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sin 2° 18' = .0401.
sin 4° 35' = .0799.

Balliol Coll. 1881. 110. Light from two exactly equal and similar small sources very close together falls on a screen. Account for the bands seen, explaining the difference in the appearance, according as the light is white or of some definite refrangibility.

The distance between the two sources of light is •184 cm., and the distance between the sources and the screen is 112 cm.; a series of bright and dark bands at a distance of .036 cm. apart is observed on the screen. Find the wave length of the light used. Ind. C. S. 1885.

111. The minimum deviation of a ray of light produced by passing through a prism of angle 60° 6' 20" is 42° 40' 20". Show how to use these results to determine the refractive index of the glass prism, and find it, having given

L sin 51° 24' = 9.89294, L sin 30° 4' =9.69984,
L sin 51° 23' =9.89284, L sin 30° 3' = 9.69963.
log 1.5610= .19340, log 1.5600= 19312.

Int. Sc. Honours 1886.

CHAPTER VII

SOUND

Velocity of Sound. - Newton proved that the velocity of sound in any medium is given by the equation V= VE/D, E denoting the elasticity and D the density of the medium.

The elasticity of a fluid is defined as being the ratio of any small increase of pressure to the proportional decrement of volume thereby produced. It can be shown that the elasticity of a perfect gas is equal to its pressure, provided that its temperature remains constant during the compression.

A geometrical proof of this important proposition is given in Maxwell's Theory of Heat. It may also be proved as follows :Let V be the volume of a given mass of gas under the pres

Now suppose the pressure to increase by a small amount p, and let v be the decrement of volume thereby produced : the pressure is now P + p, and the volume V-v. If the gas obeys Boyle's law, the product of these two quantities is equal to the product of the original pressure and volume, or

PV = (P +P)(V -v),

= PV + VP-Pv - pv. Since both the quantities P and v are small, their product may be neglected (p. 17); thus

Vp=Pv. Now the proportional decrement of volume (or decrement per

sure P.

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cm,

per sec.

unit volume) is so that the elasticity (by definition) is p

V or Vplu. But by the last equation

Volv=P, and therefore the elasticity is equal to the pressure.

We have seen (p. 70) that if the barometer stands at h cm., the atmospheric pressure is II =hpg dynes per sq. This gives us for the velocity of sound in air at oo

V=Vhpg/D, or, at the normal pressure,

V=V1,013, 226/0.001,293,

= 279,933 cm. per sec., whereas the velocity found by experiment is 332 metres

This discrepancy between the calculated and observed values remained unexplained until Laplace's time. He pointed out that the compression produced by a sound-wave takes place so rapidly that any heat which is developed by it cannot be conducted away: the elasticity cannot therefore be calculated on the supposition that the temperature remains constant. assume that no heat is allowed to escape, it can be proved that the elasticity is equal to yII, II being the

y

the ratio between the specific heat of air at constant pressure and its specific heat at constant volume. Introducing this correction (known as Laplace's correction) we have

V=Vyhpg/D, or taking y=1.4,

V=V1.4 x 1013226/0.0001 293.

= 331,221 cm. per sec., which is almost identical with the velocity found by experiment.

1. Explain, on general principles, why the velocity of sound in air increases with the temperature, but is

If we

pressure and

independent of the pressure ; and calculate its velocity at 15°. Note.The velocity of sound in air at oo is 332 metres per

second. 2. Find the temperature at which the velocity of sound in air is 350 metres per second.

3. An observer sets his watch by the sound of a signal-gun fired at a station 1500 metres off : find, to a hundredth of a second, the error due to distance, the temperature being 15°.

4. Show that the left and right hand members of the equation V= VE/D are of the same dimensions.

5. A tuning-fork is held over a tall glass jar, into which water is gradually poured until the maximum reinforcement of the sound is produced. This is found to be the case when the length of the column of air is 64.8 cm. What is the vibration number of the fork?

6. Calculate the velocity of sound in hydrogen gas, assuming its velocity in air, and having also given that i litre of hydrogen=0.0896 gm., and i litre of air = 1.293 gm.

7. Find the length of a closed organ-pipe, which when blown will give the note c (256 vibrations per second).

8. Calculate the velocity of sound in water at 10°, its coefficient of elasticity at this temperature being 2.1 x 1010. The velocity of sound in liquids is given by the same expres

sion VE/D, E denoting the coefficient of elasticity (or the reciprocal of the coefficient of compressibility) and D

the density. At 10° the density of water is sensibly equal to unity, and the required velocity is

V= 12• 1 X 106 = 144,900 cm. per sec. 9. By experiments made in the Lake of Geneva, Colladon and Sturm found that sound travelled in water at 8°. I with a velocity of 1435 metres per sec. What value does this give for the elasticity of water ?

=

10. It is found that a force equal to the weight of 3000 lbs. is required to elongate a bar of iron i sq. inch in section by 1/10,000 of its original length : calculate from this the velocity of sound in iron. [1 cub. ft. of iron = 480 lbs.] In calculating the velocity of sound in solids, E is to be

taken as denoting Young's modulus of elasticity, the value of which for iron, in poundals per square foot, is

10,000 x 3000 x 32 x 122. The density of iron, in lbs. per cub. ft., is given as 480. Thus

V=VE/D,

=V9.6 x 108 x 122/480= 17,000 ft. per sec. 11. Calculate the value of Young's modulus for steel, having given that its density is 7.8, and that sound travels in it with a velocity of 5200 metres per second.

12. What will be the pitch of the note emitted by a wire 50 cm. in length, when stretched by a weight of 25 kilogrammes, if 2 metres of the wire are found to weigh 4.79 gm. ? It can be proved that a sound-wave travels along a stretched

wire or string with a velocity v=VF/M, where F is the stretching force, and M is the mass of the wire per unit of length.

1 The modulus of elasticity (Young's modulus) is defined as follows :Suppose a bar or wire of length L and cross-section o to be stretched by a force F, and let 1 denote the elongation produced ; then the modulus of elasticity for the material of which it is composed is E=LF|lo. The modulus is frequently expressed in lbs. per sq. in., or kilogrammes per sq. cm.; in such cases it must be multiplied by g, and care should be taken to use the same units consistently in the calculation, e.g. if we wish to find the velocity in centimetres per sec., the modulus must be expressed in dynes per sq. cm. and the density in grammes per c.c.

2 If the wire be of density d, M=grld, so that if we denote by t the tension of the wire per unit of sectional area, t=F/#r2, and therefore v=vt|d. The complete expression for the number of vibrations per second produced by a wire of density d stretched by a weight P is

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