Question

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)

Answer 1

Answer:

 ΔT = $\pi \ \frac{\Delta L}{\sqrt{Lg} }$

Explanation:

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

          w = $\sqrt{\frac{g}{L} }$

angular velocity and period are related

          w = 2π / T

we substitute

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

          T = $2\pi \ \sqrt{\frac{L}{g} }$

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

          L = L₀ + ΔL

we substitute

           T = $2\pi \ \sqrt{\frac{L + \Delta L}{g} }$

            T = $2\pi \ \sqrt{\frac{L}{g} } \ \sqrt{1+ \frac{\Delta L}{L} }$

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

           $\sqrt{1+ \ \frac{\Delta L}{L} } = 1 + \frac{1}{2} \frac{\Delta L}{L} + ...$  

we keep the linear term, let's substitute

           T = $2\pi \ \sqrt{\frac{L}{g} } \ ( 1 + \frac{1}{2} \frac{\Delta L}{L} )$  

if we do

           T = T₀ + ΔT

           

           T₀ + ΔT = $2\pi \sqrt{\frac{\Delta L}{g} } + \pi \ \sqrt{\frac{L}{g} } \ \frac{\Delta L}{L}$

            T₀ + ΔT = T₀ + $\pi \sqrt{\frac{1}{Lg} } \ \Delta L$

            ΔT = $\pi \ \frac{\Delta L}{\sqrt{Lg} }$