Question

: A pendulum swings back and forth in simple harmonic motion. If the length of the pendulum is 2 m, what is the approximate time period of the pendulum?

Answer 1

The time period \( T \) of a simple pendulum can be calculated using the formula:

\[
T = 2\pi \sqrt{\frac{L}{g}}
\]

where:
- \( T \) is the time period,
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) on the surface of the Earth).

Given that the length \( L = 2 \, \text{m} \), we can substitute the values into the formula:

\[
T = 2\pi \sqrt{\frac{2}{9.81}}
\]

Calculating the square root:

\[
\sqrt{\frac{2}{9.81}} \approx \sqrt{0.20387} \approx 0.4515
\]

Now, calculate the time period:

\[
T \approx 2\pi \times 0.4515 \approx 2 \times 3.14159 \times 0.4515 \approx 2.833
\]

Thus, the approximate time period of the pendulum is:

\[
T \approx 2.83 \, \text{seconds}
\]

So, the final answer is approximately **2.83 seconds**.