A0.50 kg object is attached to an ideal spring of spring constant (force constant) 20 n/m along a horizontal, frictionless surface. the object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. at what distance from the equilibrium position are the kinetic energy and potential energy of the system the same?

0.23717 m

0.1677 m

Explanation:

m = Mass of object = 0.5 kg

y' = Speed of particle = 1.5 m/s

k = Spring constant = 20 N/m

Angular frequency is given by

Equation of motion

differentiating with respect to t

At t = 0

cos 0 = 1

Speed

The amplitude is 0.23717 m

Kinetic energy is given by

The kinetic and potential energy are conserved

This means that  

Kinetic energy = Potential energy =

In x position

At x = 0.1677 m the kinetic energy and potential energy of the system the same.