Question

The period P (in seconds) of a pendulum is given by the function P=2π√(L/32), where L

is the pendulum length (in feet). What is the length of a pendulum with a period of 3 seconds? Round your answer to the nearest hundredth of a foot.

Answer 1

We are given the formula for the period of a pendulum:
P = 2π√(L/32)
We are also given that the period is 3 seconds. We can use this information to solve for the length L:
3 = 2π√(L/32)
Squaring both sides of the equation, we get:
9 = 4π^2(L/32)
Multiplying both sides by 32/4π^2, we get:
L = (9 × 32)/(4π^2) = 2.04 feet (rounded to the nearest hundredth)
Therefore, the length of the pendulum with a period of 3 seconds is approximately 2.04 feet.

Answer 2

Explanation:

We are given the period P of a pendulum and we are asked to find its length L. We can use the formula P=2π√(L/32) to solve for L:

P = 2π√(L/32)

Squaring both sides, we get:

P^2 = 4π^2(L/32)

Simplifying, we get:

L = (P^2*32)/(4π^2)

Now, substituting P=3 into the formula, we get:

L = (3^2*32)/(4π^2) ≈ 9.58 feet

Therefore, the length of the pendulum with a period of 3 seconds is approximately 9.58 feet (rounded to the nearest hundredth).