Also divide the two equations (1) by P and Q respectively, and insert the frictional term 2k dyldt in the first of them. We then obtain day dy 1+p+Bp +2k + dt2 dt (1+B)(1+p) (1+ß) * (1+p) +2k +ay=pbz, B m3y= P m2z, (4) (6) and z=et, Solution and Frequencies.-To solve (6) and (7) let us write and, on inserting in (7), we have or Then (9) substituted in (6) gives (x2 + C) (x2 + 2kx+a) = pb, x2+2kx3+(c+a)x2 + 2kcx + ca—pb2 = 0, (9) (10) which is the auxiliary biquadratic in . Though this equation has the form of the general biquadratic, an approximate solution, presenting all the accuracy needed for our purpose, may be easily obtained by noting that k is small compared with the other constants. For, as appears from the experiments, k is of the order one-thousandth of the coefficient of 2 and of the constant term. Then we may write for the roots of x in the biquadratic (10) the values where i denotes ✔(−1), and r and s (being comparable to k) are to be treated as small quantities whose squares or products are negligible in comparison with p and q which depend upon the larger constants of the equation. Thus, with the roots from (11) we may write instead of (10) the equivalent equation or (x+r−ip)(x+r+ip)(x+s−iq)(x+s+iq)=0, x2+2 (r + s) x3 + (p2 + q2 + r2 + s2 +4rs)x2 (12) +2( p2s+q2r + r2s +rs2) x + ( p2 +r2)(q2+s2)=0. This, on omitting the negligible quantities, becomes the approximate equation sufficiently accurate for our purpose, x1+2(r+s).x3 + (p2 + q2) x2 + 2(p2s+q2r)x+p2q2=0. (13) The comparison of coefficients in (10) and (13) yields From (15) and (17) we may eliminate 92 and obtain a quadratic in p2 whose roots may be called p2 and q'. We thus find and and 2p2=c+a+√ {(a−c)2+4pb2}, 2q=c+a-{(a−c)2+4pb2}. Again, from (14) and (16) we obtain s= a−c+ √ {(a−c)2 + 4pl2} c -- a + √ {(a−c)2+4pb2} k. (19) Then, inserting the values of a, b, and c from (8) in (18), (19), and (20), we obtain Phil. Mag. S. 6. Vol. 35. No. 205. Jan. 1918. Thus, using (11) in (9) and introducing the usual constants, the general solution may be written in the form z=e¬rt (Aepit + Be ̄pit) + est (Cegit + De-git), . (25) and, omitting r2 and s2, -rt y = ( − p2 + c) 。−7 (Ae1it + Be-pit) + (−q2+c) e-st (Cerit e-st (Cegit + De-git) b e−rt (—A¿pit + Be¬pit) + 2qsi -(— Cerit+De-git). b e Or, by transformation of (25) and (26) and use of (21)(24), we may write the general solution in the form error. (F')2 — F2 ß3m2 +4(1+ B) k2 tan (p′— ¢) = −2√√/(1+B)k Bm (30) the exponential coefficient s being given by (24), and E, e, F, and being the arbitrary constants dependent on the initial conditions. In many of the experimental cases E' may be assimilated to E and F' to F without appreciable The changes (e-e) and (p'-p) of the phase angles may be distinctly appreciable for very small values of B. But in these cases the vibrations show a slow waxing and waning of amplitude and the phase is of very little importance. On the other hand, for ẞ equal to unity, we have tan (e'-e)=4k/m and tan (p'-4)=-2/2k/m. And the numerical values of these are of the order 0.020 and 0.014, hence e'-e=1° 10′ and 4'-4=0° 48′ nearly. Hence for all our present experimental cases, we may drop the four accents in equation (28). Initial Conditions. Case I.-Suppose the heavy bob of mass Q (which pP) is pulled aside and that the light one of mass P is allowed to hang at rest in its more or less displaced position according to the coupling in use. we may write : For t=0 let z=f, then it follows statically that y= also put Then dt dt Differentiating with respect to time (27) and (28) without its accents, and writing in the latter n for m/√(1+B), we find dy = dt = -pEe ̄pst [m cos (mt+e) —ps sin (mt+e)] -st: + Fe [n cos (nt +4)-s sin (nt +)]. The conditions (31) introduced in equations (27), (28), (32), and (33) give (33) f-E sine+F sin o, (34) Bpf 0=E(m cos e-ps sin e) + F(n cos -s sin ), (36) 0=-pE(m cos e—ps sin e) + F(n cos -s sin 4). (37) But, by reason of the smallness of ps in comparison with m (of the order 0.01) and of s in comparison with n (still less), we may write instead of (36) and (37) the following: 0=Em cos e + Fn cos &, 0=-pEm cos e + Fn cos p. These values inserted in (34) and (35) give Hence, for the special solution with these initial conditions, we have Thus the ratios of the amplitudes of the quick and slow components in the y and z vibrations are respectively given by -(p-1)st and -P (1+p)B 1 (1+p)Be -(p-1)st (43 a) Case II.-Suppose now that the heavy bob (of mass Q=PP) is pulled aside while the light one (of mass P) is held undisplaced. Then we have: } (44) Putting (44) in (27), (28) without accents, (32) and (33), and omitting small quantities as before, we find and putting these values in (45), we obtain } Hence, for the special solution with these initial conditions, |