Abstract

Synopsis

The primary object of the investigation is to derive a convergent series for the period of the bifilar pendulum corresponding to any feasible finite amplitude of vibration. The material is conveniently divided in three parts.

  • In part (a) attention is called to the unsatisfactory state of the literature of the problem. Then a set of rigorous equations are given which lead to a hyperelliptic integral as the formula for the period.
  • In part (b) this integral is expanded into two related series which depend upon a fundamental series the law of development of which is evident. It is shown that the second and third approximation terms of the two final series may be either independently or simultaneously positive, or zero, or negative. Numerical illustrations are then presented.
  • In part (c) the domain of convergence of the series is discussed and shown to depend upon points lying between certain straight and parabolic lines in a Cartesian diagram.

© 1923 Optical Society of America

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References

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  1. Repertorium der Physik,  23, p. 409; 1887.
  2. Nachrichten v.d. König. G. d. Wiss. zu Göttingen,  47, p. 158; 1891.
  3. The error made in the position of the decimal point introduced while taking the antilogarithm has been corrected.
  4. By using H. Andoyer’s “Nouvelles Tables Trigonométriques Fondamentales.”
  5. I do not intend to imply that the projected angle θ for the bifilar pendulum is always the precise analogue of the angular displacement of the simple pendulum.

1891 (1)

Nachrichten v.d. König. G. d. Wiss. zu Göttingen,  47, p. 158; 1891.

1887 (1)

Repertorium der Physik,  23, p. 409; 1887.

Andoyer’s, H.

By using H. Andoyer’s “Nouvelles Tables Trigonométriques Fondamentales.”

Nachrichten v.d. König. G. d. Wiss. zu Göttingen (1)

Nachrichten v.d. König. G. d. Wiss. zu Göttingen,  47, p. 158; 1891.

Repertorium der Physik (1)

Repertorium der Physik,  23, p. 409; 1887.

Other (3)

The error made in the position of the decimal point introduced while taking the antilogarithm has been corrected.

By using H. Andoyer’s “Nouvelles Tables Trigonométriques Fondamentales.”

I do not intend to imply that the projected angle θ for the bifilar pendulum is always the precise analogue of the angular displacement of the simple pendulum.

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Equations (42)

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z = L ( 1 cos ϕ ) = 2 L sin 2 1 2 ϕ
L sin ϕ = 2 D sin 1 2 θ
M d 2 z d t 2 = 2 T cos ϕ M g
M k 2 d 2 θ d t 2 = 2 D T sin ϕ cos 1 2 θ
1 2 M k 2 ( d θ d t ) 2 + 1 2 M ( d z d t ) 2 + M g z = M g z 0
d 2 θ d t 2 = 2 D 2 T L M k 2 sin θ
T = M 2 cos ϕ { g + L cos ϕ ( d ϕ d t ) 2 + L sin ϕ d 2 ϕ d t 2 }
L { sin 2 ϕ + 4 k 2 cos 2 ϕ 4 D 2 L 2 sin 2 ϕ } d 2 ϕ d t 2 + g sin ϕ + L { 1 + 4 k 2 ( L 2 4 D 2 ) ( 4 D 2 L 2 sin 2 ϕ ) 2 } sin ϕ cos ϕ ( d ϕ d t ) 2 = 0
1 2 · d d ϕ [ f ( ϕ ) ] .
f ( ϕ ) · d 2 ϕ d t 2 + 1 2 · ( d ϕ d t ) 2 · d d ϕ [ f ( ϕ ) ] + g sin ϕ = 0
f ( ϕ ) . d [ 1 2 · ( d ϕ d t ) 2 ] + 1 2 · ( d ϕ d t ) 2 . d [ f ( ϕ ) ] g . d ( cos ϕ ) = 0
f ( ϕ ) · ( d ϕ d t ) 2 2 g ( cos ϕ cos ϕ 0 ) = 0
0 P / 4 d t = 0 ϕ 0 f ( ϕ ) ¯ 2 g ( cos ϕ cos ϕ 0 ) ¯ d ϕ
P = 2 2 ¯ g ¯ 0 ϕ 0 f ( ϕ ) ¯ cos ϕ cos ϕ 0 ¯ d ϕ
sin 1 2 ϕ = sin 1 2 ϕ 0 sin ξ
P = 4 L g 0 π / 2 { [ κ 2 + 4 ( δ 2 κ 2 ) σ + 4 ( κ 2 δ 2 1 ) σ 2 + 8 σ 3 4 σ 4 ] 1 2 [ δ 2 ( 1 + δ 2 ) σ + 2 σ 2 σ 3 ] 1 2 } d ξ
a 0 + a 1 σ + a 2 σ 2 + + a j σ j + ,
δ 2 a 0 a 0 = κ 2 2 δ 2 a 0 a 1 ( 1 + δ 2 ) a 0 a 0 = 4 ( δ 2 κ 2 ) , δ 2 ( 2 a 0 a 2 + a 1 a 1 ) 2 ( 1 + δ 2 ) a 0 a 1 + a 1 + 2 a 0 a 0 = 4 ( δ 2 κ 2 + 1 ) , 2 δ 2 ( a 0 a 3 + a 1 a 2 ) ( 1 + δ 2 ) ( 2 a 0 a 2 + a 1 a 1 ) + 4 a 0 a 1 a 0 a 0 = 8 , δ 2 ( 2 a 0 a 4 + 2 a 1 a 3 + a 2 a 2 ) 2 ( 1 + δ 2 ) ( a 0 a 3 + a 1 a 2 ) + 2 ( 2 a 0 a 2 + a 1 a 1 ) 2 a 0 a 1 = 4 , 2 δ 2 ( a 0 a 5 + a 1 a 4 + a 2 a 3 ) ( 1 + δ 2 ) ( 2 a 0 a 4 + 2 a 1 a 3 + a 2 a 2 ) + 4 ( a 0 a 3 + a 1 a 2 ) ( 2 a 0 a 2 + a 1 a 1 ) = 0 , δ 2 ( 2 a 0 a 6 + 2 a 1 a 5 + 2 a 2 a 4 + a 3 a 3 ) 2 ( 1 + δ 2 ) ( a 0 a 5 + a 1 a 4 + a 2 a 3 ) + 2 ( 2 a 0 a 4 + 2 a 1 a 3 + a 2 a 2 ) 2 ( a 0 a 3 + a 1 a 2 ) = 0.
a 0 = κ δ , b 1 a 1 a 0 = ( 4 δ 4 3 δ 2 κ 2 + κ 2 ) / 2 δ 2 κ 2 , b 2 a 2 a 0 = ( 16 δ 8 + 24 δ 6 κ 2 5 δ 4 κ 4 8 δ 4 κ 2 10 δ 2 κ 4 + 3 κ 4 ) / 8 δ 4 κ 4 , b 3 a 3 a 0 = ( 64 δ 12 144 δ 10 κ 2 + 92 δ 8 κ 4 7 δ 6 κ 6 + 48 δ 8 κ 2 8 δ 6 κ 4 + 7 δ 4 κ 6 4 δ 4 κ 4 21 δ 2 κ 6 + 5 κ 6 ) / 16 δ 6 κ 6 , b 4 a 4 a 0 = ( 1280 δ 16 + 3840 δ 14 κ 2 3936 δ 12 κ 4 + 1456 δ 10 κ 6 45 δ 8 κ 8 + 1280 δ 12 κ 2 + 1344 δ 10 κ 4 176 δ 8 κ 6 + 12 δ 6 κ 8 96 δ 8 κ 4 + 16 δ 6 κ 6 + 162 δ 4 κ 8 16 δ 4 κ 6 180 δ 2 κ 8 + 35 κ 8 ) / 128 δ 8 κ 8 .
P = 4 k D L g { 0 π / 2 d ξ + b 1 c 2 0 π / 2 sin 2 ξ d ξ + b 2 c 4 0 π / 2 sin 4 ξ d ξ + + b n c 2 n 0 π / 2 sin 2 n ξ d ξ + } .
0 π / 2 sin 2 n ξ d ξ = 1 · 3 · 5 ( 2 n 1 ) 2 · 4 · 6 ( 2 n ) · π 2
P = P 0 ( 1 + 1 2 b 1 sin 2 1 2 ϕ 0 + 1 · 3 2 · 4 b 2 sin 4 1 2 ϕ 0 + 1 · 3 · 5 2 · 4 · 6 b 3 sin 6 1 2 ϕ 0 + ) ,
sin 2 1 2 ϕ 0 = 1 2 ( 1 1 4 δ 2 sin 2 1 2 θ 0 ¯ ) .
sin 2 1 2 ϕ 0 = u 2 + u 4 + 2 u 6 + 5 u 8 + 14 u 10 +
P = P 0 { 1 + 1 2 b 1 δ 2 sin 2 1 2 θ 0 + 1 2 ( b 1 + 3 4 b 2 ) δ 4 sin 4 1 2 θ 0 + ( b 1 + 3 4 b 2 + 5 1 6 b 3 ) δ 6 sin 6 1 2 θ 0 + 5 2 ( b 1 + 3 4 b 2 + 3 8 b 3 + 7 6 4 b 4 ) δ 8 sin 8 1 2 θ 0 + ( 7 b 1 + 2 1 4 b 2 + 4 5 1 6 b 3 + 3 5 3 2 b 4 + 6 3 2 5 6 b 5 ) δ 10 sin 10 1 2 θ 0 + }
δ 4 4 δ 2 + 1 = 0 ,
δ 2 = 2 ± 3 ¯ .
4 δ 4 κ 2 = 3 δ 2 1 ,
δ 2 = 2 + 3 ¯ κ 2 = 2 ( 1 + 3 ¯ ) δ 1.931 , 851 , 65 = 1 2 ( 6 ¯ + 2 ¯ ) κ 2.337 , 541 , 79 k D 1.210 , 000 , 67
κ ( 1 + Ψ ) 1 2
Ψ ( 4 κ 2 4 δ 2 + sin 2 ϕ ) sin 2 ϕ 4 κ 2 .
Ψ = ( y x + z ) z y = z ( ρ ρ 1 ) .
ρ ρ = α z ( 1 z 1 )
Ψ = z ( ρ 0 z y 0 1 ) Ψ z = ρ 0 1 2 z y 0 .
Ψ m = 1 4 y 0 ( ρ 0 1 ) 2 .
ρ 2 ρ = 1 ,
ρ ρ = 1 z 0 + 1
y 0 ( ρ 0 1 ) 2 = 4
( ρ 1 ) 2 = 4 z 0 ρ .
Ψ = 1 { 1 4 [ 4 y 0 ( ρ 0 1 ) 2 ] + 1 y 0 [ 1 2 y 0 ( ρ 0 1 ) z ] 2 }
δ 1 ( 1 σ ) 1 2 [ 1 σ ( 1 σ ) δ 2 ] 1 2 ,
ρ = ρ = ( 2 + z 0 + 2 1 + z 0 ¯ ) / z 0 .