Compound pendulum centres and minimum period

Centres of suspension and oscillation

For a compound pendulum of centre of gravity G, radius of gyration k about G and pivot O at distance l from G:

T = 2π √((k² + l²)/(g l)) = 2π √(L/g) ,   L = l + k²/l

• Centre (point) of suspension: the pivot O about which the body oscillates.
• Centre of oscillation: the point O′ on the line OG, on the other side of G, at distance L from O (the length of the equivalent simple pendulum). Its distance from G is l′ = L − l = k²/l.

Interchangeability

Suspend at O′: l′ = k²/l, so

(k² + l′²)/l′ = (k² + k⁴/l²) · l/k² = (l² + k²)/l = L

Hence T′ = 2π√(L/g) = T. The two points are interchangeable.

 O ------ G ---- O'
   <-l->  <-k²/l->
   <------ L ----->

Minimum time period

T² = (4π²/g)(k²/l + l). Differentiate with respect to l and set to zero:

d/dl (k²/l + l) = −k²/l² + 1 = 0   →   l = k

The second derivative 2k²/l³ is positive, so this is a minimum. Then

L = k + k = 2k ,   T_min = 2π √(2k/g)

So the minimum period occurs when the distance of the pivot from G equals the radius of gyration k, i.e. the length of the equivalent simple pendulum is L = 2k.