466 On a small Sphere vibrating in a resisting Medium. 

 And by integrating in the usual way to obtain the pressure on 



the whole sphere, it will be found to be — . sin $ 



3 



cos (b t — <$). This is reckoned positive in the direction con- 

 trary to that of the motion of the sphere. Hence if <r = the 

 ratio of the specific gravity of the fluid to that of the sphere, 

 the accelerative force of the resistance in the positive direction 



of the motion is ■ . sin <J> cos (bt — <p). If a = the di- 

 stance to which motion is propagated in the fluid in the time 

 of one vibration of the sphere, b = , and consequently, 



A 



tan <$> = This is an exceedingly small quantity. Hence 



A 



2 7r r b r 



very approximately sin $ = — — = , and the accelera- 



a a 



tive force of resistance = — V b <r cos b t. Again, if x = the 



distance of the centre of the sphere at the time t from the 



d x 

 mean place about which it is oscillating, — = V sin b t, 



d 2 x 

 and— r^- = V b cos bt. Hence the accelerative force of the 



d 3 x 

 resistance = — <r . -j-^- The length of the pendulum being 



I and the force of gravity g, the accelerative force of gravity, 

 taking account of the buoyancy of the fluid, is — ^ (1 — <r). 

 Hence, 



d* x gx . . d 2 x 



and consequently 



dt 2 



This is the result I obtained by my two former methods. 

 As it does not contain a, it is applicable to any resisting me- 

 dium, supposing the vibrations to be slow. Putting the 

 factor in brackets, under the form 1 — n <r, we shall have 



2 

 n = ■ - . For a brass ball of specific gravity 8, vibrating 



in air, n = 2 very nearly; and for the same vibrating in water, 





