19 Application - I

Contents

19.1 Introduction

19.2 Impulsive Start of a Bar

Keywords: Impulse, bar, wave propagation, wave reflection

19.1 Introduction

We consider the application of the traveling wave solution in studying

the transient dynamics in some practical situations. Here we consider the

dynamics of a uniform bar subjected to an impulse at one end. The motion

of the bar is understood in terms of the waves propagating in the bar.

19.2 Impulsive Start of a Bar

Consider a free-free bar of length l, initially at rest, set into motion by an im-

F0(t)

x

u(x, t) , A,E

Fig. 19.1: A free-free bar started with an impulse

2 19 Application - I

pulsive force F (x, t) = F0(t)(x), as shown in Fig. 19.1. The subsequent mo-

tion of the bar can be determined as follows. Using the impulse-momentum

equation, one can obtain the initial velocity condition

0+

0

Au,tt dt =

0+

0

F0(t)(x) dt u,t(x, 0+) = v0(x) =

F0A

(x),

(19.1)

Assuming a displacement field u(x, t) = f1(x ct) in (19.1), one can write

u,t(x, 0) = cf1(x) = v0 =

F0A

(x) f1(x) =F0Ac

[1H(x)]

u(x, t) = f1(x ct) =F0Ac

[1H(x ct)], 0 < t < l/c, (19.2)

where the constant of integration has been found from the initial zero dis-

placement condition, and H() is the Heaviside step function defined as

H(x) =

{0, x < 0

1, x 0.

When the wave reaches the right end at t = l/c, a reflected wave, denoted by

g1(x+ ct), is created. The motion of the bar can be represented as

u(x, t) =F0Ac

[1H(x ct)] + g1(x+ ct), l/c t < 2l/c. (19.3)

Applying the right end boundary condition EAu,x(l, t) = 0 yields

F0Ac

(l ct) + g1(l + ct) = 0 g1(l + ct) =

F0Ac

(l ct).

NPTEL Course: Wave Propagation in Continuous Media

19.2 Impulsive Start of a Bar 3

Defining z = l + ct, and writing l ct = 2l z in the above, we have

g1(z) =F0Ac

(2l z)] g1(z) = F0Ac

H(2l z) + CI

g1(x+ ct) = F0Ac

H(2l x ct) + CI ,

where, CI is a constant of integration. From the condition g1(x+ct)|t=l/c = 0,

we have CI = F0/Ac. Therefore, the complete solution after one reflection

is given by

u(x, t) =F0Ac

[2H(x ct)H(2l x ct)], l/c t < 2l/c. (19.4)

At t = 2l/c, the wave reaches the left end of the bar, and reflects again to

produce a reflected wave denoted by, say f2(x ct). Writing the motion of

the bar as

u(x, t) =F0Ac

[2H(xct)H(2lxct)]+f2(xct), t > 2l/c, (19.5)

and using the force-free boundary condition EAu,x(0, t) = 0 for t 2l/c

yields

F0Ac

[(ct) + (2l ct)] + f 2(ct) = 0 f2(ct) =

F0Ac

(2l ct).

Defining z = ct, we have

f 2(z) = F0Ac

(2l + z) f2(z) = F0Ac

H(2l + z) + CI ,

f2(x ct) = F0Ac

H(2l + x ct) + CI . (19.6)

NPTEL Course: Wave Propagation in Continuous Media

4 19 Application - I

Using the condition f2(x ct)|t=2l/c = 0 yields CI = F0/Ac. Therefore, the

complete solution after the second reflection is given by

u(x, t) =F0Ac

[3H(xct)H(2lxct)H(2l+xct)], 2l/c t < 3l/c.

This solution procedure can be continued indefinitely. It may be observed

that the bar continuously moves to the right. However, the ends of the bar

do not move simultaneously. The propagation of the displacement wave in

the bar is shown in Fig. 19.2. The motions of the left end and right end are

shown in Fig. 19.3. It is seen that the whole bar shifts a distance F0/Ac

in time l/c. The average velocity of the bar is, therefore, F0/Al = F0/m,

where m is the mass of the bar. This is also the velocity acquired by a rigid

bar of mass m when acted upon by an impulsive force of strength F0. Thus,

the average motion of an elastic bar is identical to that of a rigid bar.

uc/F0

0 < t < l/cx

xl/c < t < 2l/c

2l/c < t < 3l/3x

xl3l/c < t < 4l/c

Fig. 19.2: Displacement of

a bar started with an impulse

NPTEL Course: Wave Propagation in Continuous Media

19.2 Impulsive Start of a Bar 5

1 2 3

1

2

3uc/F0

ct/l

left end

mean

right end

Fig. 19.3: Motion of the ends of a free-free bar

started with an impulse

NPTEL Course: Wave Propagation in Continuous Media