576 Messrs. Morton and Vinycomb on Vibrations of 
from the mean position is equal to that done on the string in 
the inward motion. In order that this balance may be upset, 
we must have a difference of phase between the displacement 
of the string at any point and its inclination. This involves 
a progressive change of phase in the vibrations as we go along 
the string. The “nodes”? are now points of minimum 
amplitude, falling midway between the loops where the 
amplitude is greatest. There is a phase-difference 7 between 
consecutive loops; but this is now spread over the whole 
intervening segment instead of being concentrated at the 
node. The change of phase is much more rapid near the node 
than elsewhere. We shall show that the rate of change of 
phase per unit length varies directly as the energy transfer 
across that point of the string, and inversely as the square 
of the amplitude. The phase at the node itself will clearly 
be midway between the phases of the adjacent loops ; so that 
this point will be passing though the middle of its very small 
excursion when the parts of the string near the loops are in 
their extreme positions. 
Lord Rayleigh * has examined a special case of motion of 
this kind in his discussion of the steady motion of a string 
when there is a frictional term in the equation of motion. We 
may apply general considerations as follows. 
Let « be the distance of a point on the string from an 
arbitrary origin, y the lateral displacement of this point. 
Suppose that y=/(«) sin (pt+e), where ¢ is a function of « 
which we may take as vanishing at the origin. Then, if T is 
the tension of the string, and if we consider the string lying 
on the positive side of the origin, the energy leaving this part 
of the string, at the origin, is 
"an fo 
{ T Ge ay 
a 
=|\T (“) pst(O) cos pt dt ; 
e “70 
and ae =7'(0) sin pt +/(0) cos pt . (=) ’ 
aa : 
dx 
* ‘Sound,’ vol. i. p. 198. There is an error in formula (9) of § 134, 
owing to the omission of second-order terms in taking approximate 
values for « and. The term in x? in the numerator should be 
cx Tor 7 <3 Mh 
=e pee + sin B® cos 2 ; 
dap he a a a 
with a corresponding change in the denominator. 
a 
irs 
"i 
