To solve this problem we must rely on the equations of the simple harmonic movement that define the period as a function of length and gravity as
[tex]T = 2\pi \sqrt{\frac{l}{g}}[/tex]
Where
l = Length
g = Gravity
Re-arrange to find L,
[tex]L = g (\frac{T}{2\pi})^2[/tex]
Our values are given as
[tex]g = 9.81m/s[/tex]
[tex]T = 10.1s[/tex]
Replacing,
[tex]L = g (\frac{T}{2\pi})^2[/tex]
[tex]L = (9.81) (\frac{10.1}{2\pi})^2[/tex]
[tex]L = 25.348m[/tex]
Therefore the height would be 25.348m
The height of the tower is 2.51 m
The angular velocity of oscillation of the simple pendulum is :
[tex]\omega=\sqrt{\frac{g}{l}}[/tex]
We know that the time period is given by:
T = 2π/ω
So, the time period of oscillation of a simple pendulum is given by:
[tex]T = 2\pi\sqrt{\frac{l}{g}[/tex]
where, l is the length of the string, which is equal to the height of the tower.
It is given that the time period is T = 10.1 s.
So,
[tex]T = 2\pi\sqrt{\frac{l}{g}}\\\\l=\frac{Tg}{4\pi^2}\\\\l=\frac{10.1\times9.8}{4\pi^2}\\\\l=2.51\;m[/tex]
The height of the tower is 2.51 m.
Learn more about simple pendulum:
https://brainly.com/question/14759840?referrer=searchResults
