1. Verify Eqs. 1 through 5. Figure 1: mass spring-damper In class, we have studied mechanical systems of this type. Here, the main results of our in-class analysis are reviewed. The dynamic behavior of this system is deter- mined from the linear second-order ordinary differential equation: d'a + + kr = 0 dt² dt (1) where r(t) is the displacement of the mass, m is the mass, b is the damping coefficient, and k is the spring stiffness. Equations like Eq. 1 are often written in the "standard form" C M m where d'a (2) dt2 The variable wn is the natural frequency of the system and is the damping ratio. tan If the system is underdamped, i.e. < 1, and it has initial conditions (0) = o-o=0, then the solution to Eq. 2 is given by: x(t) = Lele dr +26wn+wn²x = 0 Ta xo (-) 2π Wd -Cunt sin (wat +) is the damped natural frequency. In Figure 2, the normalized plot of the response of this system reveals some useful information. Note that the amount of time Ta between peaks is constant. The elapsed time, Ta, is the period of oscillation. From Eq. 3 it can be shown that: (3) , and wa=w₁√√1-C² (4) 2T wn√√√1-(² (5)

Elements Of Electromagnetics
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1. Verify Eqs. 1 through 5.
Figure 1: mass spring damper
In class, we have studied mechanical systems of this
type. Here, the main results of our in-class analysis are
reviewed. The dynamic behavior of this system is deter-
mined from the linear second-order ordinary differential
equation:
where
(1)
where r(t) is the displacement of the mass, m is the
mass, b is the damping coefficient, and k is the spring
stiffness. Equations like Eq. 1 are often written in the
"standard form"
ď²x
dt2
r(t) =
= tan-1
d²r
dt2
m.
M
+25wn +wn²x = 0
(2)
The variable wn is the natural frequency of the system
and is the damping ratio.
If the system is underdamped, i.e. < < 1, and it has
initial conditions (0) = zot-o = 0, then the solution
to Eq. 2 is given by:
IO
√1
x(1)
T₁ =
+b+kr = 0
dt
2π
dr.
dt
ل لها
-(wat sin (wat +)
and
is the damped natural frequency.
In Figure 2, the normalized plot of the response of this
system reveals some useful information. Note that the
amount of time Ta between peaks is constant. The elapsed
time, Ta, is the period of oscillation. From Eq. 3 it can
be shown that:
Wd = W₂1
(3)
√1-5² (4)
2T
Wn√1-(2
(5)
Transcribed Image Text:1. Verify Eqs. 1 through 5. Figure 1: mass spring damper In class, we have studied mechanical systems of this type. Here, the main results of our in-class analysis are reviewed. The dynamic behavior of this system is deter- mined from the linear second-order ordinary differential equation: where (1) where r(t) is the displacement of the mass, m is the mass, b is the damping coefficient, and k is the spring stiffness. Equations like Eq. 1 are often written in the "standard form" ď²x dt2 r(t) = = tan-1 d²r dt2 m. M +25wn +wn²x = 0 (2) The variable wn is the natural frequency of the system and is the damping ratio. If the system is underdamped, i.e. < < 1, and it has initial conditions (0) = zot-o = 0, then the solution to Eq. 2 is given by: IO √1 x(1) T₁ = +b+kr = 0 dt 2π dr. dt ل لها -(wat sin (wat +) and is the damped natural frequency. In Figure 2, the normalized plot of the response of this system reveals some useful information. Note that the amount of time Ta between peaks is constant. The elapsed time, Ta, is the period of oscillation. From Eq. 3 it can be shown that: Wd = W₂1 (3) √1-5² (4) 2T Wn√1-(2 (5)
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