Question

question 9 of 10
what is the period, in seconds, of a simple pendulum of length 5 meters? use the gravitational constant $g = 9.8 \text{ m/s}^2$ and round your answer to two decimal places.

Explanation:

Step1: Recall pendulum period formula

The formula for the period $T$ of a simple pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$, where $L$ is the length of the pendulum, and $g$ is the gravitational acceleration.

Step2: Substitute given values

Substitute $L = 5\ \text{m}$ and $g = 9.8\ \text{m/s}^2$ into the formula:
$T = 2\pi\sqrt{\frac{5}{9.8}}$

Step3: Calculate the value inside the root

First compute $\frac{5}{9.8} \approx 0.5102$

Step4: Calculate the square root

$\sqrt{0.5102} \approx 0.7143$

Step5: Multiply by $2\pi$

$T \approx 2 \times 3.1416 \times 0.7143 \approx 4.49$

Answer:

4.49 seconds