# Interesting mathematical pendulum problem?

the periods of oscillations of n different pendulums (mathematical) are T1, T2, ..., Tn. The strings of these pendulums are joint into one without length loss and I got a new pendulum. Find its period of oscillation.

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• NCS
Lv 7
1 month ago

For pendulum n,

T_n = 2π√(L_n / g)

and so

L_n = (T_n / 2π)² * g

Joining the strings would result in a pendulum of length

L = Σ L_n = Σ (T_n / 2π)² * g = [g/(2π)²] * Σ (T_n)²

The resulting pendulum's period is

T = 2π√(L/g) = 2π√{[g/(2π)² * Σ (T_n)² / g} = √(Σ T_n)²

Translated to English, that means that the period of the new pendulum is equal to the square root of the sum of the squares of the original periods.

For instance, if you initially had four pendula with periods 1, 2, 3 and 4 seconds, the period of your final pendulum would be

T = √(1² + 2² + 3² + 4²) s = √30 s ≈ 5.5 s

Hope this helps!

• 1 month ago

For the i-th pendulum, Tᵢ = 2π√[Lᵢ/g]. Rearranging gives:

Lᵢ = (g/(4π²)) Tᵢ²

Total length L = L₁ + L₂ + … + Lₙ

= (g/(4π²))T₁² + (g/(4π²))T₂² + … + (g/(4π²))Tₙ² +

For neatness let S =  T₁² + T₂² +… + Tₙ²

L = (g/(4π²))S

(You could use summation notation with ‘Σ’ if you are familiar with it.)

Period of new pendulum is:

T = 2π√(L/g)

. .= 2π√[ (g/(4π²))S/g]

. .= (2π)/(2π)√[gS/g]

. .= √S

. .= √[T₁² + T₂² +… + Tₙ²]

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