequencies that are close to the natural frequency, the amplitude of π¦π π‘ becomes large. The general solution is y π‘ = π1 cosπ0π‘ + π2 sinπ0π‘ + πΉ0
π π02βπ2 cosππ‘.
Summary of general solutions The general solution of the forced vibration system ππ¦β²β² + πΎπ¦β² + ππ¦ = πΉ0 cosππ‘ is..
If..
y π‘ = π1 cosπ0π‘ + π2 sinπ0π‘ +πΉ0
2π0ππ‘ sinπ0π‘
πΎ = 0, π = π0
y π‘ = π1 cosπ0π‘ + π2 sinπ0π‘ +πΉ0
π π02 β π2 cosππ‘ πΎ = 0, π0 β π
y π‘ = π¦β π‘ +πΉ0π π02 β π2
π· cosππ‘ +πΉ0πΎππ· sinππ‘
where D = π2 π02 β π2 2 + πΎ2π2 and π¦β π‘ is the general solution of the homogeneous case: π¦β π‘ = π1ππ1π‘ + π2ππ2π‘ if the system is overdamped
π¦β π‘ = (π1+π2π‘)πβ πΎ2ππ‘ if the system is critically damped
π¦β π‘ = (π1cosππ‘ + π2 sinππ‘)πβπΎ2ππ‘ if the system is underdamped.
πΎ > 0 πΎ2 > 4ππ πΎ2 = 4ππ πΎ2 < 4ππ