Answer to Question #184399 in Physics for Jeje

Question

A pendulum that has a period of 4.45 s and that is located where the acceleration due to gravity is 9.7 m/s² is moved to a location where the acceleration due to gravity is 9.45 m/s². What is its new period?

Round your answer to 5 decimal places

Answer 1

$$T=2\pi\sqrt{\frac{l}{g}}\\T'=\sqrt{\frac{g}{g'}}T\\T'=\sqrt{\frac{9.7}{9.45}}4.45=4.51\ s$$