The time period $T$ of a simple pendulum is given by 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 $9.8 \, \text{m/s}^2$).

We are asked to find the ratio of the time periods of pendulum A and pendulum B.

Let:

• $L_A = 50 \, \text{cm} = 0.5 \, \text{m}$ (for pendulum A),
• $L_B = 2 \, \text{m}$ (for pendulum B).

The ratio of the time periods $\frac{T_A}{T_B}$ is given by:

$\frac{T_A}{T_B} = \frac{2\pi \sqrt{\frac{L_A}{g}}}{2\pi \sqrt{\frac{L_B}{g}}} = \sqrt{\frac{L_A}{L_B}}$

Substituting the values:

$\frac{T_A}{T_B} = \sqrt{\frac{0.5}{2}} = \sqrt{0.25} = 0.5$

Thus, the ratio of the time periods is:

$\frac{T_A}{T_B} = 0.5$

So, the time period of pendulum A is half that of pendulum B.

Would you like more details or have any questions?

Here are 5 related questions to further your understanding:

1. How does the length of a pendulum affect its time period?
2. What happens to the time period of a pendulum if you double its length?
3. How does the time period of a pendulum change in different gravitational fields?
4. What is the relationship between frequency and time period of a pendulum?
5. Can the mass of the bob affect the time period of a pendulum?

Tip: The time period of a pendulum is independent of the mass and only depends on the length and gravity.