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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

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

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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.

This problem has been solved

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