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7.5 Second-Order Systems and Mechanical Applications*

In this section we apply the matrix methods of Sections 7.2 and 7.3 to investigate the oscillations of typical mass-and-spring systems having two or more degrees of freedom. Our examples are chosen to illustrate phenomena that are generally characteristic of complex mechanical systems.

Figure 7.5.1 shows three masses connected to each other and to two walls by the four indicated springs. We assume that the masses slide without friction and that each spring obeys Hooke’s law—its extension or compression x and force F of reaction are related by the formula F=−kx. $F = - k x .$ We further assume that when the system is at equilibrium (so the masses are stationary) none of the springs is stretched or compressed. If the rightward displacements x1,x2, and x3 $x_{1}, x_{2}, and x_{3}$ of the three masses (from their respective equilibrium positions) are all positive, then

The first spring is stretched the distance x1 $x_{1}$ ;
The second spring is stretched the distance x2−x1 $x_{2} - x_{1}$ ;
The third spring is stretched the distance x3−x2 $x_{3} - x_{2}$ ;
The fourth spring is compressed the distance x3 $x_{3}$ .

FIGURE 7.5.1.

Three spring-coupled masses.

Therefore, application of Newton’s law F=ma $F = m a$ to the three masses (as in Example 1 of Section 7.1) yields their equations of motion:

m 1 x'' 1 m 2 x'' 2 m 3 x'' 3 = = = - k 1 x 1 - k 2 (x 2 - x 1) - k 3 (x 3 - x 2) + + - k 2 (x 2 - x 1), k 3 (x 3 - x 2), k 4 x 3 .

$\begin{aligned} m_{1} x_{1}^{″} & = & - k_{1} x_{1} & + & k_{2} (x_{2} - x_{1}), \\ m_{2} x_{2}^{″} & = & - k_{2} (x_{2} - x_{1}) & + & k_{3} (x_{3} - x_{2}), \\ m_{3} x_{3}^{″} & = & - k_{3} (x_{3} - x_{2)} & - & k_{4} x_{3} . \end{aligned}$ (1)

Although we assumed in writing these equations that the displacements of themasses are all positive, they actually follow similarly from Hooke’s and Newton’s laws, whatever the signs of these displacements.

In terms of the displacement vector x=[x1x2x3]T, $x = {[x_{1} x_{2} x_{3}]}^{T},$ the mass matrix

M = ⎡ ⎣ ⎢ m 1 00 0 m 2 0 00 m 3 ⎤ ⎦ ⎥

$M = [\begin{array}{c} m_{1} & 0 & 0 \\ 0 & m_{2} & 0 \\ 0 & 0 & m_{3} \end{array}]$ (2)

and the stiffness matrix

K = ⎡ ⎣ ⎢ - (k 1 + k 2) k 2 0 k 2 - (k 2 + k 3) k 3 0 k 3 - (k 3 + k 4) ⎤ ⎦ ⎥,

$K = [\begin{array}{c} - (k_{1} + k_{2}) & k_{2} & 0 \\ k_{2} & - (k_{2} + k_{3}) & k_{3} \\ 0 & k_{3} & - (k_{3} + k_{4}) \end{array}],$ (3)

the system in (1) takes the matrix form

M x'' = K x .

$M x^{″} = K x .$ (4)

The notation in Eqs. (1) through (4) generalizes in a natural way to the system of n spring-coupled masses shown in Fig. 7.5.2. We need only write

M = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ m 1 0 ⋮ 0 0 m 2 ⋮ 0 \dots \cdot \dots 00 ⋮ m n ⎤ ⎦ ⎥ ⎥ ⎥ ⎥

$M = [\begin{array}{c} m_{1} & 0 & \dots & 0 \\ 0 & m_{2} & \cdot & 0 \\ ⋮ & ⋮ & ⋮ \\ 0 & 0 & \dots & m_{n} \end{array}]$ (5)

FIGURE 7.5.2.

A system of `n` spring-coupled masses.

and

K = ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ - (k 1 + k 2) k 2 00 ⋮ 00 k 2 - (k 2 + k 3) k 3 0 ⋮ 00 0 k 3 - (k 3 + k 4) k 4 \dots \dots \dots \dots \dots \dots - (k n - 1 + k n) k n 0000 ⋮ k n - (k n + k n + 1) ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

$K = [\begin{array}{c} - (k_{1} + k_{2}) & k_{2} & 0 & \dots & 0 \\ k_{2} & - (k_{2} + k_{3}) & k_{3} & \dots & 0 \\ 0 & k_{3} & - (k_{3} + k_{4}) & \dots & 0 \\ 0 & 0 & k_{4} & \dots & 0 \\ ⋮ & ⋮ & ⋮ \\ 0 & 0 & \dots & - (k_{n - 1} + k_{n}) & k_{n} \\ 0 & 0 & \dots & k_{n} & - (k_{n} + k_{n + 1}) \end{array}]$ (6)

for the mass and stiffness matrices in Eq. (4).

The diagonal matrix M is obviously nonsingular; to get its inverse M−1 $M^{- 1}$ we need only replace each diagonal element with its reciprocal. Hence multiplication of each side in Eq. (4) by M−1 $M^{- 1}$ yields the homogeneous second-order system

x'' = A x,

$x^{″} = A x,$ (7)

where A=M−1K. $A = M^{- 1} K .$ There is a wide variety of frictionless mechanical systems for which a displacement or position vector x, a nonsingular mass matrix M, and a stiffness matrix K satisfying Eq. (4) can be defined.

Solution of Second-Order Systems

To seek a solution of Eq. (7), we substitute (as in Section 7.3 for a first-order system) a trial solution of the form

x (t) = v e α t,

$x (t) = v e^{α t},$ (8)

where v is a constant vector. Then x′′=α2veαt, $x^{″} = α^{2} v e^{α t},$ so substitution of Eq. (8) in (7) gives

α 2 v e α t = A v e α t,

$α^{2} v e^{α t} = A v e^{α t},$

which implies that

A v = α 2 v .

$A v = α^{2} v .$ (9)

Therefore x(t)=veαt $x (t) = v e^{α t}$ is a solution of x′′=Ax $x^{″} = A x$ if and only if α2=λ, $α^{2} = λ,$ an eigenvalue of the matrix A, and v is an associated eigenvector.

If x′′=Ax $x^{″} = A x$ models a mechanical system, then it is typical that the eigenvalues of A are negative real numbers. If

α 2 = λ = - ω 2 < 0,

$α^{2} = λ = - ω^{2} < 0,$

then α=±ωi. $α = \pm ω i .$ In this case the solution given by Eq. (8) is

x (t) = v e i ω t = v (cos ω t + i sin ω t) .

$x (t) = v e^{i ω t} = v (\cos ω t + i \sin ω t) .$

The real and imaginary parts

x 1 (t) = v cos ω t and x 2 (t) = v sin ω t

$x_{1} (t) = v \cos ω t and x_{2} (t) = v \sin ω t$ (10)

of x(t) $x (t)$ are then linearly independent real-valued solutions of the system. This analysis leads to the following theorem.

Remark

The nonzero vector v0 $v_{0}$ is an eigenvector corresponding to λ0=0 $λ_{0} = 0$ provided that Av0=0. If x(t)=(a0+b0t)v0, $A v_{0} = 0. If x (t) = (a_{0} + b_{0} t) v_{0,}$ then

x'' = 0 \cdot v 0 = (a 0 + b 0 t) \cdot 0 = (a 0 + b 0 t) \cdot (A v 0) = A x,

$x^{″} = 0 \cdot v_{0} = (a_{0} + b_{0} t) \cdot 0 = (a_{0} + b_{0} t) \cdot (A v_{0}) = A x,$

thus verifying the form in Eq. (12).

Example 1

Mass-spring system Consider the mass-and-spring system with n=2 $n = 2$ shown in Fig. 7.5.3. Because there is no third spring connected to a right-hand wall, we set k3=0. $k_{3} = 0.$ If m1=2,m2=1,k1=100, $m_{1} = 2, m_{2} = 1, k_{1} = 100,$ and k2=50, $k_{2} = 50,$ then the equation Mx′′=Kx $M x^{″} = K x$ is

[2001] x'' = [- 150 50 50 - 50] x,

$[\begin{array}{c} 2 & 0 \\ 0 & 1 \end{array}] x^{″} = [\begin{array}{r} - 150 & 50 \\ 50 & - 50 \end{array}] x,$ (13)

which reduces to x′′=Ax $x^{″} = A x$ with

A = [- 75 50 25 - 50] .

$A = [\begin{array}{r} - 75 & 25 \\ 50 & - 50 \end{array}] .$

FIGURE 7.5.3.

The mass-and-spring system of Example 1.

The characteristic equation of A is

(- 75 - λ) (- 50 - λ) - 50 \cdot 25 = = λ 2 + 125 λ + 2500 (λ + 25) (λ + 100) = 0,

$\begin{aligned} (- 75 - λ) (- 50 - λ) - 50 \cdot 25 & = & λ^{2} + 125 λ + 2500 \\ = & (λ + 25) (λ + 100) = 0, \end{aligned}$

so A has the negative eigenvalues λ1=−25 $λ_{1} = - 25$ and λ2=−100. $λ_{2} = - 100.$ By Theorem 1, the system in (13) therefore has solutions with [circular] frequencies ω1=5 $ω_{1} = 5$ and ω2=10. $ω_{2} = 10.$

Case 1: λ1=−25. $λ_{1} = - 25.$ The eigenvector equation (A−λI)v=0 $(A - λ I) v = 0$ is

[- 50 50 25 - 25] [a b] = [00],

$[\begin{array}{r} - 50 & 25 \\ 50 & - 25 \end{array}] [\begin{array}{l} a \\ b \end{array}] = [\begin{array}{l} 0 \\ 0 \end{array}],$

so an eigenvector associated with λ1=−25 is v1=[12]T. $λ_{1} = - 25 is v_{1} = {[\begin{array}{l} 1 & 2 \end{array}]}^{T} .$

Case 2: λ2=−100. $λ_{2} = - 100.$ The eigenvector equation (A−λI)v=0 $(A - λ I) v = 0$ is

[25502550] [a b] = [00],

$[\begin{array}{c} 25 & 25 \\ 50 & 50 \end{array}] [\begin{matrix} a \\ b \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}],$

so an eigenvector associated with λ2=−100 is v2=[1−1]T. $λ_{2} = - 100 is v_{2} = {[\begin{array}{l} 1 & - 1 \end{array}]}^{T} .$

By Eq. (11) it follows that a general solution of the system in (13) is given by

x (t) = (a 1 cos 5 t + b 1 sin 5 t) v 1 + (a 2 cos 10 t + b 2 sin 10 t) v 2 .

$x (t) = (a_{1} \cos 5 t + b_{1} \sin 5 t) v_{1} + (a_{2} \cos 10 t + b_{2} \sin 10 t) v_{2} .$ (14)

The two terms on the right in Eq. (14) represent free oscillations of the mass-and-spring system. They describe the physical system’s two natural modes of oscillation at its two [circular] natural frequencies ω1=5 and ω2=10. $ω_{1} = 5 and ω_{2} = 10.$ The natural mode

x 1 (t) = (a 1 cos 5 t + b 1 sin 5 t) v 1 = c 1 cos (5 t - α 1) [12]

$x_{1} (t) = (a_{1} \cos 5 t + b_{1} \sin 5 t) v_{1} = c_{1} \cos (5 t - α_{1}) [\begin{array}{l} 1 \\ 2 \end{array}]$

(with c1=a21+b21−−−−−−√,cosα1=a1/c1, $c_{1} = \sqrt{a_{1}^{2} + b_{1}^{2}}, \cos α_{1} = a_{1} / c_{1},$ and sinα1=b1/c1) $\sin α_{1} = b_{1} / c_{1})$ has the scalar component equations

x 1 (t) x 2 (t) = = c 1 2 c 1 cos (5 t - α 1), cos (5 t - α 1),

$\begin{array}{r} x_{1} (t) & = & c_{1} & \cos (5 t - α 1), \\ x_{2} (t) & = & 2 c_{1} & \cos (5 t - α 1), \end{array}$ (15)

and therefore describes a free oscillation in which the two masses move in synchrony in the same direction and with the same frequency ω1=5, $ω_{1} = 5,$ but with the amplitude of motion of m2 $m_{2}$ twice that of m1 $m_{1}$ (see Fig. 7.5.4). The natural mode

x 2 (t) = (a 2 cos 10 t + b 2 sin 10 t) v 2 = c 2 cos (10 t - α 2) [1 - 1]

$x_{2} (t) = (a_{2} \cos 10 t + b_{2} \sin 10 t) v_{2} = c_{2} \cos (10 t - α_{2}) [\begin{array}{r} 1 \\ - 1 \end{array}]$

FIGURE 7.5.4.

Oscillations in the same direction with frequency ω1=5; $ω_{1} = 5;$ the amplitude of motion of mass 2 is twice that of mass 1.

has the scalar component equations

x 1 (t) x 2 (t) = = c 2 cos (10 t - α 2), - c 2 cos (10 t - α 2),

$\begin{array}{r} x_{1} (t) & = & c_{2} \cos (10 t - α_{2}), \\ x_{2} (t) & = & - c_{2} \cos (10 t - α 2), \end{array}$ (16)

and therefore describes a free oscillation in which the two masses move in synchrony in opposite directions with the same frequency ω2=10 $ω_{2} = 10$ and with equal amplitudes of oscillation (see Fig. 7.5.5).

FIGURE 7.5.5.

Oscillations in opposite directions with frequency ω2=10; $ω_{2} = 10;$ the amplitudes of motion of the two masses are the same.

Example 2

Railway cars Figure 7.5.6 shows three railway cars connected by buffer springs that react when compressed, but disengage instead of stretching. With n=3,k2=k3=k, $n = 3, k_{2} = k_{3} = k,$ and k1=k4=0 $k_{1} = k_{4} = 0$ in Eqs. (2) through (4), we get the system

⎡ ⎣ ⎢ m 1 00 0 m 2 0 00 m 3 ⎤ ⎦ ⎥ x'' = ⎡ ⎣ ⎢ - k k 0 k - 2 k k 0 k - k ⎤ ⎦ ⎥ x,

$[\begin{array}{c} m_{1} & 0 & 0 \\ 0 & m_{2} & 0 \\ 0 & 0 & m_{3} \end{array}] x^{″} = [\begin{array}{r} - k & k & 0 \\ k & - 2 k & k \\ 0 & k & - k \end{array}] x,$ (17)

which is equivalent to

x'' = ⎡ ⎣ ⎢ - c 1 c 2 0 c 1 - 2 c 2 c 3 0 c 2 - c 3 ⎤ ⎦ ⎥ x

$x^{″} = [\begin{array}{r} - c_{1} & c_{1} & 0 \\ c_{2} & - 2 c_{2} & c_{2} \\ 0 & c_{3} & - c_{3} \end{array}] x$ (18)

with

c i = k m i (i = 1, 2, 3) .

$c_{i} = \frac{k}{m_{i}} (i = 1, 2, 3) .$ (19)

FIGURE 7.5.6.

The three railway cars of Example 2.

If we assume further that m1=m3, $m_{1} = m_{3},$ so that c1=c3, $c_{1} = c_{3},$ then a brief computation gives

- λ (λ + c 1) (λ + c 1 + 2 c 2) = 0

$- λ (λ + c_{1}) (λ + c_{1} + 2 c_{2}) = 0$ (20)

for the characteristic equation of the coefficient matrix A in Eq. (18). Hence the matrix A has eigenvalues

λ 1 = 0, λ 2 = - c 1, λ 3 = - c 1 - 2 c 2

$λ_{1} = 0, λ_{2} = - c_{1}, λ_{3} = - c_{1} - 2 c_{2}$ (21a)

corresponding to the natural frequencies

ω 1 = 0, ω 2 = c 1 - - \sqrt, ω 3 = c 1 + 2 c 2 - - - - - - - \sqrt

$ω_{1} = 0, ω_{2} = \sqrt{c_{1}}, ω_{3} = \sqrt{c_{1} + 2 c_{2}}$ (21b)

of the physical system.

For a numerical example, suppose that the first and third railway cars weigh 12 tons each, that the middle car weighs 8 tons, and that the spring constant is k=1:5 $k = 1 : 5$ tons/ft; i.e., k=3000 $k = 3000$ lb/ft. Then, using fps units with mass measured in slugs (a weight of 32 pounds has a mass of 1 slug), we have

m 1 = m 3 = 750, m 2 = 500,

$m_{1} = m_{3} = 750, m_{2} = 500,$

and

c 1 = 3000 750 = 4, c 2 = 3000 500 = 6.

$c_{1} = \frac{3000}{750} = 4, c_{2} = \frac{3000}{500} = 6.$

Hence the coefficient matrix A is

A = ⎡ ⎣ ⎢ - 4 60 4 - 12 4 06 - 4 ⎤ ⎦ ⎥,

$A = [\begin{array}{r} - 4 & 4 & 0 \\ 6 & - 12 & 6 \\ 0 & 4 & - 4 \end{array}],$ (22)

and the eigenvalue-frequency pairs given by (21a) and (21b) are λ1=0,ω1=0;λ2=−4,ω2=2; $λ_{1} = 0, ω_{1} = 0; λ_{2} = - 4, ω_{2} = 2;$ and λ3=−16,ω3=4. $λ_{3} = - 16, ω_{3} = 4.$

Case 1: λ1=0,ω1=0. $λ_{1} = 0, ω_{1} = 0.$ The eigenvector equation (A−λI)v=0 $(A - λ I) v = 0$ is

A v = ⎡ ⎣ ⎢ - 4 60 4 - 12 4 06 - 4 ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ a b c ⎤ ⎦ ⎥ = ⎡ ⎣ ⎢ 000 ⎤ ⎦ ⎥,

$A v = [\begin{array}{r} - 4 & 4 & 0 \\ 6 & - 12 & 6 \\ 0 & 4 & - 4 \end{array}] [\begin{matrix} a \\ b \\ c \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}],$ (22)

so it is clear that v1=[111]T $v_{1} = {[\begin{array}{r} 1 & 1 & 1 \end{array}]}^{T}$ is an eigenvector associated with λ1=0. $λ_{1} = 0.$ According to Theorem 1, the corresponding part of a general solution of x′′=Ax $x^{″} = A x$ is

x 1 (t) = (a 1 + b 1 t) v 1 .

$x_{1} (t) = (a_{1} + b_{1} t) v_{1} .$

Case 2: λ2=−4,ω2=2. $λ_{2} = - 4, ω_{2} = 2.$ The eigenvector equation (A−λI)v=0 $(A - λ I) v = 0$ is

(A + 4 I) v = ⎡ ⎣ ⎢ 060 4 - 8 4 060 ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ a b c ⎤ ⎦ ⎥ = ⎡ ⎣ ⎢ 000 ⎤ ⎦ ⎥,

$(A + 4 I) v = [\begin{array}{r} 0 & 4 & 0 \\ 6 & - 8 & 6 \\ 0 & 4 & 0 \end{array}] [\begin{matrix} a \\ b \\ c \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}],$

so it is clear that v2=[10−1]T $v_{2} = {[\begin{array}{r} 1 & 0 & - 1 \end{array}]}^{T}$ is an eigenvector associated with λ2=−4. $λ_{2} = - 4.$ According to Theorem 1, the corresponding part of a general solution of x′′=Ax $x^{″} = A x$ is

x 2 (t) = (a 2 cos 2 t + b 2 sin 2 t) v 2 .

$x_{2} (t) = (a_{2} \cos 2 t + b_{2} \sin 2 t) v_{2} .$

Case 3: λ3=−16,ω3=4. $λ_{3} = - 16, ω_{3} = 4.$ The eigenvector equation (A−λI)v=0 $(A - λ I) v = 0$ is

(A + 16 I) v = ⎡ ⎣ ⎢ 12604440612 ⎤ ⎦ ⎥ ⎡ ⎣ ⎢ a b c ⎤ ⎦ ⎥ = ⎡ ⎣ ⎢ 000 ⎤ ⎦ ⎥,

$(A + 16 I) v = [\begin{array}{r} 12 & 4 & 0 \\ 6 & 4 & 6 \\ 0 & 4 & 12 \end{array}] [\begin{matrix} a \\ b \\ c \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}],$

so it is clear that v3=[1−31]T $v_{3} = {[\begin{array}{r} 1 & - 3 & 1 \end{array}]}^{T}$ is an eigenvector associated with λ3=−16. $λ_{3} = - 16.$ According to Theorem 1, the corresponding part of a general solution of x′′=Ax $x^{″} = A x$ is

x 3 (t) = (a 3 cos 4 t + b 3 sin 4 t) v 3 .

$x_{3} (t) = (a_{3} \cos 4 t + b_{3} \sin 4 t) v_{3} .$

The general solution x=x1+x2+x3 of x′′=Ax $x = x_{1} + x_{2} + x_{3} of x^{″} = A x$ is therefore given by

x (t) = a 1 ⎡ ⎣ ⎢ 111 ⎤ ⎦ ⎥ + + b 1 t ⎡ ⎣ ⎢ 111 ⎤ ⎦ ⎥ + a 2 ⎡ ⎣ ⎢ 10 - 1 ⎤ ⎦ ⎥ cos 2 t b 2 ⎡ ⎣ ⎢ 10 - 1 ⎤ ⎦ ⎥ sin 2 t + a 3 ⎡ ⎣ ⎢ 1 - 3 1 ⎤ ⎦ ⎥ cos 4 t + b 3 ⎡ ⎣ ⎢ 1 - 3 1 ⎤ ⎦ ⎥ sin 4 t .

$\begin{aligned} x (t) = a_{1} [\begin{array}{r} 1 \\ 1 \\ 1 \end{array}] & + & b_{1} t [\begin{array}{r} 1 \\ 1 \\ 1 \end{array}] + a_{2} [\begin{array}{r} 1 \\ 0 \\ - 1 \end{array}] \cos 2 t \\ + & b_{2} [\begin{array}{r} 1 \\ 0 \\ - 1 \end{array}] \sin 2 t + a_{3} [\begin{array}{r} 1 \\ - 3 \\ 1 \end{array}] \cos 4 t + b_{3} [\begin{array}{r} 1 \\ - 3 \\ 1 \end{array}] \sin 4 t . \end{aligned}$ (23)

To determine a particular solution, let us suppose that the leftmost car is moving to the right with velocity v0 $v_{0}$ and at time t=0 $t = 0$ strikes the other two cars, which are together but at rest. The corresponding initial conditions are

x 1 (0) = x 2 (0) = x 3 (0) = 0,

$x_{1} (0) = x_{2} (0) = x_{3} (0) = 0,$ (24a)

x' 1 (0) = v 0, x' 2 (0) = x' 3 (0) = 0.

$x_{1}^{′} (0) = v_{0}, x_{2}^{′} (0) = x_{3}^{′} (0) = 0.$ (24b)

Then substitution of (24a) in (23) gives the scalar equations

a 1 a 1 a 1 + - a 2 a 2 + - + a 3 3 a 3 a 3 = 0, = 0, = 0,

$\begin{array}{r} a_{1} & + & a_{2} & + & a_{3} & = 0, \\ a_{1} & - & 3 a_{3} & = 0, \\ a_{1} & - & a_{2} & + & a_{3} & = 0, \end{array}$

which readily yield a1=a2=a3=0. $a_{1} = a_{2} = a_{3} = 0.$ Hence the position functions of the three cars are

x 1 (t) x 2 (t) x 3 (t) = = = b 1 t b 1 t b 1 t + - b 2 sin 2 t b 2 sin 2 t + - + b 3 sin 4 t, 3 b 3 sin 4 t, b 3 sin 4 t,

$\begin{array}{r} x_{1} (t) & = & b_{1} t & + & b_{2} \sin 2 t & + & b_{3} \sin 4 t, \\ x_{2} (t) & = & b_{1} t & - & 3 b_{3} \sin 4 t, \\ x_{3} (t) & = & b_{1} t & - & b_{2} \sin 2 t & + & b_{3} \sin 4 t, \end{array}$ (25)

and their velocity functions are

x! 1 (t) x! 2 (t) x! 3 (t) = = = b 1 b 1 b 1 + - 2 b 2 cos 2 t 2 b 2 cos 2 t + - + 4 b 3 cos 4 t, 12 b 3 cos 4 t, 4 b 3 cos 4 t .

$\begin{array}{r} x_{1}^{!} (t) & = & b_{1} & + & 2 b_{2} \cos 2 t & + & 4 b_{3} \cos 4 t, \\ x_{2}^{!} (t) & = & b_{1} & - & 12 b_{3} \cos 4 t, \\ x_{3}^{!} (t) & = & b_{1} & - & 2 b_{2} \cos 2 t & + & 4 b_{3} \cos 4 t . \end{array}$ (26)

Substitution of (24b) in (26) gives the equations

b 1 b 1 b 1 + - 2 b 2 2 b 2 + - + 4 b 3 12 b 3 4 b 3 = = = v 0, 0, 0

$\begin{array}{r} b_{1} & + & 2 b_{2} & + & 4 b_{3} & = & v_{0}, \\ b_{1} & - & 12 b_{3} & = & 0, \\ b_{1} & - & 2 b_{2} & + & 4 b_{3} & = & 0 \end{array}$

that readily yield b1=38v0, b2=14v0, $b_{1} = \frac{3}{8} v_{0}, b_{2} = \frac{1}{4} v_{0},$ and b3=132v0. $b_{3} = \frac{1}{32} v_{0} .$ Finally, the position functions in are

x 1 (t) x 2 (t) x 3 (t) = = = 1 32 v 0 (12 t 1 32 v 0 (12 t 1 32 v 0 (12 t + - 8 sin 2 t 8 sin 2 t + - + sin 4 t), 3 sin 4 t), sin 4 t) .

$\begin{array}{r} x_{1} (t) & = & \frac{1}{32} v_{0} (12 t & + & 8 \sin 2 t & + & \sin 4 t), \\ x_{2} (t) & = & \frac{1}{32} v_{0} (12 t & - & 3 \sin 4 t), \\ x_{3} (t) & = & \frac{1}{32} v_{0} (12 t & - & 8 \sin 2 t & + & \sin 4 t) . \end{array}$ (27)

But these equations hold only so long as the two buffer springs remain compressed; that is, while both

x 2 - x 1 < 0 and x 3 - x 2 < 0.

$x_{2} - x_{1} < 0 and x_{3} - x_{2} < 0.$

To discover what this implies about t, we compute

x 2 (t) - x 1 (t) = = = 1 32 v 0 (- 8 sin 2 t - 4 sin 4 t) - 1 32 v 0 (8 sin 2 t + 8 sin 2 t cos 2 t) - 1 4 v 0 (sin 2 t) (1 + cos 2 t)

$\begin{aligned} x_{2} (t) - x_{1} (t) & = & \frac{1}{32} v_{0} (- 8 \sin 2 t - 4 \sin 4 t) \\ = & - \frac{1}{32} v_{0} (8 \sin 2 t + 8 \sin 2 t \cos 2 t) \\ = & - \frac{1}{4} v_{0} (\sin 2 t) (1 + \cos 2 t) \end{aligned}$

and, similarly,

$x_{3} (t) - x_{2} (t) = - \frac{1}{4} v_{0} (\sin 2 t) (1 - \cos 2 t) .$

It follows that $x_{2} - x_{1} < 0 and x_{3} - x_{2} < 0 until t = π / 2 \approx 1.57$ (seconds), at which time the equations in (26) and (27) give the values

$\begin{aligned} x_{1} (\frac{π}{2}) & = & x_{2} (\frac{π}{2}) = x_{3} (\frac{π}{2}) = \frac{3 π v_{0}}{16}, \\ x_{1}^{′} (\frac{π}{2}) & = & x_{2}^{′} (\frac{π}{2}) = 0, x_{3}^{′} (\frac{π}{2}) = v_{0} . \end{aligned}$

We conclude that the three railway cars remain engaged and moving to the right until disengagement occurs at time $t = π / 2.$ Thereafter, cars 1 and 2 remain at rest (!), while car 3 continues to the right with speed $v_{0} .$ If, for instance, $v_{0} = 48$ feet per second (about 33 miles per hour), then the three cars travel a distance of $9 π \approx 28.27$ (ft) during their 1:57 seconds of engagement, and

$x_{1} (t) = x_{2} (t) = 9 π, x_{3} (t) = 48 t - 15 π$ (27′)

for $t > π / 2.$ Figure 7.5.7 illustrates the “before” and “after” situations, and Fig. 7.5.8 shows the graphs of the functions $x_{1} (t), x_{2} (t), and x_{3} (t)$ in Eqs. (27) and (27′).

FIGURE 7.5.8.

Position functions of the three railway cars of Example 2.

Forced Oscillations and Resonance

Suppose now that the ith mass of the mass-and-spring system in Fig. 7.5.2 is subject to an external force $F_{i} (i = 1, 2, \dots, n)$ in addition to the forces exerted by the springs attached to it. Then the homogeneous equation $M x ″ = K x$ is replaced with the nonhomogeneous equation

$M x^{″} = K x + F,$ (28)

where $F = {[\begin{array}{r} F_{1} & F_{2} & \dots & F_{n} \end{array}]}^{T}$ is the external force vector for the system. Multiplication by $M^{- 1}$ yields

$x^{″} = A x + f,$ (29)

where f is the external force vector per unit mass. We are especially interested in the case of a periodic external force

$f (t) = F_{0} \cos ω t$ (30)

(where $F_{0}$ is a constant vector). We then anticipate a periodic particular solution

$x_{p} (t) = c \cos w t$ (31)

with the known external frequency $ω$ and with a coefficient vector c yet to be determined. Because $x_{p}^{″} = - ω^{2} c \cos ω t,$ substitution of (30) and (31) in (29), followed by cancellation of the common factor $\cos ω t$ , gives the linear system

$(A + ω^{2} I) c = - F_{0}$ (32)

to be solved for c.

Observe that the matrix $A + ω^{2} I$ is nonsingular—in which case Eq. (32) can be solved for c—unless $- ω^{2} = λ$ , an eigenvalue of A. Thus a periodic particular solution of the form in Eq. (31) exists provided that the external forcing frequency does not equal one of the natural frequencies $ω_{1}, ω_{2}, \dots, ω_{n}$ of the system. The case in which $ω$ is a natural frequency corresponds to the phenomenon of resonance discussed in Section 5.6.

Example 3

Mass-spring resonance Suppose that the second mass in Example 1 is subjected to the external periodic force $50 \cos ω t$ . Then with $m_{1} = 2, m_{2} = 1, k_{1} = 100, k_{2} = 50$ , and $F_{0} = 50$ in Fig. 7.5.9, Eq. (29) takes the form

$x^{″} = [\begin{array}{r} - 75 & 25 \\ 50 & - 50 \end{array}] x + [\begin{array}{r} 0 \\ 50 \end{array}] \cos ω t,$ (33)

FIGURE 7.5.9.

The forced mass-and-spring system of Example 3.

and the substitution $x = c \cos ω t$ leads to the equation

$[\begin{array}{c} ω^{2} - 75 & 25 \\ 50 & ω^{2} - 50 \end{array}] c = [\begin{array}{r} 0 \\ - 50 \end{array}]$ (34)

for the coefficient vector $c = {[\begin{array}{r} c_{1} & c_{2} \end{array}]}^{T} .$ This system is readily solved for

$c_{1} = \frac{1250}{(ω^{2} - 25) (ω^{2} - 100)}, c_{2} = \frac{50 (ω^{2} - 75)}{(ω^{2} - 25) (ω^{2} - 100)} .$ (35)

For instance, if the external squared frequency is $ω^{2} = 50$ , then (35) yields $c_{1} = - 1, c_{2} = - 1$ . The resulting forced periodic oscillation is described by

$x_{1} (t) = - \cos ω t, x_{2} (t) = - \cos ω t .$

Thus the two masses oscillate in synchrony with equal amplitudes and in the same direction.

If the external squared frequency is $ω^{2} = 125$ , then (35) yields $c_{1} = \frac{1}{2}, c_{2} = - 1.$ The resulting forced periodic oscillation is described by

$x_{1} (t) = \frac{1}{2} \cos ω t, x_{2} (t) = - \cos ω t .$

and now the two masses oscillate in synchrony in opposite directions, but with the amplitude of motion of $m_{2}$ twice that of $m_{1}$ .

It is evident from the denominators in (35) that $c_{1}$ and $c_{2}$ approach $+ \infty$ as $ω$ approaches either of the two natural frequencies $ω_{1} = 5$ and $ω_{2} = 10$ (found in Example 1). Figure 7.5.10 shows a plot of the amplitude $\sqrt{c_{1}^{2} + c_{2}^{2}}$ of the forced periodic solution $x (t) = c \cos ω t$ as a function of the forced frequency $ω$ . The peaks at $ω_{2} = 5$ and $ω_{2} = 10$ exhibit visually the phenomenon of resonance.

FIGURE 7.5.10.

Frequency–amplitude plot for Example 3.

Periodic and Transient Solutions

It follows from Theorem 4 of Section 7.2 that a particular solution of the forced system

$x ″ = A x + F_{0} \cos ω t$ (36)

will be of the form

$x (t) = x_{c} (t) + x_{p} (t),$ (37)

where $x_{p} (t)$ is a particular solution of the nonhomogeneous system and $x_{c} (t)$ is a solution of the corresponding homogeneous system. It is typical for the effects of frictional resistance in mechanical systems to damp out the complementary function solution $x_{c} (t)$ , so that

$x_{c} (t) \to 0 as t \to + \infty .$ (38)

Hence $x_{c} (t)$ is a transient solution that depends only on the initial conditions; it dies out with time, leaving the steady periodic solution $x_{p} (t)$ resulting from the external driving force:

$x (t) \to x_{p} (t) as t \to + \infty .$ (39)

As a practical matter, every physical system includes frictional resistance (however small) that damps out transient solutions in this manner.

7.5 Problems

Problems 1 through 7 deal with the mass-and-spring system shown in Fig. 7.5.11 with stiffness matrix

$K = [\begin{array}{c} - (k_{1} + k_{2}) & k_{2} \\ k_{2} & - (k_{2} + k_{3}) \end{array}]$

and with the given mks values for the masses and spring constants. Find the two natural frequencies of the system and describe its two natural modes of oscillation.

FIGURE 7.5.11.

The mass-and-spring system for Problems 1 through 6.

$m_{1} = m_{2} = 1; k_{1} = 0, k_{2} = 2, k_{3} = 0$ (no walls)
$m_{1} = m_{2} = 1; k_{1} = 1, k_{2} = 4, k_{3} = 1$
$m_{1} = 1, m_{2} = 2; k_{1} = 1, k_{2} = k_{3} = 2$
$m_{1} = m_{2} = 1; k_{1} = 1, k_{2} = 2, k_{3} = 1$
$m_{1} = m_{2} = 1; k_{1} = 2, k_{2} = 1, k_{3} = 2$
$m_{1} = 1, m_{2} = 2; k_{1} = 2, k_{2} = k_{3} = 4$
$m_{1} = m_{2} = 1; k_{1} = 4, k_{2} = 6, k_{3} = 4$

In Problems 8 through 10 the indicated mass-and-spring system is set in motion from rest $(x_{1}^{′} (0) = x_{2}^{′} (0) = 0$ ) in its equilibrium position ( $x_{1} (0) = x_{2} (0) = 0$ ) with the given external forces $F_{1} (t)$ and $F_{2} (t)$ acting on the masses $m_{1}$ and $m_{2}$ , respectively. Find the resulting motion of the system and describe it as a superposition of oscillations at three different frequencies.

The mass-and-spring system of Problem 2, with $F_{1} (t) = 96 \cos 5 t, F_{2} (t) \equiv 0$
The mass-and-spring system of Problem 3, with $F_{1} (t) \equiv 0, F_{2} (t) = 120 \cos 3 t$
The mass-and-spring system of Problem 7, with $F_{1} (t) = 30 \cos t, F_{2} (t) = 60 \cos t$
Consider a mass-and-spring system containing two masses $m_{1} = 1$ and $m_{2} = 1$ whose displacement functions x(t) and y(t) satisfy the differential equations

$\begin{array}{r} x ″ & = & - 40 x & + & 8 y, \\ y ″ & = & 12 x & - & 60 y . \end{array}$

(a) Describe the two fundamental modes of free oscillation of the system. (b) Assume that the two masses start in motion with the initial conditions

$x (0) = 19, x^{′} (0) = 12$

and

$y (0) = 3, y^{′} (0) = 6$

and are acted on by the same force, $F_{1} (t) = F_{2} (t) = - 195 \cos 7 t$ . Describe the resulting motion as a superposition of oscillations at three different frequencies.

In Problems 12 and 13, find the natural frequencies of the three-mass system of Fig. 7.5.1, using the given masses and spring constants. For each natural frequency $ω$ , give the ratio $a_{1} : a_{2} : a_{3}$ of amplitudes for a corresponding natural mode $x_{1} = a_{1} \cos ω t, x_{2} = a_{2} \cos ω t, x_{3} = a_{3} \cos ω t$ .

$m_{1} = m_{2} = m_{3} = 1; k_{1} = k_{2} = k_{3} = k_{4} = 1$
$m_{1} = m_{2} = m_{3} = 1; k_{1} = k_{2} = k_{3} = k_{4} = 2$ (Hint: One eigenvalue is $λ = - 4$ .)
In the system of Fig. 7.5.12, assume that $m_{1} = 1, k_{1} = 50, k_{2} = 10$ , and $F_{0} = 5$ in mks units, and that $ω = 10$ . Then find $m_{2}$ so that in the resulting steady periodic oscillations, the mass $m_{1}$ will remain at rest(!). Thus the effect of the second mass-and-spring pair will be to neutralize the effect of the force on the first mass. This is an example of a dynamic damper. It has an electrical analogy that some cable companies use to prevent your reception of certain cable channels.

FIGURE 7.5.12.

The mechanical system of Problem 14.
Suppose that $m_{1} = 2, m_{2} = \frac{1}{2}, k_{1} = 75, k_{2} = 25, F_{0} = 100$ , and $ω = 10$ (all in mks units) in the forced mass-and-spring system of Fig. 7.5.9. Find the solution of the system $M x ″ = K x + F$ that satisfies the initial conditions $x (0) = x ′ (0) = 0$ .

In Problems 16 through 19 we apply the analysis of Example 2 to a system of two railway cars.

Figure 7.5.13 shows two railway cars with a buffer spring. We want to investigate the transfer of momentum that occurs after car 1 with initial velocity $v_{0}$ impacts car 2 at rest. The analog of Eq. (18) in the text is

$x^{″} = [\begin{array}{r} - c_{1} & c_{1} \\ c_{2} & - c_{2} \end{array}] x$

with $c_{i} = k / m_{i}$ for $i = 1, 2$ . Show that the eigenvalues of the coefficient matrix A are $λ_{1} = 0$ and $λ_{2} = - c_{1} - c_{2}$ , with associated eigenvectors $v_{1} = {[\begin{array}{r} 1 & 1 \end{array}]}^{T}$ and $v_{2} = {[\begin{array}{r} c_{1} & - c_{2} \end{array}]}^{T} .$

FIGURE 7.5.13.

The two railway cars of Problems 16 through 19.
If the two cars of Problem 16 both weigh 16 tons (so that $m_{1} = m_{2} = 1000$ (slugs)) and $k = 1$ ton/ft (that is, 2000 lb/ft), show that the cars separate after $π / 2$ seconds, and that $x_{1}^{′} (t) = 0$ and $x_{2}^{′} (t) = v_{0}$ thereafter. Thus the original momentum of car 1 is completely transferred to car 2.
If cars 1 and 2 weigh 8 and 16 tons, respectively, and $k = 3000$ lb/ft, show that the two cars separate after $π / 3$ seconds, and that

$x_{1}^{′} (t) = - \frac{1}{3} v_{0} and x_{2}^{′} (t) = + \frac{2}{3} v_{0}$

thereafter. Thus the two cars rebound in opposite directions.
If cars 1 and 2 weigh 24 and 8 tons, respectively, and $k = 1500$ lb/ft, show that the cars separate after $π / 2$ seconds, and that

$x_{1}^{′} (t) = + \frac{1}{2} v_{0} and x_{2}^{′} (t) = + \frac{3}{2} v_{0}$

thereafter. Thus both cars continue in the original direction of motion, but with different velocities.

Problems 20 through 23 deal with the same system of three railway cars (same masses) and two buffer springs (same spring constants) as shown in Fig. 7.5.6 and discussed in Example 2. The cars engage at time $t = 0$ with $x_{1} (0) = x_{2} (0) = x_{3} (0) = 0$ and with the given initial velocities (where $v_{0} = 48$ ft/s). Show that the railway cars remain engaged until $t = π / 2$ (s), after which time they proceed in their respective ways with constant velocities. Determine the values of these constant final velocities $x_{1}^{′} (t), x_{2}^{′} (t)$ , and $x_{3}^{′} (t)$ of the three cars for $t > π / 2$ . In each problem you should find (as in Example 2) that the first and third railway cars exchange behaviors in some appropriate sense.

$x_{1}^{′} (0) = v_{0}, x_{2}^{′} (0) = 0, x_{3}^{′} (0) = - v_{0}$
$x_{1}^{′} (0) = 2 v_{0}, x_{2}^{′} (0) = 0, x_{3}^{′} (0) = - v_{0}$
$x_{1}^{′} (0) = v_{0}, x_{2}^{′} (0) = v_{0}, x_{3}^{′} (0) = - 2 v_{0}$
$x_{1}^{′} (0) = 3 v_{0}, x_{2}^{′} (0) = 2 v_{0}, x_{3}^{′} (0) = 2 v_{0}$
In the three-railway-car system of Fig. 7.5.6, suppose that cars 1 and 3 each weigh 32 tons, that car 2 weighs 8 tons, and that each spring constant is 4 tons/ft. If $x_{1}^{′} (0) = v_{0}$ and $x_{2}^{′} (0) = x_{3}^{′} (0) = 0$ , show that the two springs are compressed until $t = π / 2$ and that

$x_{1}^{′} (t) = - \frac{1}{9} v_{0} and x_{2}^{′} (t) = x_{3}^{′} (t) = + \frac{8}{9} v_{0}$

thereafter. Thus car 1 rebounds, but cars 2 and 3 continue with the same velocity.

The Two-Axle Automobile

In Example 4 of Section 5.6 we investigated the vertical oscillations of a one-axle car—actually a unicycle. Now we can analyze a more realistic model: a car with two axles and with separate front and rear suspension systems. Figure 7.5.14 represents the suspension system of such a car. We assume that the car body acts as would a solid bar of mass m and length $L = L_{1} + L_{2}$ . It has moment of inertia I about its center of mass C, which is at distance $L_{1}$ from the front of the car. The car has front and back suspension springs with Hooke’s constants $k_{1}$ and $k_{2}$ , respectively. When the car is in motion, let x(t) denote the vertical displacement of the center of mass of the car from equilibrium; let $θ (t)$ denote its angular displacement (in radians) from the horizontal. Then Newton’s laws of motion for linear and angular acceleration can be used to derive the equations

$\begin{aligned} m x ″ & = & - (k_{1} + k_{2}) x + (k_{1} L_{1} - k_{2} L_{2}) θ, \\ I θ ″ & = & (k_{1} L_{1} - k_{2} L_{2}) x - (k_{1} L_{1}^{2} - k_{2} L_{2}^{2}) θ . \end{aligned}$ (40)

FIGURE 7.5.14.

Model of the two-axle automobile.

Suppose that $m = 75$ slugs (the car weighs 2400 lb), $L_{1} = 7$ ft, $L_{2} = 3$ ft (it’s a rear-engine car), $k_{1} = k_{2} = 2000$ lb/ft, and $I = 1000$ $ft \cdot lb \cdot s^{2}$ . Then the equations in (40) take the form

$\begin{array}{r} 75 x^{″} & + & 4000 x & - & 8000 θ & = & 0, \\ 1000 θ^{″} & - & 8000 x & + & 116, 000 θ & = & 0. \end{array}$

(a) Find the two natural frequencies $ω_{1}$ and $ω_{2}$ of the car. (b) Now suppose that the car is driven at a speed of v feet per second along a washboard surface shaped like a sine curve with a wavelength of 40 ft. The result is a periodic force on the car with frequency $ω = 2 π v / 40 = π v / 20$ . Resonance occurs when $ω = ω_{1}$ or $ω = ω_{2}$ . Find the corresponding two critical speeds of the car (in feet per second and in miles per hour).
Suppose that $k_{1} = k_{2} = k$ and $L_{1} = L_{2} = \frac{1}{2} L$ in Fig. 7.5.14 (the symmetric situation). Then show that every free oscillation is a combination of a vertical oscillation with frequency

$ω_{1} = \sqrt{2 k / m}$

and an angular oscillation with frequency

$ω_{2} = \sqrt{k L^{2} / (2 I)} .$

In Problems 27 through 29, the system of Fig. 7.5.14 is taken as a model for an undamped car with the given parameters in fps units. (a) Find the two natural frequencies of oscillation (in hertz). (b) Assume that this car is driven along a sinusoidal washboard surface with a wavelength of 40 ft. Find the two critical speeds.

$m = 100, I = 800, L_{1} = L_{2} = 5, k_{1} = k_{2} = 2000$
$m = 100, I = 1000, L_{1} = 6, L_{2} = 4, k_{1} = k_{2} = 2000$
$m = 100, I = 800, L_{1} = L_{2} = 5, k_{1} = 1000, k_{2} = 2000$

7.5 Application Earthquake-Induced Vibrations of Multistory Buildings

In this application you are to investigate the response to transverse earthquake ground oscillations of the seven-story building illustrated in Fig. 7.5.15. Suppose that each of the seven (above-ground) floors weighs 16 tons, so the mass of each is $m = 1000$ (slugs). Also assume a horizontal restoring force of $k = 5$ (tons per foot) between adjacent floors. That is, the internal forces in response to horizontal displacements of the individual floors are those shown in Fig. 7.5.16. It follows that the free transverse oscillations indicated in Fig. 7.5.15 satisfy the equation $M x ″ = K x$ with $n = 7, m_{i} = 1000$ (for each i), and $k_{i} = 10, 000$ (lb/ft) for $1 ≦ i ≦ 7$ . The system then reduces to the form $x ″ = A x$ with

$A = [\begin{array}{r} - 20 & 10 & 0 & 0 & 0 & 0 & 0 \\ 10 & - 20 & 10 & 0 & 0 & 0 & 0 \\ 0 & 10 & - 20 & 10 & 0 & 0 & 0 \\ 0 & 0 & 10 & - 20 & 10 & 0 & 0 \\ 0 & 0 & 0 & 10 & - 20 & 10 & 0 \\ 0 & 0 & 0 & 0 & 10 & - 20 & 10 \\ 0 & 0 & 0 & 0 & 0 & 10 & - 10 \end{array}] .$ (1)

FIGURE 7.5.15.

The seven-story building.

Once the matrix A has been entered, the TI-Nspire command eigVl(A) instantly computes the seven eigenvalues shown in the $λ$ -column of the table in Fig. 7.5.17. Alternatively, you can use the Maple command eigenvals(A), the Matlab command eig(A), or the Mathematica command Eigenvalues[A].

FIGURE 7.5.16.

Forces on the `i`th floor.

FIGURE 7.5.17.

Frequencies and periods of natural oscillations of the seven-story building.

	Eigenvalue	Frequency	Period
`i`	$λ$	$ω = \sqrt{- λ}$	$P = \frac{2 π}{ω} (\sec)$
1	$- 38.2709$	6.1863	1.0157
2	$- 33.3826$	5.7778	1.0875
3	$- 26.1803$	5.1167	1.2280
4	$- 17.9094$	4.2320	1.4847
5	$- 10.0000$	3.1623	1.9869
6	$- 3.8197$	1.9544	3.2149
7	$- 0.4370$	0.6611	9.5042

Then calculate the entries in the remaining columns of the table showing the natural frequencies and periods of oscillation of the seven-story building. Note that a typical earthquake producing ground oscillations with a period of 2 seconds is uncomfortably close to the fifth natural frequency (with period 1.9869 seconds) of the building.

A horizontal earthquake oscillation $E \cos ω t$ of the ground, with amplitude E and acceleration $a = - E ω^{2} \cos ω t$ , produces an opposite inertial force $F = m a = m E ω^{2} \cos ω t$ on each floor of the building. The resulting nonhomogeneous system is

$x^{″} = A x + (E ω^{2} \cos ω t) b,$ (2)

where $b = {[\begin{array}{r} 1 & 1 & 1 & 1 & 1 & 1 & 1 \end{array}]}^{T}$ and A is the matrix of Eq. (1). Figure 7.5.18 shows a plot of maximal amplitude (for the forced oscillations of any single floor) versus the period of the earthquake vibrations. The spikes correspond to the first six of the seven resonant frequencies. We see, for instance, that whereas an earthquake with period 2 (s) likely would produce destructive resonance vibrations in the building, it probably would be unharmed by an earthquake with period 2.5 (s). Different buildings have different natural frequencies of vibration, and so a given earthquake may demolish one building but leave untouched the one next door. This seeming anomaly was observed in Mexico City after the devastating earthquake of September 19, 1985.

For your personal seven-story building to investigate, let the weight (in tons) of each story equal the largest digit of your student ID number and let k (in tons/ft) equal the smallest nonzero digit. Produce numerical and graphical results like those illustrated in Figs. 7.5.17 and 7.5.18. Is your building susceptible to likely damage from an earthquake with period in the 2- to 3-second range?

FIGURE 7.5.18.

Resonance vibrations of a seven-story building—maximal amplitude as a function of period.

You might like to begin by working manually the following warm-up problems.

Find the periods of the natural vibrations of a building with two above-ground floors, each weighing 16 tons and with each restoring force being $k = 5 t o n s / f t .$
Find the periods of the natural vibrations of a building with three above-ground floors, with each weighing 16 tons and with each restoring force being $k = 5 t o n s / f t .$
Find the natural frequencies and natural modes of vibration of a building with three above-ground floors as in Problem 2, except that the upper two floors weigh 8 tons each instead of 16. Give the ratios of the amplitudes A, B, and C of oscillations of the three floors in the form A:B:C with $A = 1$ .
Suppose that the building of Problem 3 is subject to an earthquake in which the ground undergoes horizontal sinusoidal oscillations with a period of 3 s and an amplitude of 3 in. Find the amplitudes of the resulting steady periodic oscillations of the three above-ground floors. Assume the fact that a motion $E \sin ω t$ of the ground, with acceleration $a = - E ω^{2} \sin ω t$ , produces an opposite inertial force $F = - m a = m E ω^{2} \sin ω t$ on a floor of mass m.

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Table of Contents for
7.5 Second-Order Systems and Mechanical Applications*

7.5 Second-Order Systems and Mechanical Applications*

FIGURE 7.5.1.

FIGURE 7.5.2.

Solution of Second-Order Systems

Remark

Example 1

FIGURE 7.5.3.

FIGURE 7.5.4.

FIGURE 7.5.5.

Example 2

FIGURE 7.5.6.

FIGURE 7.5.7.

FIGURE 7.5.8.

Forced Oscillations and Resonance

Example 3

FIGURE 7.5.9.

FIGURE 7.5.10.

Periodic and Transient Solutions

7.5 Problems

FIGURE 7.5.11.

FIGURE 7.5.12.

FIGURE 7.5.13.

The Two-Axle Automobile

FIGURE 7.5.14.

7.5 Application Earthquake-Induced Vibrations of Multistory Buildings

FIGURE 7.5.15.

FIGURE 7.5.16.

FIGURE 7.5.17.

FIGURE 7.5.18.

Table of Contents for 7.5 Second-Order Systems and Mechanical Applications*

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Table of Contents for
7.5 Second-Order Systems and Mechanical Applications*