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

Answer:

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

$\omega=\sqrt{\frac{k}{m}}\\\Rightarrow \omega=\sqrt{\frac{20}{0.5}}\\\Rightarrow \omega=6.3245\ rad/s$

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

$y'=A\times 6.3245\\\Rightarrow A=\frac{y'}{6.3245}\\\Rightarrow A=\frac{1.5}{6.3245}\\\Rightarrow A=0.23717\ m$

The amplitude is 0.23717 m

Kinetic energy is given by

$K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 0.5\times 1.5^2\\\Rightarrow K=0.5625\ J$

The kinetic and potential energy are conserved

This means that  

Kinetic energy = Potential energy = $\frac{0.5625}{2}=0.28125$

In x position

$\frac{1}{2}kx^2=0.28125\\\Rightarrow x=\sqrt{\frac{2\times 0.28125}{20}}\\\Rightarrow x=0.1677\ m$

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