538 

 1 



■E{kin) == — mv'^ 



Li 



the velocit}^ v is known . The mass m of a part dy of the string is 

 expressed by : 



m =: Ttr^g dy 



in which r is lialf the diameter, g the specific weight of the string. 

 We obtain finally for the kinetical energy of the part dy 



an 



1 / 2 \ 2 ■ ■ 



Integrating over half the length of the string, and by multiplying 

 by 2, the total kinetical energy of the string gives at its passage 

 through the position of equilibrium 



}-l n:a jra 



E^kin) = 2 r i- (A- '^\ oi'jir'gs ^<^dy = — jir'glojVre ^to 







This value should have been equal to the value obtained for the 

 potential energy, if the movement had been undamped. The damping 

 makes the amount of the kinetical energy too small. The exact 



amount is obtained by multiplication by f"^^. 



As the expression of the potential enei'gy is expressed in ergs, 

 this must likewise be done with the kinetical energy, which causes 

 the introduction of the factor 1.0197. 



We obtain tinallv after the introduction of i\ = — : 



2-r 



— Hllh = 1.0197 X — :^r'gIoj'P 

 3 15 ^^ 



and hence : 



N^hjir^g 



/i — 32 2 



I 



2"^ method. 



We can calculate the lateral pressure on the string in its deviated 

 position in different ways. Above we had already mentioned the 

 expression HII for it. 



Proceeding from the graphostatical construction, discussed before, 

 we see that the slope of the string at the point of suspension B 

 corresponds with the inclination of the line MC (vide fig. 4). This 

 slope is given by the tangent at the point, i.e. by the magnitude of 



