LONGITUDINAL PART 2

7.2 A Mass on a Light Spring

Let us return to the system that we studied at the very beginning of the book, the harmonic

oscillator constructed by putting a mass at the end of a light spring. We are now in a position

to understand precisely what “light” means for this system, because we can now allow the

spring to have a nonzero linear mass density, ½L, and find the normal modes of this system.

We will then be able to see what happens as ½L ! 0.

To be specific, consider a spring with equilibrium length ` and spring constant K, fixed at

x = 0 and constrained to oscillate only in the x direction (that is longitudinally). Now attach

a mass, m, to the free end (with equilibrium position x = `). The spring, for 0 < x < `, can

be regarded as part of a space translation invariant system. To find the normal modes for this

system, we look for linear combination of the modes of the infinite spring (for a given !) that

reproduces the physics at x = 0 and x = `. The fixed end at x = 0 is easy. This fixes the

form of the modes to be proportional to

sin knx (7.16)

with frequency

!n =

s

K`

½L

kn : (7.17)

As always, kn and !n are related by the dispersion relation, (7.4). Now to determine the

possible values of kn, we require that F = ma be satisfied for the mass. Suppose, for

example, that the amplitude of the oscillation is A (a length). Then the displacement of the

point on the spring with equilibrium position x is

Ã(x; t) = Asin knx cos !nt ; (7.18)

158 CHAPTER 7. LONGITUDINAL OSCILLATIONS AND SOUND

and the displacement of the mass is determined by the displacement of the end of the spring,

x(t) ´ Ã(`; t) = Asin kn` cos !nt : (7.19)

The acceleration is

a(t) = @2

@t2Ã(`; t) = ¡!2

n Asin kn` cos !nt (7.20)

q q q ..................

...

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...

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...

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... i..................

...

..................

...

..................

...

...................

...

m

q q q .......

................

.......

................

.......

................

.......

................ i.......

................

.......

................

.......

................

.......

................

m

equilibrium

stretched

j j

` ¡ a `

j j j j

Ã(` ¡ a; t) Ã(`; t)

Figure 7.2: The stretching of the last spring is Ã(`; t) ¡ Ã(` ¡ a; t).

To find the force on the mass, consider the massive spring as the continuum limit as

a ! 0 of masses connected by massless springs of equilibrium length a, as at the beginning

of the chapter. Then the force on the mass at the end is determined by the stretching of the last

spring in the series. This, in turn, is the difference between the displacement of the system at

x = ` and x = ` ¡ a, as illustrated in figure 7.2. Thus the force is

F = ¡Ka [Ã(`; t) ¡ Ã(` ¡ a; t)] : (7.21)

In order to take the limit, a ! 0, rewrite this as

F = ¡Kaa

Ã(`; t) ¡ Ã(` ¡ a; t)

a

: (7.22)

Now in the continuum limit,Kaa isK`, and the last factor goes to a derivative, @

@x Ã(x; t)jx=`.

The final result for the force is therefore2

F = ¡K`

@

@x

Ã(x; t)jx=` = ¡K` kn Acos kn` cos !nt : (7.23)

2Note that we can use this to give an alternate derivation of the boundary condition for a free end, (7.14).

7.2. A MASS ON A LIGHT SPRING 159

Note that the units work. K` is a force. @

@xà is dimensionless.

Putting (7.20) and (7.23) into F = ma and canceling a factor of ¡Acos !nt on both

sides gives,

K`kn cos kn` = m!2

n sin kn` : (7.24)

Using the dispersion relation to eliminate !2

n , we obtain

kn` tan kn` = ½L`

m

: (7.25)

We have multiplied both sides of (7.25) by ` in order to deal with the dimensionless

variables kn` (which is 2¼ times the number of wavelengths that fit onto the spring) and the

dimensionless number

² ´

½L`

m

(7.26)

(which is the ratio of the mass of the spring, ½L`, to the mass, m). The spring is light if ² is

much smaller than one.

The important point is that (7.25) has only one solution for kn` that goes to zero as ² ! 0.

Because tan k` ¼ k` for small k`, it is

k0` ¼

p

² : (7.27)

For all the other solutions, the smallness of the left-hand side of (7.25) must come because

tan kn` is very small,

kn` ¼ n¼ for n = 1 to 1: (7.28)

But (7.28) implies

x(t) ´ Ã(`; t) = Asin kn` cos !nt ¼ 0 for n = 1 to 1: (7.29)

In other words, in all the solutions except k0, the mass is hardly moving at all, and the spring

is doing almost all the oscillating, looking very much like a system with two fixed ends.

Furthermore, the frequencies of all the modes except the k0 mode are large,

!n ¼ n¼

s

K

½L`

for n = 1 to 1; (7.30)

while the frequency of the k0 mode is

!0 ¼

s

K

m

: (7.31)

For small ² (large mass), the k0 mode is associated primarily with the oscillation of the mass,

and has about the frequency we found for the case of the massless spring. The other modes

are in an entirely different range of frequencies. They are associated with the oscillations of

the spring. This is an important example of the way in which a single system can behave in

very different ways in different regimes of frequency.

160 CHAPTER 7. LONGITUDINAL OSCILLATIONS AND SOUND

7.3 The Speed of Sound

z = 0

z = `

6

z

Figure 7.3: An organ pipe.

The physics of sound waves is obviously a three-dimensional problem. However, we can

learn a lot about sound by considering motion of air in only one-dimension. Consider, for

example, standing waves in the air in a long narrow tube like an organ pipe, shown in cartoon

form in figure 7.3. Here, we will ignore the motion of the air perpendicular to the length of

the pipe, and consider only the one-dimensional motion along the pipe. As we will see later,

when we can deal with three-dimensional problems, this is a sensible thing to do for low

frequencies, at which the transverse modes of oscillation cannot be excited. If we consider

only one-dimensional motion, we can draw an analogy between the oscillations of the air in

the pipe and the longitudinal waves in a massive spring.

It is clear what the analog of ½L is. The linear mass density of the air in the tube is

½L = ½A (7.32)

where A is the cross-sectional area of the tube. The question then is what is K` for a tube of

air?

Consider putting a piston at the top of the tube, as shown in figure 7.4. With the piston at

the top of the tube, there is no force on the piston, because the pressure of the air in the tube

is the same as the pressure of the air in the room outside. However, if the piston is moved in

a distance dz, as shown figure 7.5, the volume of the air in the tube is decreased by

¡ dV = Adz : (7.33)

7.3. THE SPEED OF SOUND 161

z = 0

z = `

6

z

Figure 7.4: The organ pipe with a piston at the top. The air in the tube acts like a spring.

z = 0

z = `

6

z

dz

¡

¡

Figure 7.5: Pushing in the piston changes the volume of the air in the tube.

If the piston were moved in slowly enough for the temperature of the gas to stay constant,

162 CHAPTER 7. LONGITUDINAL OSCILLATIONS AND SOUND

then the pressure would simply be inversely proportional to the volume. However, in a sound

wave, the motion of the air is so rapid that almost no heat has a chance to flow in or out of

the system. Such a change in the volume is called “adiabatic.” When the volume is decreased

adiabatically, the temperature goes up (because the force on the piston is doing work) and the

pressure increases faster than 1=V , like

p / V ¡° (7.34)

where ° is a positive constant that depends on the thermodynamic properties of the gas. More

precisely, ° is the ratio of the specific heat at constant pressure to the specific heat at constant

volume:3

CP =CV (7.35)

In air, at standard temperature and pressure

°air ¼ 1:40 (7.36)

Now we can write from (7.34),

dp

p

= ¡°

dV

V

(7.37)

or

dp = ¡°p

dV

V

¼

°Ap0

V

dz = °p0

`

dz (7.38)

where p0 is the equilibrium (room) pressure. Then the force on the piston is

dF = Adp = ° A2 p0

V

dz = ° Ap0

`

dz (7.39)

so that

K = dF

dz

= ° Ap0

`

(7.40)

and K` is

K` = ° Ap0 : (7.41)

Thus we expect the dispersion relation to be

!2 = v2

soundk2 = K`

½L

k2 = ° p0

½

k2 (7.42)

where we have defined the “speed of sound”, vsound, as

v2

sound = ° p0

½

(7.43)

3See, for example, Halliday and Resnick.

7.3. THE SPEED OF SOUND 163

For air at standard temperature and pressure,

vsound ¼ 332

m

s : (7.44)

As we will see in the next chapter, this is actually the speed at which sound waves travel. For

now, it is just a parameter in our calculation of the normal modes.

In the pipe shown in (7.3), the displacement of the air, which we will call Ã(z; t), must

vanish at z = 0, because the bottom of the tube is closed and there is nowhere for the gas to

go.

The z derivative of à must vanish at z = `, because the excess pressure is proportional

to ¡ @

@zÃ. The pressure is proportional to the force in our analogy with longitudinal waves in

the massive spring. Using (7.41) and (7.23), we expect the longitudinal force to be

§ °Ap0

@

@z

à (7.45)

or the excess pressure to be

p ¡ p0 = ¡° p0

@

@z

à : (7.46)

We want the negative sign because for @

@zà > 0, the air is spreading out and has lower

pressure.

Thus for a standing wave in the pipe, (7.3), we expect the boundary conditions

Ã(0; t) = 0 ;

@

@z

Ã(z; t)jz=` = 0 ; (7.47)

for which the solution is

Ã(z; t) = sin kz cos !t (7.48)

k =

(n + 1=2)¼

`

; ! = vk ; (7.49)

where v = vsound, for nonnegative integer n. In particular, the lowest frequency mode of the

tube corresponds to n = 0,

! = v¼

2`

; º = !

2¼

= v

4`

: (7.50)