constant. Hence just as much water must enter through DK in each second as leaves through BL. Thus

or

(u − v)(d+a) = − (u+v)(d− a),

a=ud ... (1).

Next, if we consider the motion of a small element of volume which moves from D to B, we shall have, as in the last section, by equating the decrease in kinetic energy to the increase in potential energy,

Since the pressure at the surface of the liquid is everywhere the same, the pressure at D and B is the same, and so no work is done against an opposing hydrostatic pressure as the element of volume of the fluid moves along the surface from D to B. If, however, we considered an element of volume which moves from K to L along the bottom, no work will be done against gravity, for the element will be at the same height at L and K. The hydrostatic pressure at L, due to the head d+a, is greater than that at K, due to the head d-a, and so work will have been done in moving the element against this opposing hydrostatic pressure, and this work will be the equivalent of the decrease in kinetic energy. In the case of a particle between the surface and the bottom, work will be done against both gravity and hydrostatic pressure.

The velocity of the wave being equal to the square root of the product of the depth into the acceleration of gravity, and the velocity acquired by a body falling freely through a height d/2 being √gh, it follows that the wave moves with the velocity a body would acquire in falling through a height equal to half the depth of the water.

279*. Velocity of a Wave of Compression or Dilatation in an Elastic Fluid.-We have next to consider the velocity with which a longitudinal wave (§ 266) moves. In such a wave the particles of the medium execute harmonic motions along straight lines which are parallel to the direction in which the wave is moving. By the forward motion of the particles the medium in front will be compressed, as shown in Fig. 216, so that owing to the elasticity of the medium the pressure will be increased, while by the backward motion the medium in front will be rarefied, and the pressure reduced.

Suppose that the medium is contained in a tube of which the cross section is unity, and that the velocity with which the wave moves is v, and that by imparting a velocity to the medium, we bring all the waves to rest. Consider two imaginary partitions, A and B (Fig. 232), placed across the tube; then, since the waves are, by the motion of the

medium, rendered fixed in position, the phase of the wave at each of these partitions is always the same, and they always include the same number

A

B

122

FIG. 232.

of waves between them. Thus we may suppose, if we like, that A is a crest or point where the compression is a maximum, and B a trough or point where the rarefaction is a maximum, and that there is a quarter wave between, so that we should be dealing with a similar case to that shown for water waves in Fig. 231. This being so, the pressure, and therefore also the volume of unit mass of the medium, and the velocity with which the particles of the medium are moving, due to their to-and-fro motion, remain constant both at A and at B. Let 1, v1, u1 and P, V, u be the pressure, volume of unit mass, and the velocity of the particles (supposing the medium were at rest) at A and B respectively.

Now since the space intercepted between A and B always contains the same number of waves, and the density of the medium at each point is constant, the mass of the medium intercepted between a and B is always the same. Hence the quantity, Q, of the medium which, owing to the motion supposed to be imparted to the medium and to the vibration of the particles, enters through B in a given time must be equal to that which leaves through A.

Now the velocity of the particles at A with reference to A is u1- v, so that the volume of the medium which passes through A in unit time is (u,v), for the cross section of the tube is unity. Since the volume of unit mass of the medium at A is v1, the mass of the medium which crosses A in unit time is

In the same way the mass of the medium which enters through B in unit time is

Now the momentum lost through A due to this passage of the medium in unit time is the product of the mass lost by the velocity, or

Thus the gain of momentum within the space included between A and B, owing to the passage of the medium, is

Q(u2 — u1),

or, substituting for u, and u, the values given by (1) and (2), the gain is

Now the momentum contained between A and B must remain constant throughout, for the state of the medium remains the same throughout. There must, therefore, be some cause which causes a loss of momentum exactly equal to the gain we have found above to be produced by the passage of the medium. This cause is the external forces, namely, the pressures at the two planes, which act on the medium contained between A and B. Now the pressure acting forwards on the medium contained within A and B, that is, the pressure at B, is 2, while the pressure acting backwards on the front a is 1. Hence the loss of momentum during unit time by the portion of the medium included between A and B, due to the external forces, is equal to the difference of the pressures,1 that is,

If is greater than 2, there will be a loss of momentum owing to the effect of the pressures, for the resultant force opposes the motion of the medium. But p, being greater than p, must be less than 7, for at the greater pressure the volume of unit mass will be smaller. If v1 is less than 2, then Q2(2-v1) is positive, so that this quantity does really represent a gain of momentum, and we see that our signs are right. Equating the gain and loss of momentum, we get

Now we are considering the change in volume produced by a change in pressure. Hence if p, and p are two pressures, and v1 and v1⁄2 the corresponding volumes, we have the stress is - and the strain is (V-V1), for the strain is the change in volume per unit volume. Hence we get カーカッ

Therefore from (4)

v1E=

Q2=v1E. . . . (5).

Now, in the case of all waves such as we are considering, the changes of pressure are very small compared to the whole pressure, so that v1 is very nearly the same as 7%, the volume of unit mass of the medium when

1 The resultant force acting on a body is measured by the change in momentum it produces in unit time (Newton's Second Law, § 60). The force acting on the plane B is the pressure p, multiplied by the area, which is unity, while the force acting in the opposite direction on the plane A is in the same way 1.

undisturbed by the passage of the wave. Also, if we consider a partition across the tube at the part of a wave where there is neither compression or rarefaction, that is, half-way between a crest and a trough, then, as is at once apparent from a consideration of the arrows in Fig. 216, the particles of the medium are at rest as far as their vibratory motion is concerned, and so the velocity of the medium at such a place is v, the velocity which has been impressed on the whole medium. Therefore

Now vo is the volume of unit mass of the medium ; and if p is the density of the medium, we have

That is, the velocity of wave of compression and rarefaction in a medium is equal to the square root of the elasticity of the medium divided by its density. This is Newton's expression for the velocity of a longitudinal wave in a homogeneous medium.

Since the expression for the velocity does not involve the wavelength of the wave, it follows that waves of all wave-lengths or frequencies travel with the same velocity.

PART II-SOUND

CHAPTER II

PRODUCTION AND PROPAGATION OF sound

280. Sounding Body.-All bodies which are the source of that particular disturbance which, when it strikes our ear and affects the auditory nerves, we call sound, are in a state of vibration. This can

be easily proved, for if a sounding body, such as a bell, is touched, the vibratory movement can be felt, and as under the influence of the resistance offered by the finger the vibrations die out, the sound emitted will also die out. The vibrations of a stretched string, which, when plucked, gives out a sound, are visible to the eye, and as they decrease in amplitude, the intensity of the sound also decreases.

281. Conveyance of Sound to the Ear.-In order that we may perceive a sound by our ears, it is necessary that the sounding body should be connected with our ear by an uninterrupted series of portions of elastic matter. The physical state of the matter, whether it is gaseous, liquid, or solid, is immaterial.

In order to show that this is the case, we may make the well-known experiment of suspending a bell by a thin string within the receiver of an air-pump. As the receiver is exhausted, the intensity of the sound heard when the bell is struck diminishes, till finally, when a fairly good vacuum is produced, no sound at all can be heard. If air, or even a few drops of a liquid which will form a vapour in the receiver, is introduced the sound is again heard, as is also the case if the sounding bell is allowed to dip into a vessel containing mercury or other liquid standing on the plate of the pump, or to touch a solid rod which is connected to the receiver or the plate. This experiment, therefore, shows that sound cannot be transmitted through a vacuum, but that it is transmitted through gases, vapours, liquids, and solids.

The ease with which sound travels through some solids, such as wood, is very clearly shown by holding one end of a long wooden rod against the ear, when even a very light scratch with a pin at the far end will be heard with great distinctness.

Since sound requires the presence of matter for its transmission, we are at once led to inquire what is the mechanism by which this transference takes place. There are two distinct methods by means of which