Let's address each part of the question separately with detailed solution steps.
### Part 1: Finding the period of the pendulum
The period [tex]\( T \)[/tex] of a simple pendulum is given by the formula:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
where:
- [tex]\( L \)[/tex] is the length of the pendulum (in meters),
- [tex]\( g \)[/tex] is the acceleration due to gravity (in [tex]\( m/s^2 \)[/tex]).
Given:
- The length of the pendulum, [tex]\( L = 7.81 \)[/tex] m,
- Acceleration due to gravity, [tex]\( g = 9.8 \)[/tex] m/s².
Now, substitute the given values into the formula:
[tex]\[ T = 2\pi \sqrt{\frac{7.81}{9.8}} \][/tex]
First, calculate the fraction inside the square root:
[tex]\[ \frac{7.81}{9.8} \approx 0.7969 \][/tex]
Then, take the square root of this value:
[tex]\[ \sqrt{0.7969} \approx 0.8927 \][/tex]
Now, multiply by [tex]\( 2\pi \)[/tex]:
[tex]\[ T \approx 2 \times 3.14159 \times 0.8927 \approx 5.607 \][/tex]
So, the period [tex]\( T \)[/tex] of the pendulum is approximately [tex]\( 5.61 \)[/tex] seconds.
### Part 2: Finding the length to double the period
To double the period [tex]\( T \)[/tex], let's denote the new period as [tex]\( T' = 2T \)[/tex].
From the formula for the period of a pendulum:
[tex]\[ T' = 2\pi \sqrt{\frac{L'}{g}} \][/tex]
Given that [tex]\( T' = 2T \)[/tex], we can write:
[tex]\[ 2T = 2\pi \sqrt{\frac{L'}{g}} \][/tex]
We already know [tex]\( T = 2\pi \sqrt{\frac{7.81}{9.8}} \)[/tex], so:
[tex]\[ 2 \times (2\pi \sqrt{\frac{7.81}{9.8}}) = 2\pi \sqrt{\frac{L'}{9.8}} \][/tex]
Divide both sides by [tex]\( 2\pi \)[/tex]:
[tex]\[ 2 \sqrt{\frac{7.81}{9.8}} = \sqrt{\frac{L'}{9.8}} \][/tex]
Square both sides to eliminate the square root:
[tex]\[ \left( 2 \sqrt{\frac{7.81}{9.8}} \right)^2 = \frac{L'}{9.8} \][/tex]
[tex]\[ 4 \left( \frac{7.81}{9.8} \right) = \frac{L'}{9.8} \][/tex]
Now, solve for [tex]\( L' \)[/tex]:
[tex]\[ 4 \times 7.81 = L' \][/tex]
[tex]\[ L' = 31.24 \][/tex]
So, to double the period, the pendulum must have a length of approximately [tex]\( 31.24 \)[/tex] meters.
