Question

A 220g mass is on a frictionless horizontal surface at the end of a spring that has a force constant of 7.0N/m The mass is displaced 5 2m from its equilibrium position and then released to undergo simple harmonic motion.

At what displacement from the equilibrium position is the potential energy equal to kinetic energy?​

Answer 1

Step-by-step explanation:

Your numbers seem wonky, so I'll just assume that the initial displacement is a distance A (Amplitude) from the equilibrium position. Spring constant = k

Initial potential energy is

PE = ½kA²

As potential energy and kinetic energy are constantly exchanging in SHM,

the position x where half of the original spring potential exists is found where

½kx² = ½(½kA²)

x² = ½A²

x = (√0.5)A

x ≈ 0.707A

just plug in your actual starting position A

With A = 5.2 cm

x = 3.67695... 3.7 cm