in a spring mass system with m=1kg,k=60N/m,the energy of the system decays to 1/e of its initial value in 5 seconds. The damping constant and Q factor are
Answers
Answer:
Explanation:
In order to represent energy dissipation with linear damping it is important to first
understand how a single degree of freedom system behaves. Once understood, the
knowledge of the system and damping in a linear system can be used to define an
effective measure for damping based on other system quantities. Essentially the linear
system is a model for the work to be completed throughout. Knowing the intricacies
of a linear system gives background knowledge on more difficult nonlinear problems.
The model considered is a single degree of freedom system with a spring,
mass, and damper. The spring and damper are in parallel attached between the mass
and ground and there is an applied force acting on the mass itself. The equations of
motion for this particular system are
mx¨ + bx˙ + kx = f(t), (3.1)
where m is the mass, b the damping constant, k the spring constant, and f(t) the
applied force. The applied force is taken to be harmonic
f(t) = f0 sin(ωt). (3.2)
Eqn. 3.1 can be scaled and written as
x¨ + 2ζωnx˙ + ω 2
nx =f0msin(ωt), 10 (3.3)whereζ =b2√km,andω2n =km
.
ζ is known as the non-dimensional damping ratio and ω2n as the square of the natural
frequency of oscillation. λ, an alternative damping parameter that will be used later
considerations is defined as
λ = 2ζωn=bm
. (3.4)
Eqn. 3.3 must now be solved for the system response. It is an ordinary,
inhomogenous differential equation, thus the solution is a linear combination of the
homogenous and particular solutions. The homogenous solution to this particular
differential equation is found by solving the characteristic polynomial generated after the general solution of x = Aeαx has been substituted in and factored. The
characteristic polynomial is
α2 + 2ζωnα + ω2n = 0, (3.5)and has rootsα+,− = −ωnζ ±pζ2 − 1
, (3.6)
thus the solution to the homogenous problem is
xH(t) = C1e
α+t + C2e
α−t
, (3.7)
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where C1 and C2 are arbitrary constants that are found when initial conditions are
applied. Eqn. 3.6 has three distinct cases. The discriminate of the square root can
either be greater than zero, the over damped case, less than zero, the underdamped
case, or exactly zero, the critically damped case. Only the underdamped case will be
considered. Eqn. 3.7 can be written as
xH(t) = e−ζωnt
C1 cos(ωdt) + C2 sin(ωdt)m , (3.8)
where ωd is the damped natural frequency, given by ωd =
p1 − ζ2, for 0 < ζ < 1.
Eqn. 3.8 can be rewritten in a more useful form by combining the sine and cosine
terms into a single sine term with a phase shift. Doing so gives
xH(t) = A1 e−ζωntsin(ωdt + φ), (3.9)where A1 =qC21 + C22 , (3.10)
and φ is a phase angle given by
tan(φ) = C1C2 . (3.11)
The method of undetermined coefficients can be used to solve for the particular solution. The right hand side of Eqn. 3.3 is used to construct an attempted
solution of
xP (t) = B1sin(ωt) + B2cos(ωt), (3.12)
which is substituted into Eqn. 3.3. After setting like parts equal, B1 and B2 are
determined to give equality between the left hand side and right hand side of Eqn.
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3.3. The particular solution, written in the same useful form as Eqn. 3.9, is found to
be
xP (t) = A2 sin(ωt − ψ), (3.13)
where
A2 =f0mpn − ω22 + (2ζωnω)2 , (3.14)
and ψ is another phase angle given by
tan(ψ) = 2ζωnωω2n − ω2 . (3.15)
The response of Eqn. 3.3 is thus
x(t) = A1 e−ζωntsin(ωdt + φ) + A2 sin(ωt − ψ). (3.16)
Initial conditions can now be applied to determine the unknowns C1 and C2.
The initial conditions are
x(0) = x0
x˙(0) = ˙x0. (3.17)
Using Eqn. 3.8 and the initial conditions we find
C1 = x0 +2f0ζωnωm(ω2n − ω2)2 + (2ζωnω)2 (3.18)
and
C2 =x˙ 0ωd+ζωnωdx0 +2f0ζωnωm(ω2n − ω22 + (2ζωnω)
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The dynamics of a linear system are now known. This information will be
used in helping to develop an effective damping measure and will serve as the model
for the forced systems to come. For convenience the response is summarized here:
x(t) =A1 e
−ζωnt sin(ωdt + φ) + A2 sin(ωt − ψ),
A1 =qC21 + C22,C1 =x0 +2f0ζωnω(ω2n − ω2)2 + (2ζωnω)2,C2 =x˙ 0ωd+ζωnωd x0 +2f0ζωnωm(ω2n − ω2)2 + (2ζωnω)2
−
f0ω(ω2n − ω2)m ωd(ω2n − ω2)2 + (2ζωnω)2, (3.20)
A2 =f0mp(ω2− ω2)2 + (2ζωnω)2,φ=ArctanC1C2,ψ=Arctan2ζωnωω2n − ω2