hat is the period of a simple pendulum that is 1.13-m long in each situation? In an elevator accelerating 2.41 m/s22.41 m/s2 downward
Question
What is the period of a simple pendulum that is 1.13-m long in each situation?
In an elevator accelerating 2.41 m/s² downward.
Solution
The period of a simple pendulum is given by the formula:
T = 2π √(L/g)
where: T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
Normally, g is approximately 9.81 m/s² on the surface of the Earth. However, in this case, the elevator is accelerating downward at 2.41 m/s². This effectively increases the acceleration due to gravity to 9.81 + 2.41 = 12.22 m/s².
Substituting these values into the formula gives:
T = 2π √(1.13/12.22) ≈ 0.60 seconds
So, the period of the pendulum in the accelerating elevator is approximately 0.60 seconds.
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