Kinetic Theory of Solids. 219 



This is the case treated by Zoppritz, whose solution in our 

 symbols is 



. 9 9 6ml 3 __> 9'17<7 ???£ 2 



4,,v rg75P =6B ±-8"-r' 



which differs only slightly from ours in the numerical co- 

 efficient of the small term. 



Let wj n 2 be the numbers of vibrations per second when the 

 bar vibrates with the loaded end up and down respectively, then 



and the effect of gravity is eliminated. This, then, is the equa- 

 tion to use with Kupffer's method. 

 WhenM = 0, 



but 1 B = qk 2 cr, 



If the bar is of circular section, P = ?* 2 /4 and o-=77T 2 ; and 

 then 



Equation (B) can be confirmed by Kupffer's experiments. 

 For one and the same bar the left-hand side is to be constant 

 for all values of the load. For a steel bar 49 "66 inches long 

 and weighing 1*5848 Eussian pounds, we have the following 

 values of the left-hand side for the following values of M :— 



M in Eussian pounds... 0-6852 1-0644 1-8173 



5-669 5-650 5-630 5-600 



Again, for a piece of the same bar 40 '08 inches long and 

 weighing 1*2792 Eussian pounds, we have the values: — 



M... 0-6852 1-0644 1-8173 3-2970 4-0566 



10-63 10-60 10-58 10-52 10-45 10-39 



In both cases the number which ought to be constant shows 

 a slight diminution with increasing load, which is easily ac- 

 counted for, and the approximation to constancy is close 

 enough to warrant the use of the formula for the determina- 

 tion of Young's modulus. To satisfy myself of the complete 



