17.8. THE FOUCAULT PENDULUM∗ 367

where T , the tension in the string of the pendulum, is directed towards the point at whichthe pendulum is supported, and m is the mass of the weight at the end of the pendulum.The pendulum can be thought of as the position vector from (0,0, l) to the surface of thesphere x2 + y2 +(z− l)2 = l2. Therefore,

T =−Txli−T

ylj+T

l − zl

k

and consequently, the differential equations of relative motion are

x′′ =−T xml +2ωy′ cosφ ,

y′′ =−T yml −2ω (x′ cosφ + z′ sinφ) ,

z′′ = T l−zml −g+2ωy′ sinφ .

If the vibrations of the pendulum are small so that for practical purposes, z′′ = z = 0, thelast equation may be solved for T to get gm− 2ωy′ sin(φ)m = T.Therefore, the first twoequations become

x′′ =−(gm−2ωmy′ sinφ

) xml

+2ωy′ cosφ

andy′′ =−

(gm−2ωmy′ sinφ

) yml

−2ω(x′ cosφ + z′ sinφ

).

All terms of the form xy′ or y′y can be neglected because it is assumed x and y remain small.Also, the pendulum is assumed to be long with a heavy weight so that x′ and y′ are alsosmall. With these simplifying assumptions, the equations of motion become

x′′+g xl = 2ωy′ cosφ ,

y′′+g yl =−2ωx′ cosφ .

These equations are of the form

x′′+a2x = by′, y′′+a2y =−bx′ (17.20)

where a2 = gl and b = 2ω cosφ . There are systematic ways to solve the above linear system

of ordinary differential equations, but for the purposes here, it is fairly tedious but routineto verify that for each constant c,

x = csin(

bt2

)sin

(√b2 +4a2

2t

), y = ccos

(bt2

)sin

(√b2 +4a2

2t

)(17.21)

yields a solution to 17.20 along with the initial conditions,

x(0) = 0,y(0) = 0,x′ (0) = 0,y′ (0) =c√

b2 +4a2

2. (17.22)

It is clear from experiments with the pendulum that the earth does indeed rotate out fromunder it causing the plane of vibration of the pendulum to appear to rotate. The purposeof this discussion is not to establish this obvious fact but to predict how long it takes forthe plane of vibration to make one revolution. There will be some instant in time at whichthe pendulum will be vibrating in a plane determined by k and j. (Recall k points awayfrom the center of the earth and j points East. ) At this instant in time, defined as t = 0,

17.8. THE FOUCAULT PENDULUM* 367where T’, the tension in the string of the pendulum, is directed towards the point at whichthe pendulum is supported, and m is the mass of the weight at the end of the pendulum.The pendulum can be thought of as the position vector from (0,0,/) to the surface of thesphere x7 + y?+(z—1 )? =[?. Therefore,XxT=-Tll—i—T> j+T—kl land consequently, the differential equations of relative motion arex= TS +2ay'coso,y’ =-T —2@ (x cos +z'sing),2! =T3— 94 2@y’sing.If the vibrations of the pendulum are small so that for practical purposes, z’” = z = 0, thelast equation may be solved for T to get gm — 2@y’sin(@)m = T.Therefore, the first twoequations becomeNx" = —(gm—2omy' sing) = +2ay' cos @mandiy’ =—(gm—2amy' sing) = —2@ (x'cos@ +z’ sing).mAll terms of the form xy’ or y’y can be neglected because it is assumed x and y remain small.Also, the pendulum is assumed to be long with a heavy weight so that x’ and y’ are alsosmall. With these simplifying assumptions, the equations of motion becomex! +g% = 2ay'cos¢@,y" +87 = —20x' cos 9.These equations are of the formxX" +a-x= by’, y" +a’y = —bx’ (17.20)where a* = £ and b = 2@cos @. There are systematic ways to solve the above linear systemof ordinary differential equations, but for the purposes here, it is fairly tedious but routineto verify that for each constant c,_ (bt\ . [ vb?+4a2 bt\ . [ Vb? +4a2x=csin > sin —z , Y=ccos > sin —z7,—# (17.21)yields a solution to 17.20 along with the initial conditions,\/b2 2x(0) =0,y(0) =0,x' (0) =0,9" (0) = we (17.22)It is clear from experiments with the pendulum that the earth does indeed rotate out fromunder it causing the plane of vibration of the pendulum to appear to rotate. The purposeof this discussion is not to establish this obvious fact but to predict how long it takes forthe plane of vibration to make one revolution. There will be some instant in time at whichthe pendulum will be vibrating in a plane determined by k and 7. (Recall k points awayfrom the center of the earth and 7 points East. ) At this instant in time, defined as t = 0,