To determine the period \(\( T \)\) of a simple pendulum with a length of 50 cm, we can use the well-known formula for the period of a simple pendulum:
\(\[ T = 2\pi \sqrt{\frac{L}{g}} \]\)
where:
- \(\( T \)\) is the period of the pendulum,
- \(\( L \)\) is the length of the pendulum,
- \(\( g \)\) is the acceleration due to gravity,
- \(\( \pi \)\) 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.
\(\[ L = 50 \, \text{cm} = 50 \, \text{cm} \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 0.5 \, \text{m} \]\)
2. Acceleration due to gravity: We are given \(\( g = 9.8 \, \text{m/s}^2 \)\).
3. Substitute the values into the formula:
\(\[ T = 2\pi \sqrt{\frac{L}{g}} \]\)
\(\[ T = 2\pi \sqrt{\frac{0.5 \, \text{m}}{9.8 \, \text{m/s}^2}} \]\)
4. Calculate the value inside the square root:
\(\[ \frac{0.5 \, \text{m}}{9.8 \, \text{m/s}^2} \approx 0.05102 \, \text{s}^2 \]\)
5. Take the square root:
\(\[ \sqrt{0.05102 \, \text{s}^2} \approx 0.2258 \, \text{s} \]\)
6. Multiply by \(\( 2\pi \)\):
\(\[ T = 2 \times 3.14159 \times 0.2258 \, \text{s} \approx 1.418 \, \text{s} \]\)
Therefore, the period of the simple pendulum with a length of 50 cm (0.5 m) is approximately \(\( 1.418 \, \text{s} \)\).
