To determine the period [tex]\( T \)[/tex] of a simple pendulum with a length of 50 cm, we can use the well-known formula for the period of a simple pendulum:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
where:
- [tex]\( T \)[/tex] is the period of the pendulum,
- [tex]\( L \)[/tex] is the length of the pendulum,
- [tex]\( g \)[/tex] is the acceleration due to gravity,
- [tex]\( \pi \)[/tex] is a constant approximately equal to 3.14159.
Let's break this down into specific steps:
1. Convert the length to meters: The length is given in centimeters (cm), but we need it in meters (m) for the formula.
[tex]\[ L = 50 \, \text{cm} = 50 \, \text{cm} \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 0.5 \, \text{m} \][/tex]
2. Acceleration due to gravity: We are given [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex].
3. Substitute the values into the formula:
[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]
[tex]\[ T = 2\pi \sqrt{\frac{0.5 \, \text{m}}{9.8 \, \text{m/s}^2}} \][/tex]
4. Calculate the value inside the square root:
[tex]\[ \frac{0.5 \, \text{m}}{9.8 \, \text{m/s}^2} \approx 0.05102 \, \text{s}^2 \][/tex]
5. Take the square root:
[tex]\[ \sqrt{0.05102 \, \text{s}^2} \approx 0.2258 \, \text{s} \][/tex]
6. Multiply by [tex]\( 2\pi \)[/tex]:
[tex]\[ T = 2 \times 3.14159 \times 0.2258 \, \text{s} \approx 1.418 \, \text{s} \][/tex]
Therefore, the period of the simple pendulum with a length of 50 cm (0.5 m) is approximately [tex]\( 1.418 \, \text{s} \)[/tex].
