Answer to Question #340894 in Physics for Lloyd

Question

A pendulum that has a period of 7.01 s and that is located where the acceleration due to gravity is 9.44 m/s2 is moved to a location where the acceleration due to gravity is 9.31 m/s2. What is its new period?

Round your answer to 5 decimal places.

Pls dont forget the unit

Answer 1

$$T_1=2\pi\sqrt{\frac{l}{g_1}}$$

$$T_2=2\pi\sqrt{\frac{l}{g_2}}$$

$$T_2=T_1\sqrt{\frac{g_1}{g_2}}=7.01\sqrt{\frac{9.44}{9.31}}=7.06\:\rm s$$