Respuesta :
The period of the pendulum allows to find the result for how much it advances when the length is reduced is:
Δt = 1550 s
Simple harmonic motion is an oscillatory motion where the restoring force is proportional to the displacement.
In the case of the simple pendulum, this is fulfilled for small angles minus 15º, the angular velocity of the pendulum is
w = [tex]\frac{g}{L}[/tex]
Angular velocity and period are related
w = [tex]\frac{2pi}{T}[/tex]
We substitute
T = [tex]2\pi \ \frac{L}{g}[/tex]
They indicate that for the initial length L₀ and the pendulum marks the exact time, how much time changes if the length is 3.97 feet, therefore the initial length is L₀ = 3.90 feet.
[tex]T_o^2 = 4 \pi ^2 \ \frac{L_o}{g}[/tex]
The period for the reduced length is:
[tex]T'^2 = 4\pi ^2 \ \frac{L}{g}[/tex]
The relationship between the periods is:
[tex]( \frac{T}{T'}^2 = \frac{L_o}{L} \\T' = \sqrt{\frac{L_o}{L} } \ T[/tex]
Let's calculate
T ’= [tex]T \ \sqrt{\frac{3.97}{3.9} }[/tex]
T ’= T 1.01795
In the total time of a day.
T = 24 hours (3600 s / 2h) = 86 400 s
We calculate
T ’= 86400 1.01795
T ’= 87,950 s
Therefore the pendulum moves forward in a time of:
ΔT = T'- T
ΔT = 87950 - 86400
ΔT = 1550 s
In conclusion, using the period of the pendulum we can find the result for how much it advances when the length is change is:
ΔT = 1550 s
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