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4.2 Transverse Vibration of Cables

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The first continuous member which we will study is a string or cable in tension. This is a line structure

whose geometric configuration can be completely defined by the position of its axial line with reference

to a fixed coordinate line. We will study the transverse (lateral) vibration problem; that is, the vibration in

a direction perpendicular to its axis and in a single plane. Applications will include stringed musical

instruments, overhead transmission lines (of electric power or telephone signals), drive systems (belt

drives, chain drives, pulley ropes, etc.), suspension bridges, and structural cables carrying cars (e.g., ski

lifts, elevators, overhead sightseeing systems, and cable cars).

4-2 Vibration and Shock Handbook

As usual, we will make some simplifying assumptions for analytical convenience. However, the results

and insight obtained in this manner will be useful in understanding the behavior of more complex

systems containing cable-like structures. The main assumptions are:

1. The system is a line structure. The lateral dimensions are much smaller compared with the

longitudinal dimension (normally in the x direction).

2. The structure stays in a single plane and the motion of every element of the structure will be in a

fixed transverse direction ð yÞ:

3. The cable tension ðTÞ remains constant during motion. In other words, the initial tension is

sufficiently large that the variations during motion are negligible.

4. Variations in slope ðuÞ along the structure are small. Hence, for example, u ø sin u ø tan u ¼

›v

›x

A general configuration of a cable (or string) is shown in Figure 4.1(a). Consider a small element of

length dx of the cable at location x; as shown in Figure 4.1(b). The equation (Newton’s Second Law) of

motion (transverse) of this element is given by

f ðx; tÞdx 2 T sin u þ T sinðu þ duÞ ¼ mðxÞdx

›2vðx; tÞ

›t2 ð4:1Þ

in which

vðx; tÞ ¼ transverse displacement of the cable

f ðx; tÞ ¼ lateral force per unit length of cable

mðxÞ ¼ mass per unit length of cable

T ¼ cable tension

u ¼ cable slope at location x:

Note that the dynamic loading f ðx; tÞ may arise due to such causes as aerodynamic forces, fluid drag,

and electromagnetic forces, depending on the specific application.

(a)

(b)

0 x x+dx

Mass per unit length = m(x)

Force per unit length = f(x,t)

∂x

θ + ∂q

x x+dx

m.dx

f.dx

FIGURE 4.1 (a) Transverse vibration of a cable in tension; (b) motion of a general element.

Distributed-Parameter Systems 4-3

Using the small slope assumption we have sin u ø u and sinðu þ duÞ ø u þ du with u ¼ ›v=›x and

du ¼ ð›2v=›x2Þdx as dx ! 0: On substitution of these approximations into Equation 4.1 and canceling

out dx, we obtain

mðxÞ

›2vðx; tÞ

›t2 ¼ T

›2vðx; tÞ

›x2 þ f ðx; tÞ ð4:2Þ

Now consider the case of free vibration where f ðx; tÞ ¼ 0: We have

›2vðx; tÞ

›t2 ¼ c2 ›2vðx; tÞ

›x2 ð4:3Þ

with

c ¼

ffiffiffiffiffiffi

T=m p ð4:4Þ

Also, assume that the cable is uniform so that m is constant.

4.2.1 Wave Equation

The solution to any equation of the form (Equation 4.3) will appear as a wave, traveling either in the

forward (positive x) or in the backward (negative x) direction at speed c: Hence, Equation 4.3 is called the

wave equation and c is the wave speed. To prove this fact, first, we show that a solution to Equation 4.3 can

take the form

vðx; tÞ ¼ v1ðx 2 ctÞ ð4:5Þ

First, let x 2 ct ¼ z: Hence, v1ðx 2 ctÞ ¼ v1ðzÞ: Then,

›v1

›x ¼

dv1

›z

›x

and

›v1

›t ¼

dv1

›z

›t

with

›z

›x ¼ 1 and

›z

›t ¼ 2c

It follows that

›2v1

›x2 ¼ v00 1 and

›2v1

›t2 ¼ c2v00 1

where

v00 1 ¼

d2v1

dz2

Clearly, then, v1 satisfies Equation 4.3.

Now, let us examine the nature of the solution v1ðx 2 ctÞ: It is clear that v1 will be constant when

x 2 ct ¼ constant: However, the equation x 2 ct ¼ constant corresponds to a point moving along the x

axis in the positive direction at speed c: What this means is that the shape of the cable at t ¼ 0 will

“appear” to travel along the cable at speed c: This is analogous to the waves we observe in a pond when

excited by dropping a stone. Note that the particles of the cable do not travel along x: it is the deformation

“shape” (the wave) that travels.

Similarly, it can be shown that

vðx; tÞ ¼ v2ðx þ ctÞ ð4:6Þ

4-4 Vibration and Shock Handbook

is also a solution to Equation 4.3 and this corresponds to a wave that travels backward (negative x

direction) at speed c: The general solution, of course, will be of the form

vðx; tÞ ¼ v1ðx 2 ctÞ þ v2ðx þ ctÞ ð4:7Þ

which represents two waves, one traveling forward and the other backward.

4.2.2 General (Modal) Solution

As usual, we look for a separable solution of the form

vðx; tÞ ¼ Y ðxÞqðtÞ ð4:8Þ

for the cable/string vibration problem given by the wave equation 4.3. If a solution in the form of

Equation 4.8 is obtained, it will be essentially a modal solution. This should be clear from the separability

itself of the solution. Specifically, at any given time t; the time function qðtÞ will be fixed and the structure

will have a shape given by Y ðxÞ: Hence, at all times the structure will maintain a particular shape Y ðxÞ and

this will be a mode shape. Also, at a given point x of the structure, the space function Y ðxÞ will be fixed and

the structure will vibrate according to the time response qðtÞ: It will be shown that qðtÞ will obey the

simple harmonic motion of a specific frequency. This is the natural frequency of vibration corresponding

to that particular mode. Note that, for a continuous system, there will be an infinite number of solutions

of the form of Equation 4.8 with different natural frequencies. The corresponding functions Y ðxÞ will be

“orthogonal” in some sense. Hence, they are called normal modes (normal meaning perpendicular). The

systems will be able to move independently in each mode and this collection of solutions in the form of

Equation 4.8 will be a complete set. With this qualitative understanding, let us now seek a solution of the

form of Equation 4.8 for the system Equation 4.3.

Substitute Equation 4.8 in Equation 4.3. We obtain

Y ðxÞ

d2qðtÞ

dt2 ¼ c2 d2Y ðaÞ

dx2 qðtÞ

Y ðxÞ

d2Y ðxÞ

dx2 ¼

c2qðtÞ

d2qðtÞ

dt2 ¼ 2l2 ð4:9Þ

In Equation 4.9, since the left-hand terms are a function of x only and the right-hand terms are a function

of t only, for the two sides to be equal in general, each function should be a constant (that is independent

of both x and t). This constant is denoted by 2l2, which is called the separation constant and is

designated to be negative. There are two reasons for this. If this common constant were positive, the

function qðtÞ would be nonoscillatory and transient, which is contrary to the nature of undamped

vibration. Furthermore, it can be shown that a nontrivial solution for Y ðxÞ would not be possible if the

common constant were positive.

The unknown constant l is determined by solving the space equation (mode shape equation) of

Equation 4.9; specifically

d2Y ðxÞ

dx2 þ l2Y ðxÞ ¼ 0 ð4:10Þ

and then applying the BCs of the problem. There will be an infinite number of solutions for l; with

corresponding natural frequencies v and mode shapes Y ðxÞ:

The characteristic equation of 4.10 is

p2 þ l2 ¼ 0 ð4:11Þ

which has the characteristic roots (or eigenvalues)

p ¼ ^jl ð4:12Þ

Distributed-Parameter Systems 4-5

The general solution is

Y ðxÞ ¼ A1e jlx þ A2e2jlx ¼ C1 cos lx þ C2 sin lx ð4:13Þ

Note that, since Y ðxÞ is a real function representing a geometric shape, the constants A1 and A2 have

to be complex conjugates and C1 and C2 have to be real. Specifically, in view of the fact that

cos lx ¼ ðejlx þ e2jlx = Þ2 and sin lx ¼ ðejlx 2 e2jlx =2j Þ, we can show that

A1 ¼

2 ðC1 2 jC2Þ and A2 ¼

2 ðC1 þ jC2Þ

For analytical convenience, we will use the real-parameter form of Equation 4.13.

Note that we cannot determine both constants C1 and C2 using BCs. Only their ratio is determined

and the constant multiplier is absorbed into qðtÞ in Equation 4.8 and then determined using the

appropriate initial conditions (at t ¼ 0). It follows that the ratio of C1 and C2 and the value of l are

determined using the BCs. Two BCs will be needed. Some useful situations and appropriate relations are

given in Table 4.1.

4.2.3 Cable with Fixed Ends

Let us obtain the complete solution for the free vibration of a taut cable that is fixed at both ends. The

applicable BCs are

Y ð0Þ ¼ Y ðlÞ ¼ 0 ð4:14Þ

where l is the length of the cable. Substitution into Equation 4.13 gives

C1 £ 1 þ C2 £ 0 ¼ 0

C1 cos ll þ C2 sin ll ¼ 0

Hence, we have

C1 ¼ 0 and C2 sin ll ¼ 0 ð4:15Þ

A possible solution for Equation 4.15 is C2 ¼ 0: However, this is the trivial solution, which corresponds to

Y ðxÞ ¼ 0 (i.e., a stationary cable with no vibration). It follows that the applicable, nontrivial solution is

sin ll ¼ 0

which produces an infinite number of solutions for l given by

li ¼

with i ¼ 1; 2; …; 1 ð4:16Þ

As mentioned earlier, the corresponding infinite number of mode shapes is given by

YiðxÞ ¼ Ci sin

ipx

l ð4:17Þ

Note: If we had used a positive constant l2 instead of 2l2 in Equation 4.9, only a trivial solution (with

C1 ¼ 0 and C2 ¼ 0) would be possible for Y ðxÞ: This further justifies our decision to use 2l2: Substitute

Equation 4.16 into Equation 4.9 to determine the corresponding time response (generalized coordinates)

qiðtÞ; thus

d2qiðtÞ

dt2 þ v2i

qiðtÞ ¼ 0 ð4:18Þ

in which

vi ¼ lic ¼

ffiffiffiffi

for i ¼ 1; 2; …; 1 ð4:19Þ

4-6 Vibration and Shock Handbook

TABLE 4.1 Some Useful Boundary Conditions for the Cable Vibration Problem

Type of End Condition Nature of End x ¼ x0 Boundary Condition Modal Boundary Condition

Fixed

vðx0 ; tÞ ¼ 0 Yi ðx0Þ ¼ 0

Free

∂x

∂v

T T

›vðx0; tÞ

›x ¼ 0

dYi ðx0 Þ

dx ¼ 0

Flexible

∂x

∂v

›vðx0 ; tÞ

›x

2 kvðx0 ; tÞ ¼ 0 T

dYi ðx0 Þ

2 kYi ðx0Þ ¼ 0

Flexible and inertial

∂x

T ∂v

›vðx0; tÞ

›x

2 kvðx0 ; tÞ ¼ M

›2 vðx0 ; tÞ

›t2

dYi ðx0Þ

2 ðk 2v2i

MÞYi ðx0Þ¼0

Distributed-Parameter Systems 4-7

Equation 4.18 represents a simple harmonic motion with the modal natural frequencies vi given by

Equation 4.19. It follows that there are an infinite number of natural frequencies, as mentioned earlier.

The general solution of Equation 4.19 is given by

qiðtÞ ¼ ci sinðvit þ fiÞ ð4:20Þ

where the amplitude parameter ci and the phase parameter fi are determined using two of the initial

conditions of the system. It should be clear that it is redundant to use a separate constant Ci for YiðxÞ in

Equation 4.17, and that this may be absorbed into the amplitude constant in Equation 4.20 to express the

general free response of the cable as

vðx; tÞ ¼

ci sin

ipx

sinðvit þ fiÞ ð4:21Þ

In this manner, the complete solution has been expressed as a summation of the modal solutions. This is

known as the modal series expansion. Such a solution is quite justified because of the fact that the mode

shapes are orthogonal in some sense, and what we obtained above were a complete set of normal modes

(normal in the sense of perpendicular or orthogonal). The system is able to move in each mode

independently, with a unique spatial shape, at the corresponding natural frequency, because each modal

solution is separable into a space function, YiðxÞ, and a time function (generalized coordinate), qiðtÞ: Of

course, the system will be able to simultaneously move in a linear combination of two modes (say,

C1Y1ðxÞq1ðtÞ þ C2Y2ðxÞq2ðtÞ), since this combination satisfies the original system Equation 4.3 because of

its linearity and because each modal component satisfies the equation. However, clearly, this solution,

with two modes, is not separable into a product of a space function and a time function. Hence, it is not a

modal solution. In this manner, it can be argued that the infinite sum of modal solutions

ciYiðxÞqiðtÞ is

the most general solution to the system (Equation 4.3). The orthogonality of mode shapes plays a key role

in this argument and, furthermore, it is useful in the analysis of the system, as we shall see. In particular,

in Equation 4.21, the unknown constants ci and fi are determined using the system initial conditions,

and the orthogonality property of modes is useful in that procedure.

4.2.4 Orthogonality of Natural Modes

A cable can vibrate at frequency vi while maintaining a unique natural shape YiðxÞ; called the mode shape

of the cable. We have shown that, for the fixed-ended cable, the natural mode shapes are given by

sinðipx=lÞ with the corresponding natural frequencies, vi: It can be easily verified that

ðl

sin

ipx

sin

jpx

dx ¼

0 for i – j

for i ¼ j

8><

ð4:22Þ

In other words, the natural modes are orthogonal. Equation 4.22 represents the principle of

orthogonality of natural modes in this case.

Orthogonality makes the modal solutions independent and the corresponding mode shapes “normal.”

It also makes the infinite set of modal solutions a complete set, or a basis, so that any arbitrary response

can be formed as a linear combination of these normal mode solutions.

Orthogonality holds for other types of BCs as well. To show this, we observe from Equation 4.9 that

d2YiðxÞ

dx2 þ l2i

YiðxÞ ¼0 for mode i ð4:23Þ

d2YjðxÞ

dx2 þ l2j

YjðxÞ ¼0 for mode j ð4:24Þ

4-8 Vibration and Shock Handbook

Multiply Equation 4.23 by YjðxÞ; Equation 4.24 by YiðxÞ; subtract that second result from the first,

and integrate with respect to x along the cable length from x ¼ 0 to l: We obtain

ðl

d2Yi

dx2 2 Yi

d2Yj

dx2

" #

dx þ ðl2i

2 l2j

ðl

YiYj dx ¼ 0 ð4:25Þ

Integrating by parts, we obtain the results

ðl

d2Yi

dx2 dx ¼ Yj

dYi

􀀈 􀀈 􀀈 􀀈 􀀈

ðl

dYi

dYj

ðl

d2Yj

dx2 dx ¼ Yi

dYj

􀀈 􀀈 􀀈 􀀈 􀀈

ðl

dYi

dYj

Hence, the first term of Equation 4.25 becomes

dYi

2 Yi

dYj

􀀒 􀀓l

which will vanish for common BCs. Then, since li – lj for i – j, we have

ðl

YiðxÞYjðxÞdx ¼0 for i – j

We can pick the value of the multiplication constant in the general solution for Y ðxÞ; given by Equation

4.13, so as to normalize the mode shapes such that

ðl

Y 2

i ðxÞdx ¼

which is consistent with the result 4.22. Hence, the general condition of orthogonality of natural modes

may be expressed as

ðl

YiðxÞYjðxÞdx ¼

0 for i – j

for i ¼ j

8><

ð4:26Þ

4.2.4.1 Nodes

When vibrating in a particular mode, one or more points of the system (cable) that are not physically

fixed may remain stationary at all times. These points are called the nodes of that mode. For example, in

the second mode of a cable with its ends fixed, there will be a node at the midspan. This should be clear

from the fact that the mode shape of the second mode is sin 2px=l which becomes zero at x ¼ l=2:

Similarly, in the third mode, with mode shape sin 3px=l; there will be nodes at x ¼ l=3 and 2l=3:

Example 4.1

If the cable tension varies along the length x; what is the corresponding equation of free lateral vibration?

A hoist mechanism has a rope of freely hanging length l in a particular equilibrium configuration and

carrying a load of mass M; as shown in Figure 4.2(a). Determine the equation of lateral vibration and the

applicable BCs for the rope segment.

Solution

With reference to Figure 4.1(b), Equation 4.1 may be modified for the case of variable T as

2T sin u þ ðT þ dTÞ sinðu þ duÞ ¼ m dx

›2v

›t2 ð4:27Þ

Distributed-Parameter Systems 4-9

where f ðx; tÞ ¼ 0 for free vibration. Now, with the

assumption of small u; and by neglecting the

second-order product term dT du; we obtain

T du þ u dT ¼ m

›2v

›x2 dx

Next, using

u ¼

›v

›x

; du ¼

›2v

›x2

; and dT ¼

›T

›x

and canceling dx; we obtain the equation of lateral

vibration of a cable as

›2v

›t2 ¼ T

›2v

›x2 þ

›T

›x

›v

›x ð4:28Þ

Longitudinal (axial) dynamics of the rope are

negligible for the case of a stationary hoist. Then,

longitudinal equilibrium (in the x direction) of the

small element of rope shown in Figure 4.2(b) gives

ðT þ dTÞ cosðu þ duÞ 2 T cos u 2 mg dx ¼ 0

For small u; we have cos u ø 1 and cosðu þ

duÞ ø 1 up to the first-order term in the Taylor

series expansion. Hence,

dT ¼ mg dx ð4:29Þ

Integration gives

T ¼ T0 þ mgx ð4:30Þ

with the end condition

T ¼ Mg at x ¼ 0

Hence,

T ¼ Mg þ mgx ð4:31Þ

Note from Equation 4.29 that ›T=›x ¼ dT=dx ¼ mg for this problem. Substitute in (Equation 4.28) this

fact and Equation 4.31 to obtain

›2v

›t2 ¼ ðM þ mxÞg

›2v

›x2 þ mg

›v

›x

›2v

›t2 ¼

m þ x

􀀏 􀀐

›2v

›x2 þ g

›v

›x ð4:32Þ

The BC at x ¼ 0 is obtained by applying Newton’s Second Law to the end mass in the lateral ðyÞ direction.

This gives

›vð0; tÞ

›x ¼ M

›2vð0; tÞ

›t2

Now, using the fact that T0 ¼ Mg, we have the boundary condition

›vð0; tÞ

›x ¼

›2vð0; tÞ

›t2

(a)

(b)

x+dx

mgdx

θ+dθ

T+dT

FIGURE 4.2 (a) Free segment of a stationary hoist;

(b) a small element of the rope.

4-10 Vibration and Shock Handbook

For mode i:

›vð0; tÞ

›x ¼

dYið0Þ

qiðtÞ

and

›2vð0; tÞ

›t2 ¼ Yið0Þ

d2qiðtÞ

dt2 ¼ 2v2i

Yið0ÞqiðtÞ

which holds for all t and where vi is the ith natural frequency of vibration. Hence, the modal BC at x ¼ 0

dYið0Þ

dx þ v2i

Yið0Þ ¼0 for i ¼ 1; 2; … ð4:33Þ

The BC at x ¼ l is

vðl; tÞ ¼ 0 ð4:34Þ

which holds for all t: Hence, the corresponding modal BC is

YiðlÞ ¼0 for i ¼ 1; 2; … ð4:35Þ

4.2.5 Application of Initial Conditions

The general solution to the cable vibration problem is given by

vðx; tÞ ¼

ciYiðxÞ sinðvit þ fiÞ ð4:36Þ

where YiðxÞ are the normalized mode shapes which satisfy the orthogonality property (Equation 4.26).

The unknown constants ci and fi are determined using the initial conditions

vðx; 0Þ ¼ dðxÞ ð4:37Þ

›vðx; 0Þ

›t ¼ sðxÞ ð4:38Þ

By substituting Equation 4.36 into Equation 4.37 and Equation 4.38, we obtain

dðxÞ ¼

ciYiðxÞ sin fi ð4:39Þ

sðxÞ ¼

civiYiðxÞ cos fi ð4:40Þ

Multiply Equation 4.39 and Equation 4.40 by YjðxÞ and integrate with respect to x from 0 to l; making use

of the orthogonality condition (Equation 4.26). We obtain

ðl

dðxÞYjðxÞdx ¼ cj

sin fj

ðl

sðxÞYjðxÞdx ¼ cjvj

cos fj

Solving these two equations, we obtain

tan fj ¼ vj

ðl

dðxÞYjðxÞdx

ðl

sðxÞYjðxÞdx

for j ¼ 1; 2; 3; … ð4:41Þ

Distributed-Parameter Systems 4-11

Once fj is determined in this manner, we can obtain cj by using

cj ¼

l sin fj

ðl

dðxÞYjðxÞdx for j ¼ 1; 2; 3; … ð4:42Þ

Example 4.2

Consider a taut horizontal cable of length l and

mass m per unit length, as shown in Figure 4.3,

excited by a transverse point force f0 sin vt at

location x ¼ a; where v is the frequency of

(harmonic) excitation and f0 is the forcing

amplitude. Determine the resulting response of

the cable under general end conditions and initial

conditions. For the special case of fixed ends, what

is the steady-state response of the cable?

Solution

We have shown that the forced transverse response of a cable is given by Equation 4.2:

›2vðx; tÞ

›t2 ¼ c2 ›2vðx; tÞ

›x2 þ

f ðx; tÞ

m ð4:2Þ

where vðx; tÞ is the transverse displacement and f ðx; tÞ is the external force per unit length of the cable.

For the point force F at x ¼ a; an analytical representation of the equivalent distributed force per unit

length is

f ðx; tÞ ¼ Fdðx 2 aÞ ð4:43Þ

where the Dirac delta function (unit impulse function) dðxÞ is such that

ða2

gðxÞdðx 2 aÞdx ¼ gðaÞ ðiÞ

for an arbitrary function gðxÞ; provided that the point a is within the interval of integration ½a1; a2􀀉: We

seek a “modal superposition” solution of the form

vðx; tÞ ¼

YiðxÞq􀀊iðtÞ ð4:44Þ

where q􀀊iðtÞ are the generalized coordinates of the forced response solution (which are generally different

from those for the free solution; i.e., qiðtÞ).

Substitute the solution (Equation 4.44) into the system Equation 4.2 and make use of the governing

equation of the mode shapes (see Equation 4.10)

d2YiðxÞ

dx2 ¼ 2l2i

YiðxÞ ð4:45Þ

we obtain

YiðxÞ€􀀊

qiðtÞ ¼ 2T

l2i

YiðxÞq􀀊iðtÞ þ f0 sin vtdðx 2 aÞ ðiiÞ

Multiply Equation (ii) by YjðxÞ; and integrate from x ¼ 0 to l using the orthogonality property (Equation

4.26) and also Equation (i). We obtain

€􀀊

qiðtÞ ¼ 2T

l2j

q􀀊jðtÞ þ f0YjðaÞ sin vt

0 x

fo sinwt

a l

FIGURE 4.3 A cable excited by a point harmonic force.

4-12 Vibration and Shock Handbook

Now since vj ¼ lj

ffiffiffiffiffiffi

T=m p (see Equation 4.19), we obtain

€􀀊

qjðtÞ þ v2j

q􀀊jðtÞ ¼

2f0

YjðaÞ sin vt for j ¼ 1; 2; 3; … ð4:46Þ

This has the familiar form of a simple oscillator excited by a harmonic force and its solution is well

known. The initial conditions 􀀊qjð0Þ and _

􀀊

qjð0Þ are needed. Suppose that the initial transverse displacement

and the speed of the cable are

vðx; 0Þ ¼ dðxÞ and v_ðx; 0Þ ¼ sðxÞ

Then, in view of Equation 4.45, we can write

YiðxÞq􀀊ið0Þ ¼ dðxÞ ð4:47Þ X

YiðxÞ_􀀊

qið0Þ ¼ sðxÞ ð4:48Þ

Multiply Equation 4.47 and Equation 4.48 by YjðxÞ; and integrate from x ¼ 0 to l using the orthogonality

property 4.26. We obtain the necessary initial conditions

q􀀊jð0Þ ¼

ðl

dðxÞYjðxÞdx ð4:49Þ

_􀀊

qjð0Þ ¼

ðl

sðxÞYjðxÞdx ð4:50Þ

which will provide the complete solution for Equation 4.46 and hence will completely determine

Equation 4.44.

For a fixed-ended cable, we have

YiðxÞ ¼ sin

ipx

l ðiiiÞ

and, at steady state, the time response q􀀊jðtÞ will be harmonic at the same frequency as the excitation

frequency v: Hence, we have

q􀀊jðtÞ ¼ q0j sinðvt þfjÞ ð4:51Þ

We see that, for Equation 4.51 to satisfy Equation 4.46 in this undamped problem, we must have fj ¼ 0:

Direct substitution gives

½2v2 þ v2j 􀀉q0j ¼

2f0

YjðaÞ

which determines q0j: Hence, from Equation 4.45, the complete solution for the fixed-ended problem, at

steady state, is

vðx; tÞ ¼

2f0

sin vt

X sin ipa=l

ðv2i

2 v2Þ

sin

ipx

l ð4:52Þ

Some important results for transverse vibration of strings and cables are summarized in Box 4.1.

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4.2 Transverse Vibration of Cables

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