The period, T, of a pendulum can be approximated by the formula β‰ˆ 2πœ‹βˆšπΏπ‘”, where L is the

length of the pendulum and g is the gravitational constant. What is the approximate length of
the pendulum if it has a period of 2 s? Note: On Earth the gravitational constant is 9.8 m/s2
.

The formula Tβ‰ˆ2πœ‹βˆšπΏπ‘” can be rearranged to solve for L:

L = (T / (2πœ‹))^2 * g

Substituting T = 2 s and g = 9.8 m/sΒ² gives:

L = (2 / (2πœ‹))^2 * 9.8 β‰ˆ 0.10 m

We can rearrange the formula for the period of a pendulum to solve for the length, L. The formula is T β‰ˆ 2πœ‹βˆšπΏπ‘”

Given that the period, T, is 2 seconds and the gravitational constant, g, is 9.8 m/s^2 on Earth, we can substitute these values into the formula:

2 β‰ˆ 2πœ‹βˆšπΏ(9.8)

Dividing both sides of the equation by 2πœ‹βˆšπΏ(9.8):

2/(2πœ‹βˆš(9.8)) β‰ˆ √𝐿

Squaring both sides of the equation:

(2/(2πœ‹βˆš(9.8)))^2 β‰ˆ L

Evaluating the expression on the left-hand side:

(2/(2πœ‹βˆš(9.8)))^2 β‰ˆ L

0.10162 β‰ˆ L

Therefore, the approximate length of the pendulum is 0.10162 meters.

To find the approximate length of the pendulum given its period of 2 seconds, we can rearrange the formula:

T β‰ˆ 2πœ‹βˆšπΏπ‘”

Let's substitute the known values:

2 β‰ˆ 2πœ‹βˆšπΏ(9.8)

Next, we can simplify the equation by dividing both sides by 2πœ‹ and then squaring both sides:

1 β‰ˆ (√𝐿(9.8))Β²

1 β‰ˆ 9.8𝐿

Now, we can solve for L by dividing both sides by 9.8:

1/9.8 β‰ˆ L

L β‰ˆ 0.102 meters (rounded to three decimal places)

Therefore, the approximate length of the pendulum is approximately 0.102 meters.