Longitudinal
Oscillations and Sound
Transverse oscillations of a continuous system are
easy to visualize because you can see
directly the function that describes the
displacement. The mathematics of longitudinal oscillations
of a continuous linear space translation invariant
system is the same. It must be,
because it is completely determined by the space
translation invariance. But the physics is
different.
Preview
In this chapter, we introduce two physical systems
with longitudinal oscillations: massive
springs and organ pipes.
i. We describe the massive spring as the continuum
limit of a system of masses connected
by massless springs and study its normal modes for
various boundary conditions.
ii. We discuss in some detail the system of a mass
at the end of a massive spring. When
the spring is “light,” this is an important example
of physics with two different “scales.”
iii. We discuss the physics of sound waves in a
tube, by analogy with the oscillations of
the massive spring. We also introduce the “Helmholtz”
approximation for the lowest
mode of a bottle.
Longitudinal
Modes in a Massive Spring
So far, in our extensive discussions of waves in
systems of springs and blocks, we have
assumed that the only degrees of freedom are those
associated with the motion of the blocks.
This is a reasonable assumption at low frequencies,
when the blocks are very heavy compared
to the springs, because the blocks move so slowly
that the springs have time to readjust and are
153
LONGITUDINAL OSCILLATIONS AND SOUND
always nearly uniform.1 In this case, the dispersion relation for the
longitudinal oscillations
of the blocks is just the dispersion relation for
coupled pendulums, (5.35), in the limit in
which we ignore gravity, and keep only the coupling
between the masses produced by the
spring constant, K.
In other words, we take the limit of (5.35) as g=` ! 0. The result can be
written as
!2 =
4Ka
m
sin2 ka
2
(7.1)
where Ka is the spring constant of the
springs, m
the mass of the
blocks, and a the
equilibrium
separation. We have put a subscript a on Ka because we will want to vary the
spring
constant as we vary the separation between the
blocks in the discussion below.
Now what happens when the blocks are absent, but the
spring is massive? We can find
this out by considering the limit of (7.1) as a ! 0. In this limit, the massive
blocks and the
massless spring melt into one another, so that the
result looks like a uniform, massive spring.
In order to take the limit, however, we must
understand what variables describe the massive
spring, and have a finite limit as a ! 0. One such variable is the linear
mass density,
½L = lim
a!0
m
a
: (7.2)
We must take the masses of the blocks to zero as a ! 0 in order to keep ½L finite.
To understand what happens to Ka as a ! 0, consider what happens when you
cut a
spring in half. When a spring is stretched, each
half contributes half the displacement. But
the tension is uniform throughout the stretched
spring. Thus the spring constant of half a
spring is twice as great as that of the full spring,
because half the displacement gives the
same force. This relation is illustrated in figure
7.1. The spring in the center is unstretched.
The spring on top is stretched by x to the right. The bottom shows
the same stretched spring,
still stretched by x,
but now symmetrically. Comparing top and bottom, you can see that the
return force from stretching the spring by x is the same as from stretching
half the spring by
x=2.
The diagram in figure 7.1 is an example of the
following result. In general, the spring
constant, Ka, depends not just on what the
spring is made of, it depends on how long the
spring is. But the quantity Kaa, where a is the length of the spring, is
actually independent
of a,
for a spring made of uniform material. Thus we should take the limit a ! 0 holding
Kaa fixed.
This implies that the dispersion relation for the
massive spring is
!2 = Kaa
½L
k2; (7.3)
where we have used the Taylor series expansion of sin x, (1.58), and kept only the first
term.
1We will say this much more
formally below.
7.1. LONGITUDINAL MODES IN A MASSIVE SPRING 155
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Ãx!
Ã!
x
2
Ã!
x
2
Figure 7.1: Half a spring has twice the spring
constant.
According to the discussion above, we can rewrite
this as
!2 = K`
½L
k2 (7.4)
where ` is
the length of the spring and K is
the spring constant of the spring as a whole.
Note that in longitudinal oscillations in a
continuous material in the x direction,
the equilibrium
position, x,
doesn’t actually describe the x position
of the material. Because the
displacement is longitudinal, the actual x position of the point on the
spring with equilibrium
position x is
x + Ã(x; t) ; (7.5)
where à is
the displacement. You will need this to do problem (7.1).
7.1.1 Fixed Ends
.
........................................... . .................... . . . . . .
. . . 7-1
Suppose that we have a massive spring with length ` and its ends fixed at x = 0 and x = `.
Then the displacement, Ã(x; t) must vanish at the ends,
Ã(0; t) = 0 ; Ã(`; t) = 0 : (7.6)
The modes of the system are the same as for any
other space translation invariant system.
The linear combinations of the complex exponential
modes of the infinite system that satisfy
(7.6) are
An(x) = sin n¼x
`
; (7.7)
156 CHAPTER 7. LONGITUDINAL OSCILLATIONS AND SOUND
with angular wave number
kn = n¼
`
(7.8)
and frequency (from the dispersion relation, (7.4))
!n =
s
K`
½L
kn =
s
K`
½L
n¼
`
: (7.9)
However, because the oscillations are longitudinal,
the modes look very different from the
transverse modes of the string that we studied in
the previous chapter. The position of the
point on the string whose equilibrium position is x, in the nth normal mode, has the general
form (from (7.5))
x + ² sin n¼x
`
cos(!nt + Á) (7.10)
where ² and
Á are the amplitude and phase of
the oscillation.
The lowest 9 modes in (7.10) are animated in program
7-1. Compare these with the modes
animated in program 6-1. The mathematics is the
same, but the physics is very different
because of (7.5). Stare at these two animations
until you can visualize the relation between
the two. Then you will have understood (7.5).
7.1.2 Free Ends
.
........................................... . .................... . . . . . .
. . . 7-2
Now let us look at the situation in which the end of
the spring at x =
0 is fixed, but the
end at
x = ` is free. The boundary conditions
in this case are analogous to the normal modes of the
string with one fixed end. The displacement at x = 0 must vanish because the end is
fixed.
Also, the derivative of the displacement at x = ` must vanish. You can see this by
looking at
the continuous spring as the limit of discrete
masses coupled by springs. As we saw in (5.43),
the last real mass must have the same displacement
as the first “imaginary” mass,
Ã(`; t) = Ã(` + a; t) : (7.11)
Therefore, for the finite system with a free end at `, we have the relation
Ã(`; t) ¡ Ã(` + a; t)
a
= 0 for
all a. (7.12)
In the limit that the distance between masses goes
to zero, this becomes the condition that the
derivative of the displacement, Ã, with respect to x vanishes at x = `,
@
@x
Ã(x; t)jx=` = 0 : (7.13)
7.2. A MASS ON A LIGHT SPRING 157
Thus the boundary conditions on the displacement are
the same as in (6.11) for the transverse
oscillation of a continuous string with x = 0 fixed and x = ` free,
Ã(0; t) = 0 ;
@
@x
Ã(x; t)jx=` = 0 : (7.14)
This, in turn, implies that the normal modes are the
same as for the transversely oscillating
string, (6.15),
An(x) = sin
µ
(2n + 1)¼x
2`
¶
for n = 0 to
1: (7.15)
However, again because of (7.5), these modes look
very different from those of the string.
The first nine are animated in program 7-2 (compare
with program 6-2).
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