Question

A 0.50 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?

Answer 1

0.23717 m

0.1677 m

Step-by-step 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

`\omega=\sqrt{(k)/(m)}\\\Rightarrow \omega=\sqrt{(20)/(0.5)}\\\Rightarrow \omega=6.3245\ rad/s`

Equation of motion

`y=Asin(\omega t)`

differentiating with respect to t

`y'=A\omega cos(\omega t)`

At t = 0

cos 0 = 1

Speed

`y'=A* 6.3245\\\Rightarrow A=(y')/(6.3245)\\\Rightarrow A=(1.5)/(6.3245)\\\Rightarrow A=0.23717\ m`

The amplitude is 0.23717 m

Kinetic energy is given by

`K=(1)/(2)mv^2\\\Rightarrow K=(1)/(2)* 0.5* 1.5^2\\\Rightarrow K=0.5625\ J`

The kinetic and potential energy are conserved

This means that

Kinetic energy = Potential energy =
`(0.5625)/(2)=0.28125`

In x position

`(1)/(2)kx^2=0.28125\\\Rightarrow x=\sqrt{(2* 0.28125)/(20)}\\\Rightarrow x=0.1677\ m`

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