A simple pendulum and a spring-block system are calibrated to have the exact same time period T on the surface of the Earth. The entire setup is then taken to a new planet whose mass is twice that of the Earth and whose radius is half that of the Earth. What will be the ratio of the new time period of the simple pendulum to the new time period of the spring-block system on this planet?

Step-by-step solution

The acceleration due to gravity on the surface of a planet is given by g = GM R^2 . For the new planet, the mass is M' = 2M and the radius is R' = R 2 . The new acceleration due to gravity is g' = G(2M) ( R 2 )^2 = 8 GM R^2 = 8g . The time period of a simple pendulum is T_p = 2 l g . Since g' = 8g , the new time period of the simple pendulum is T_p' = 2 l 8g = T 8 = T 2 2 . The time period of a spring-block system is T_s = 2 m k . This time period is independent of the acceleration due to gravity, so it remains unc