Answer :
To determine the length of a pendulum that takes [tex]\(2.4\pi\)[/tex] seconds to swing back and forth, we will use the formula for the period of a pendulum:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
Where:
- [tex]\( T \)[/tex] is the period (time for one complete cycle).
- [tex]\( L \)[/tex] is the length of the pendulum.
- [tex]\( g \)[/tex] is the acceleration due to gravity, which is approximately [tex]\( 32 \, \text{ft/s}^2 \)[/tex] in this problem.
We are given [tex]\( T = 2.4\pi \)[/tex] seconds. We need to find [tex]\( L \)[/tex].
First, we rearrange the formula to solve for [tex]\( L \)[/tex]:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
Divide both sides by [tex]\( 2\pi \)[/tex]:
[tex]\[ \frac{T}{2\pi} = \sqrt{\frac{L}{g}} \][/tex]
Square both sides to get rid of the square root:
[tex]\[ \left( \frac{T}{2\pi} \right)^2 = \frac{L}{g} \][/tex]
Multiply both sides by [tex]\( g \)[/tex] to isolate [tex]\( L \)[/tex]:
[tex]\[ L = g \left( \frac{T}{2\pi} \right)^2 \][/tex]
Substitute the known values [tex]\( T = 2.4\pi \)[/tex] and [tex]\( g = 32 \)[/tex]:
[tex]\[ L = 32 \left( \frac{2.4\pi}{2\pi} \right)^2 \][/tex]
Simplify the expression inside the parentheses:
[tex]\[ L = 32 \left( \frac{2.4\pi}{2\pi} \right)^2 = 32 \left( \frac{2.4}{2} \right)^2 = 32 \left( 1.2 \right)^2 \][/tex]
Calculate [tex]\( (1.2)^2 \)[/tex]:
[tex]\[ (1.2)^2 = 1.44 \][/tex]
Then multiply by 32:
[tex]\[ L = 32 \cdot 1.44 = 46.08 \][/tex]
Therefore, the approximate length of the pendulum that takes [tex]\( 2.4\pi \)[/tex] seconds to swing back and forth is:
[tex]\[ \boxed{46.08 \, \text{ft}} \][/tex]
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
Where:
- [tex]\( T \)[/tex] is the period (time for one complete cycle).
- [tex]\( L \)[/tex] is the length of the pendulum.
- [tex]\( g \)[/tex] is the acceleration due to gravity, which is approximately [tex]\( 32 \, \text{ft/s}^2 \)[/tex] in this problem.
We are given [tex]\( T = 2.4\pi \)[/tex] seconds. We need to find [tex]\( L \)[/tex].
First, we rearrange the formula to solve for [tex]\( L \)[/tex]:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
Divide both sides by [tex]\( 2\pi \)[/tex]:
[tex]\[ \frac{T}{2\pi} = \sqrt{\frac{L}{g}} \][/tex]
Square both sides to get rid of the square root:
[tex]\[ \left( \frac{T}{2\pi} \right)^2 = \frac{L}{g} \][/tex]
Multiply both sides by [tex]\( g \)[/tex] to isolate [tex]\( L \)[/tex]:
[tex]\[ L = g \left( \frac{T}{2\pi} \right)^2 \][/tex]
Substitute the known values [tex]\( T = 2.4\pi \)[/tex] and [tex]\( g = 32 \)[/tex]:
[tex]\[ L = 32 \left( \frac{2.4\pi}{2\pi} \right)^2 \][/tex]
Simplify the expression inside the parentheses:
[tex]\[ L = 32 \left( \frac{2.4\pi}{2\pi} \right)^2 = 32 \left( \frac{2.4}{2} \right)^2 = 32 \left( 1.2 \right)^2 \][/tex]
Calculate [tex]\( (1.2)^2 \)[/tex]:
[tex]\[ (1.2)^2 = 1.44 \][/tex]
Then multiply by 32:
[tex]\[ L = 32 \cdot 1.44 = 46.08 \][/tex]
Therefore, the approximate length of the pendulum that takes [tex]\( 2.4\pi \)[/tex] seconds to swing back and forth is:
[tex]\[ \boxed{46.08 \, \text{ft}} \][/tex]
