Question 3.1: A mass of 0.5 kg is suspended in a vertical plane by a sprin...

The differential equation of motion is given by m\ddot{x} +kx = 0. Since m = 0.5 kg and k = 300 N/m, the differential equation can be written as
0.5 \ddot{x}+ 300x = 0

The natural frequency ω is given by
ω = \sqrt{\frac{k}{m}} = \sqrt{\frac{300}{0.5}}= 24.495 rad/s

The frequency f is given by
f = \frac{ω}{2π} = \frac{24.495}{2π} = 3.898 Hz

Using Eq. 3.9, the response of the system is given by x = X sin(ωt + Φ), where

X = \sqrt{ x_{0}^{2} + (\frac{\dot{x_{0}}}{ω})^{2} } = \sqrt{(0.01)² + 0} = 0.01

Φ = tan^{−1} (\frac{x_{0}}{\dot{x_{0}}/ω}) = tan^{−1}(∞) = \frac{π}{2}

that is,
x = 0.01 cos 24.495t

For this simple conservative system, the system total energy is equal to the maximum kinetic energy or the maximum strain energy. One therefore has

E = \frac{1}{2} kX² = \frac{1}{2} (300)(0.01)² = 0.015 N · m

This is the same as the maximum kinetic energy given by
E = \frac{1}{2} m(ωX)² = \frac{1}{2}(0.5)(24.495 × 0.01)² = 0.015 N · m