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Four collinear points of compound pendulum

Short answer

Bar pendulum uses gravity; torsional pendulum uses wire elasticity. From T = 2π√((k²+l²)/(gl)) the quadratic l² − (gT²/(4π²))l + k² = 0 gives two roots l₁, l₂ on each side of CG, giving four collinear points. l₁ + l₂ = L, the equivalent simple pendulum length; these are interchangeable centers of suspension and oscillation.

Differentiate between bar pendulum and torsional pendulum. Prove that there exist four collinear points in a bar (compound) pendulum having the same time period. Give their physical significance.

Difference between bar pendulum and torsional pendulum

Bar pendulumTorsional pendulum
Swings about a horizontal axisTwists about a vertical wire
Restoring torque from gravity, mgl sinθRestoring torque from wire's elasticity, Cθ
T = 2π√((k² + l²)/gl)T = 2π√(I/C)
Depends on gIndependent of g
Used to measure gUsed to measure η and I

Four collinear points

Let G be the centre of gravity, k the radius of gyration about G, and the body be suspended at a point distance l from G.

T = 2π √((k² + l²)/(g l))
  1. Square and rearrange: 4π² (k² + l²) = g T² l.
  2. Quadratic in l:
l² − (gT²/4π²) l + k² = 0
  1. For a given T it has two roots l₁ and l₂ with
l₁ + l₂ = gT²/4π² ,    l₁ l₂ = k²

So l₁ < k < l₂ (their product is k²). 4. Each root is a distance from G, and the same equation holds on either side of G. Hence there are four points with the same period: two on each side.

 D        B      G      A          C
 x--------x------o------x----------x
 GD = GC = l2 ,  GB = GA = l1  (l1 < k < l2)

Points A and B are at distance l₁ from G, one on each side; C and D are at distance l₂ from G, one on each side. A, C lie on one side of G and B, D on the other.

  1. The distance between a point at l₁ on one side and a point at l₂ on the opposite side (for example B and C, or A and D) is l₁ + l₂ = gT²/4π² = L.

Physical significance

  • The distance l₁ + l₂ between two such non-symmetrical points on opposite sides equals the length L of the equivalent simple pendulum with the same period.
  • These points are the centres of suspension and oscillation, which are interchangeable (l₂ = k²/l₁).
  • Used in Kater's pendulum: when the periods about the two knife edges are equal, g = 4π² L/T² with L measured as the distance between the knife edges. This needs no knowledge of k.

✓ Answer checked by Claude Opus 5.5 (high effort)

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