An uncertainty no less troublesome than this arises from lack of knowledge of the distribution of sound intensity about the room. Were the source of sound out of doors and upon the ground, and were the ground a perfect reflector, the energy would stream uniformly through the surface of a hemisphere. A whole hemisphere is not available in this room-perhaps only half a hemisphere as the sound first travels away. Also the reflecting power of the walls and floor before me are unknown. So it appears that my hearers will not all hear the same sound, and perhaps none of them just the particular sound to which the figures apply. But I shall consider the energy uniformly distributed over the surface of a hemisphere and shall make computations for a point six meters (6 m) from the source. By comparison with a standard fork the frequency of the tone--the first possible overtone of the pipe—is 780 vibrations per sec. We will let V=34000 cm. per sec. represent the velocity of sound. Since 1=V/N, = the wave length is 34000/780=43.55 cms. The tone I get is very nearly a pure one. Therefore the displacements of the layers of air from their positions of rest may be represented by the ordinates of a sine curve. In figure I (P. 13) the lines 1, 2, 3 etc. represent the air layers in their positions of equilibrium. In figure 3 they are shown with displacements proportional to the several ordinates of the sine curve of figure 2. that layers 5 and 13 are not displaced, and represent the centers of condensations, and that layers 9 and 17 are not displaced and represent the centers of rarefactions. Also any condensation corresponds, not to a crest, but to a region of negative slope along the sine curve,-e. g. from 3 to 8– and a rarefaction corresponds to a region of positive slope-e. g. from 7 to 11. If the curve of figure 2 were detached and could slide bodily with uniform motion in the positive direction, the appropriate back and forth movements of the layers in figure 3 would represent the behavior of the air in transmitting the plane sound wave of a pure tone. In our case the distance A C, figure 2, is 43-55 cms. We want first to find the distance KS, figure 3, corresponding to HT in figure 2, the amplitude of vibration, which we will call r. When the pipe is blown with a pressure of 15 cms. of water—the particular value employed—it is found that the volume of air used was 2 liters in 4 secs., or the rate is 500 c.c. per sec. Then p=15.1.980 dynes per sq. cm., and the energy supplied per sec. is PV/t or 15.980.500/1= 15.980.500 ergs. per sec. Now we consider a sphere whose radius is 6 m, and suppose all the energy flow is distributed uniformly over one half its surface. Then the energy resident at one time in a layer of thickness S over the hemisphere is S/V of the energy sent out per sec., where V is the velocity of sound. s Therefore the energy in the shell at one time is 15.980.500 34000 Ergs. The volume of the shell is 211R-S C.C. 211.600'S =7211.10'S C.C. Then the energy per unit mass of the shell is the total energy divided by the mass, or S:15:980.500/34000.7211Siod=75.98/34.11.72.104d. Now taking S very small compared to d(=AC), the energy of the shell is alternately all kinetic as at 5, 9, 13 or 17, and all potential, as at 3, 7, 11 or 15. Since the sum of the energies in the two forms is constant, the maximum kinetic energy (which is accompanied by zero potential energy) equals the total energy. Therefore the total energy of unit mass equals the maximum kinetic, energy MV?, where V is the maximum velocity. Here M=1. Therefore K.E.=1/2V2=75.98/34.72.II.10*d. Since the air layers have simple harmonic motion the maximum velocity Vm= 211r/T where T is the period and r the amplitude sought, or what is the same thing Vm= 2IIrn, where n is the frequency, 780 vibrations per sec. Therefore, as above, KE=1202 = 1/2 47no = 75.98/34.72.10*7 d from which at once pe=2.75.98/473.7802 34.72 104d Taking d, the density of the air, to be 0.001293, or nearly 0.0013 gms. per C.c. we have 2.75.98 15.9.8 473.7802.34.72.13 5278.78234:72 from which r=7.82-10-5 cm, or about 8/100000 of a centimeter, which is KS, Fig. 3. This gives the ratio AB/HT, figure 2, as 43.5/7.82.10-5 5.57.10%, or about 560000. It is easy to get from r the value of Vm, the maximum velocity, for Vm=27r n=27 7.8:10-6.780 =0.38 cms. per sec., or about 0.0085 miles per hour. It is to be remembered that these are large values for a rather loud sound. Lord Rayleigh estimates that a sound of high frequency might, under favorable circumstances, be audible with an amplitude of 1.10-8 cms. Such figures give one an increased respect for his ears. The corresponding volume and pressure changes may be found as follows: Since the wave is plane we may consider a cylinder whose axis coincides with the direction of propogation of the sound. In any section of this cylinder the change in length divided by the original length equals the change in volume of the mass of air divided by its original volume. To get the maximum change consider a cylinder whose axis along A x is 0.001 d. At A the layer is at its position of equilibrium. The equation of the curve is y=rsin 24 x/1. Therefore the displacement at x=0.001, is y=0.000078 sin 27 0.001 M/A =0.000078 / 500, since the angle is small. Hence y=0.00000049, which is the change in length. The change in length per unit length is therefore 0.00000049/0.001.43.55=0.0000 1 125=11.25.10. This is also the change in volume per unit volume, or the maximum condensation due to the wave. To get the pressure change we consider that the conditions are adiabatic. Therefore any transformation takes place according to the equation PV'=const., where r is 1.41 for air. This may be written PV'=PV.', or (P. + AP) (V.- AV)'=P.V.. Therefore (P. + AP)/P,=V*/(V. - AV)'=V,/(V.- AV)". Since Av is very small compared to V (the ratio is 11.25.10-) we may write V/(V.-Av)=(V. + A2)/V=I + Av/V, from which I + AP/P. = (1 + AVIV.)'=Ito AV/V, if we neglect terms involving the square and higher powers of Av/V. Therefore AP/P,=r AV/V.=1.41•11.25.10-6 / = = = 15.8.10-6, the relative pressure change. 1012630 dynes per cm2, or one atmosphere, AP= 1012630-15.8-10-6 = 16 dynes per sq. cm., the maximum pressure change due to the wave. 0 0 If P. SOME USEFUL LABORATORY DEVICES. PROFESSOR F. R. GORTON, STATE NORMAL COLLEGE. A Self-filling Barometer.—The apparatus shown in Figure I consists of a glass tube one meter long which ends at the lower exeremity in an oblong bulb B. The long tube is closed at the top and attached to a meter stick by means of brass clips D, A, and C. The stick is mounted on a pedestal by the screw A in such a manner as to turn with considerable friction. A stop is set in the standard at D, so that the tube cannot be turned in the wrong direction. By turning the tube to the horizontal position, mercury can be poured into the bulb through the opening E. A little tilting of the tube will cause the mercury to run into the long tube until it is completely filled. A small excess of mercury is left in the bulb, so that it prevents any air from entering the long tube when it is restored to the vertical position. The apparatus is designed for the purpose of permitting the pupil to study fully the barometer principle; for he can rotate the tube until the bulb is uppermost and observe the filling of the tube. He can then slowly restore the tube to the vertical position while observing the formation of the vacuum at the top, precisely as in the Torricellian experiment. Furthermore, the device may be used as a laboratory barometer; for the air will become so completely removed from the tube by tilting it repeatedly that the mecurial column will stand very nearly as high as that of a good barometer. Another important use has been found for the instrument when made in the form shown in the figure. If the opening E is connected to an airpump, the difference in level of the mercury on the two sides of the meter stick shows at once the pressure of the air left in the instument. As a manometer, the apparatus will be found useful in many experiments. A Simple Mounting for a Boyle's Law Tube.—Figure 2 shows a common barometer tube whose inside diameter does not exceed three millimeters mounted at the side of a meter stick by means of three brass clips. The stick is then attached at its center to a standard, as shown, in such a manner as to turn with friction enough to hold it in any position. A thread of mercury about 30 cm. long is then introduced into the tube to such a point that the enclosed air space is about 35 cm. long. By means of a fine iron (not copper) wire, it will be found that the mercury can beadjusted to any desired position in the tube. Both tube and mercury should be perfectly clean and dry. In using the device to illustrate Boyle's law, the tube is placed in a horizontal position, and readings are then made to ascertain the length of the air space in the closed end and the length of the thread of mercury. Since the tube is horizontal, the thread of mercury has no influence upon the pressure of the gas; hence the pressure on the enclosed gas in one atmosphere. In other positions, as the one shown in the figure, the distances AB and CD are measured and the effective pressure of the thread of mercury ED is computed. This amount is then added to, or subtracted from, the atmospheric pressure according as the closed end of the tube is below or above the horizontal line passing through the axis. A Self-Starting Siphon.—The siphon shown in Fig. 3 was described in a scientific journal a few years ago and has been modified only slightly in this instance. An ordinary glass siphon is provided with a small opening at A about 2 inches from the end. This hole is about the size of a pin. A bottle which has been cut in two at the middle is fitted to the end of the - siphon with a rubber stopper as shown. By placing the end of the apparatus in a vessel of water, the liquid begins immediately to flow over the bend and out at the lower end. This is caused by the fact that the compressed air in chamber A is driven through the pin hole, thus forming air bubbles in the tube. These bubbles together with the water which is carried along between them pass over the bend in the tube in sufficient number to start the siphon. While the instrument is an interesting curiosity, it will be found useful in removing liquids from battery jars and in other cases of a similar nature. |