Un reloj de péndulo de largo L y período T, aumenta su largo en ΔL (ΔL << L). Demuestre que su período aumenta en: ΔT = π ΔL /√(L g)

 ΔT =

Explanation:

In a simple harmonic motion, specifically in the simple pendulum, the angular velocity

          w =

angular velocity and period are related

          w = 2π / T

we substitute

          2π / T = \sqrt{\frac{g}{L} }

          T =

In this exercise indicate that for a long Lo the period is To, then and increase the long

          L = L₀ + ΔL

we substitute

           T =

            T =

in general the length increments are small ΔL/L «1, let's use a series expansion

we keep the linear term, let's substitute

           T =  

if we do

           T = T₀ + ΔT

           T₀ + ΔT =

            T₀ + ΔT = T₀ +

            ΔT =