16
2 Single Degree of Freedom (SDOF) Systems
x h (t) = e
−ζ ω t a
x 0
a
cos(ω d t) +
˙
x 0 + ζ ω x 0
ω d a
sin(ω d t)
= a e
−ζ ω t
sin(ϕ) cos(ω d t) + cos(ϕ) sin(ω d t)
= a e
−ζ ω t sin
ω d t + ϕ
.
The notations sin(ϕ) = x 0 /a and cos(ϕ) =
˙
x 0 + ζ ω x 0
/
ω d a
are meaningful
since the absolute values of these ratios are smaller than unity and the sum of their
squares is unity. The latter equality results by using a trigonometric identity.
We summarize some useful features of the amplitude-phase representation of the
free vibration solution.
1. The largest value of x h (t) and its derivatives are readily available. They are a,
a ω d , and a ω 2
d for x h (t), ˙
x h (t), and ¨
x h (t).
2. The graphical representation of x h (t), ˙
x h (t), and ¨
x h (t) is simple as they are
modulated periodic functions. For example, x h (t) is a sine wave shifted by the
phase ϕ which is modulated by the exponential amplitude a exp
− ζ ω t
.
3. For undamped systems, x h (t) = a sin(ω t + ϕ) so that the elastic and kinetic
energies are SE(t) = (1/2) k x h (t) 2 = (1/2) k a 2 sin
2 (ω t + ϕ) and KE(t) =
(1/2) m ˙
x h (t) 2 = (1/2) m a 2 ω 2 cos 2 (ω t + ϕ) so that, since m ω 2 = k, we
have SE(t) + KE(t) = k a 2 /2 = constant, as expected since the system is
conservative.
4. The velocity ˙
x h (t) and acceleration ¨
x h (t) are out of phase by π/2 and π relative
to x h (t). These relationships follow from properties of trigonometric functions.
We have
˙
x h (t) = a ω cos(ω t + ϕ) = a ω sin(π/2 − ω t − ϕ)
= −a ω sin(ω t + ϕ − π/2)
= a ω sin(ω t + ϕ + π/2)
and
¨
x h (t) = −a ω
2 sin(ω t + ϕ) = a ω
2 sin(ω t + ϕ + π)
so that ˙
x h (t) and ¨
x h (t) are out of phase relative to x h (t) by π/2 and π . Their
amplitudes are a ω and a ω 2 . The solid, dashed, and dotted lines of Fig. 2.4 show
the solutions x h (t), ˙
x h (t), and ¨
x h (t) over a few periods for a = 1, ω = π , and
ϕ = π/4. The plots illustrate the phase difference between these three functions.
The initial values of these solutions are x h (0) = sin(π/4) = 0.7071, ˙
x h (0)/ω =
sin(π/4 + π/2) = 0.7071, and ¨
x h (0) = sin(π/4 + π) = −0.7071.
2 Single Degree of Freedom (SDOF) Systems
x h (t) = e
−ζ ω t a
x 0
a
cos(ω d t) +
˙
x 0 + ζ ω x 0
ω d a
sin(ω d t)
= a e
−ζ ω t
sin(ϕ) cos(ω d t) + cos(ϕ) sin(ω d t)
= a e
−ζ ω t sin
ω d t + ϕ
.
The notations sin(ϕ) = x 0 /a and cos(ϕ) =
˙
x 0 + ζ ω x 0
/
ω d a
are meaningful
since the absolute values of these ratios are smaller than unity and the sum of their
squares is unity. The latter equality results by using a trigonometric identity.
We summarize some useful features of the amplitude-phase representation of the
free vibration solution.
1. The largest value of x h (t) and its derivatives are readily available. They are a,
a ω d , and a ω 2
d for x h (t), ˙
x h (t), and ¨
x h (t).
2. The graphical representation of x h (t), ˙
x h (t), and ¨
x h (t) is simple as they are
modulated periodic functions. For example, x h (t) is a sine wave shifted by the
phase ϕ which is modulated by the exponential amplitude a exp
− ζ ω t
.
3. For undamped systems, x h (t) = a sin(ω t + ϕ) so that the elastic and kinetic
energies are SE(t) = (1/2) k x h (t) 2 = (1/2) k a 2 sin
2 (ω t + ϕ) and KE(t) =
(1/2) m ˙
x h (t) 2 = (1/2) m a 2 ω 2 cos 2 (ω t + ϕ) so that, since m ω 2 = k, we
have SE(t) + KE(t) = k a 2 /2 = constant, as expected since the system is
conservative.
4. The velocity ˙
x h (t) and acceleration ¨
x h (t) are out of phase by π/2 and π relative
to x h (t). These relationships follow from properties of trigonometric functions.
We have
˙
x h (t) = a ω cos(ω t + ϕ) = a ω sin(π/2 − ω t − ϕ)
= −a ω sin(ω t + ϕ − π/2)
= a ω sin(ω t + ϕ + π/2)
and
¨
x h (t) = −a ω
2 sin(ω t + ϕ) = a ω
2 sin(ω t + ϕ + π)
so that ˙
x h (t) and ¨
x h (t) are out of phase relative to x h (t) by π/2 and π . Their
amplitudes are a ω and a ω 2 . The solid, dashed, and dotted lines of Fig. 2.4 show
the solutions x h (t), ˙
x h (t), and ¨
x h (t) over a few periods for a = 1, ω = π , and
ϕ = π/4. The plots illustrate the phase difference between these three functions.
The initial values of these solutions are x h (0) = sin(π/4) = 0.7071, ˙
x h (0)/ω =
sin(π/4 + π/2) = 0.7071, and ¨
x h (0) = sin(π/4 + π) = −0.7071.