The period of a 1.30-m long simple pendulum in an elevator that is 2.52 s.The elevator is accelerating uniformly.What is the acceleration of the elevator? A)1.20 m/s² upward B)1.20 m/s² downward C)9.81 m/s² downward D)1.73 m/s² upward E)1.73 m/s² downward

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The period of a simple pendulum is given by:$$T = 2\pi \sqrt{\frac{L}{g}}$$where:- T is the period of the pendulum- L is the length of the pendulum- g is the acceleration due to gravityIn this case, the period of the pendulum is 2.52 s and the length of the pendulum is 1.30 m. We can use this information to solve for the acceleration due to gravity:$$2.52 s = 2\pi \sqrt{\frac{1.30 m}{g}}$$$$g = 9.81 m/s^2$$The acceleration of the elevator is equal to the difference between the acceleration due to gravity and the acceleration of the pendulum:$$a = g - \frac{4\pi^2 L}{T^2}$$$$a = 9.81 m/s^2 - \frac{4\pi^2 (1.30 m)}{(2.52 s)^2}$$$$a = 1.73 m/s^2$$Therefore, the acceleration of the elevator is 1.73 m/s^2 downward.

The Period of a 1