Respuesta :
Answer:
ΔT = [tex]\pi \ \frac{\Delta L}{\sqrt{Lg} }[/tex]
Explanation:
In a simple harmonic motion, specifically in the simple pendulum, the angular velocity
w = [tex]\sqrt{\frac{g}{L} }[/tex]
angular velocity and period are related
w = 2π / T
we substitute
2π / T = \sqrt{\frac{g}{L} }
T = [tex]2\pi \ \sqrt{\frac{L}{g} }[/tex]
In this exercise indicate that for a long Lo the period is To, then and increase the long
L = L₀ + ΔL
we substitute
T = [tex]2\pi \ \sqrt{\frac{L + \Delta L}{g} }[/tex]
T = [tex]2\pi \ \sqrt{\frac{L}{g} } \ \sqrt{1+ \frac{\Delta L}{L} }[/tex]
in general the length increments are small ΔL/L «1, let's use a series expansion
[tex]\sqrt{1+ \ \frac{\Delta L}{L} } = 1 + \frac{1}{2} \frac{\Delta L}{L} + ...[/tex]
we keep the linear term, let's substitute
T = [tex]2\pi \ \sqrt{\frac{L}{g} } \ ( 1 + \frac{1}{2} \frac{\Delta L}{L} )[/tex]
if we do
T = T₀ + ΔT
T₀ + ΔT = [tex]2\pi \sqrt{\frac{\Delta L}{g} } + \pi \ \sqrt{\frac{L}{g} } \ \frac{\Delta L}{L}[/tex]
T₀ + ΔT = T₀ + [tex]\pi \sqrt{\frac{1}{Lg} } \ \Delta L[/tex]
ΔT = [tex]\pi \ \frac{\Delta L}{\sqrt{Lg} }[/tex]
