Following equation (2.26) with the values given here, the system to be solved is
m x ¨ + c x ˙ + k x = F 0 cos ω t m \ddot{x}+c \dot{x}+k x=F_0 \cos \omega t m x ¨ + c x ˙ + k x = F 0 cos ω t
100 x ¨ ( t ) + 100 x ˙ ( t ) + 910 x ( t ) = 10 cos 3 t , x 0 = 0.001 m , v 0 = 0.02 m / s 100 \ddot{x}(t)+100 \dot{x}(t)+910 x(t)=10 \cos 3 t, x_0=0.001 m , v_0=0.02\ m / s 1 0 0 x ¨ ( t ) + 1 0 0 x ˙ ( t ) + 9 1 0 x ( t ) = 1 0 cos 3 t , x 0 = 0 . 0 0 1 m , v 0 = 0 . 0 2 m / s
where the units (and thus numerical values) of the initial conditions have been changed to agree with the equation of motion per the previous example . Dividing by the mass yields the vibration properties
f 0 = F 0 m = 10 100 = 0.1 m s 2 , ω n = k m = 910 100 = 3.017 r a d s , ζ = c 2 m k = 100 2 100 ⋅ 910 = 0.166 , ω d = ω n 1 − ζ 2 = 3.017 1 − 0.16 6 2 = 2.975 r a d / s \begin{aligned} f_0 & =\frac{F_0}{m}=\frac{10}{100}=0.1 \frac{ m }{ s ^2}, \omega_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{910}{100}}=3.017 \frac{ rad }{ s }, \\ \zeta & =\frac{c}{2 \sqrt{m k}}=\frac{100}{2 \sqrt{100 \cdot 910}}=0.166, \\ \omega_d & =\omega_n \sqrt{1-\zeta^2}=3.017 \sqrt{1-0.166^2}=2.975\ rad / s \end{aligned} f 0 ζ ω d = m F 0 = 1 0 0 1 0 = 0 . 1 s 2 m , ω n = m k = 1 0 0 9 1 0 = 3 . 0 1 7 s r a d , = 2 m k c = 2 1 0 0 ⋅ 9 1 0 1 0 0 = 0 . 1 6 6 , = ω n 1 − ζ 2 = 3 . 0 1 7 1 − 0 . 1 6 6 2 = 2 . 9 7 5 r a d / s
Since ω = 3 rad/s, the system is near resonance. Computing the amplitude and phase for the particular solution from the values given in Window 2.3 yields
Window 2.3F 0 F_0 F 0
The general response for the undamped case has the form
x ( t ) = A sin ( ω n t + ϕ ) + X cos ω t x(t)=A \sin \left(\omega_n t+\phi\right)+X \cos \omega t x ( t ) = A sin ( ω n t + ϕ ) + X cos ω t
where for the free response
ϕ = tan − 1 ω n x 0 v 0 , A = x 0 2 + v 0 2 ω n 2 , X = 0 \phi=\tan ^{-1} \frac{\omega_n x_0}{v_0}, \quad A=\sqrt{x_0^2+\frac{v_0^2}{\omega_n^2}}, \quad X=0 ϕ = tan − 1 v 0 ω n x 0 , A = x 0 2 + ω n 2 v 0 2 , X = 0
and for the forced response:
ϕ = tan − 1 ω n ( x 0 − X ) v 0 , A = ( v 0 ω n ) 2 + ( x 0 − X ) 2 , X = f 0 ω n 2 − ω 2 \phi=\tan ^{-1} \frac{\omega_n\left(x_0-X\right)}{v_0}, A=\sqrt{\left(\frac{v_0}{\omega_n}\right)^2+\left(x_0-X\right)^2}, X=\frac{f_0}{\omega_n^2-\omega^2} ϕ = tan − 1 v 0 ω n ( x 0 − X ) , A = ( ω n v 0 ) 2 + ( x 0 − X ) 2 , X = ω n 2 − ω 2 f 0
The general response in the underdamped case has the form
x ( t ) = A e − ζ ω n t sin ( ω d t + ϕ ) + X cos ( ω t − θ ) x(t)=A e^{-\zeta \omega_n t} \sin \left(\omega_d t+\phi\right)+X \cos (\omega t-\theta) x ( t ) = A e − ζ ω n t sin ( ω d t + ϕ ) + X cos ( ω t − θ )
where the free response:
ϕ = tan − 1 x 0 ω d v 0 + ζ ω n x 0 , A = 1 ω d ( v 0 + ζ ω n x 0 ) 2 + ( x 0 ω d ) 2 , X = 0 \phi=\tan ^{-1} \frac{x_0 \omega_d}{v_0+\zeta \omega_n x_0}, A=\frac{1}{\omega_d} \sqrt{\left(v_0+\zeta \omega_n x_0\right)^2+\left(x_0 \omega_d\right)^2}, X=0 ϕ = tan − 1 v 0 + ζ ω n x 0 x 0 ω d , A = ω d 1 ( v 0 + ζ ω n x 0 ) 2 + ( x 0 ω d ) 2 , X = 0
and for the forced response
θ = tan − 1 2 ζ ω n ω ω n 2 − ω 2 , X = f 0 ( ω n 2 − ω 2 ) 2 + ( 2 ζ ω n ω ) 2 ϕ = tan − 1 ω d ( x 0 − X cos θ ) v 0 + ( x 0 − X cos θ ) ζ ω n − ω X sin θ and A = x 0 − X cos θ sin ϕ \begin{gathered} \theta=\tan ^{-1} \frac{2 \zeta \omega_n \omega}{\omega_n^2-\omega^2}, \quad X=\frac{f_0}{\sqrt{\left(\omega_n^2-\omega^2\right)^2+\left(2 \zeta \omega_n \omega\right)^2}} \\ \phi=\tan ^{-1} \frac{\omega_d\left(x_0-X \cos \theta\right)}{v_0+\left(x_0-X \cos \theta\right) \zeta \omega_n-\omega X \sin \theta} \text { and } A=\frac{x_0-X \cos \theta}{\sin \phi} \end{gathered} θ = tan − 1 ω n 2 − ω 2 2 ζ ω n ω , X = ( ω n 2 − ω 2 ) 2 + ( 2 ζ ω n ω ) 2 f 0 ϕ = tan − 1 v 0 + ( x 0 − X cos θ ) ζ ω n − ω X sin θ ω d ( x 0 − X cos θ ) and A = sin ϕ x 0 − X cos θ
X = f 0 ( ω n 2 − ω 2 ) 2 + ( 2 ζ ω n ω ) 2 = 0.1 ( 3.01 7 2 − 3 2 ) 2 + ( 2 ⋅ 0.166 ⋅ 3.017 ⋅ 3 ) 2 = 0.033 m X=\frac{f_0}{\sqrt{\left(\omega_n^2-\omega^2\right)^2+\left(2 \zeta \omega_n \omega\right)^2}}=\frac{0.1}{\sqrt{\left(3.017^2-3^2\right)^2+(2 \cdot 0.166 \cdot 3.017 \cdot 3)^2}}=0.033\ m X = ( ω n 2 − ω 2 ) 2 + ( 2 ζ ω n ω ) 2 f 0 = ( 3 . 0 1 7 2 − 3 2 ) 2 + ( 2 ⋅ 0 . 1 6 6 ⋅ 3 . 0 1 7 ⋅ 3 ) 2 0 . 1 = 0 . 0 3 3 m
θ = tan − 1 2 ζ ω n ω ω n 2 − ω 2 = tan − 1 2 ⋅ 0.166 ⋅ 3.017 ⋅ 3 3.01 7 2 − 3 2 = 1.537 r a d \theta=\tan ^{-1} \frac{2 \zeta \omega_n \omega}{\omega_n^2-\omega^2}=\tan ^{-1} \frac{2 \cdot 0.166 \cdot 3.017 \cdot 3}{3.017^2-3^2}=1.537\ rad θ = tan − 1 ω n 2 − ω 2 2 ζ ω n ω = tan − 1 3 . 0 1 7 2 − 3 2 2 ⋅ 0 . 1 6 6 ⋅ 3 . 0 1 7 ⋅ 3 = 1 . 5 3 7 r a d
Next, compute phase for the transient response
ϕ = tan − 1 ω d ( x 0 − X cos θ ) v 0 + ( x 0 − X cos θ ) ζ ω n − ω X sin θ = 2.975 ( 0.001 − 0.033 ⋅ 0.033 ) 0.02 + ( 0.001 − 0.033 ⋅ 0.033 ) 0.166 ⋅ 3.017 − 3 ⋅ 0.033 ⋅ 0.999 = 4.089 × 1 0 − 3 r a d \begin{aligned} \phi & =\tan ^{-1} \frac{\omega_d\left(x_0-X \cos \theta\right)}{v_0+\left(x_0-X \cos \theta\right) \zeta \omega_n-\omega X \sin \theta} \\ & =\frac{2.975(0.001-0.033 \cdot 0.033)}{0.02+(0.001-0.033 \cdot 0.033) 0.166 \cdot 3.017-3 \cdot 0.033 \cdot 0.999}=4.089 \times 10^{-3}\ rad \end{aligned} ϕ = tan − 1 v 0 + ( x 0 − X cos θ ) ζ ω n − ω X sin θ ω d ( x 0 − X cos θ ) = 0 . 0 2 + ( 0 . 0 0 1 − 0 . 0 3 3 ⋅ 0 . 0 3 3 ) 0 . 1 6 6 ⋅ 3 . 0 1 7 − 3 ⋅ 0 . 0 3 3 ⋅ 0 . 9 9 9 2 . 9 7 5 ( 0 . 0 0 1 − 0 . 0 3 3 ⋅ 0 . 0 3 3 ) = 4 . 0 8 9 × 1 0 − 3 r a d
The amplitude of the transient is
A = x 0 − X cos θ sin ϕ = 0.001 − 0.033 ⋅ 0.0333 4.089 × 1 0 − 3 = − 0.027 m A=\frac{x_0-X \cos \theta}{\sin \phi}=\frac{0.001-0.033 \cdot 0.0333}{4.089 \times 10^{-3}}=-0.027\ m A = s i n ϕ x 0 − X c o s θ = 4 . 0 8 9 × 1 0 − 3 0 . 0 0 1 − 0 . 0 3 3 ⋅ 0 . 0 3 3 3 = − 0 . 0 2 7 m
The total response is then written from equation (2.37) as
x ¨ + 2 ζ ω n x ˙ + ω n 2 x = f 0 cos ω t \ddot{x}+2 \zeta \omega_n \dot{x}+\omega_n^2 x=f_0 \cos \omega t x ¨ + 2 ζ ω n x ˙ + ω n 2 x = f 0 cos ω t
x ( t ) = A e − ζ ω n t sin ( ω d t + ϕ ) + X cos ( ω t – θ ) x(t) = Ae^{-ζω_nt} \sin (ω_dt + ϕ) + X \cos (ωt – θ) x ( t ) = A e − ζ ω n t sin ( ω d t + ϕ ) + X cos ( ω t – θ )
= − 0.027 e − 0.5 t sin ( 2.975 t + 4.089 × 1 0 − 3 ) + 0.033 cos ( 3 t – 1.537 ) m = -0.027e^{-0.5t}\sin(2.975t + 4.089 \times 10^{-3}) + 0.033 \cos(3t – 1.537)\ m = − 0 . 0 2 7 e − 0 . 5 t sin ( 2 . 9 7 5 t + 4 . 0 8 9 × 1 0 − 3 ) + 0 . 0 3 3 cos ( 3 t – 1 . 5 3 7 ) m