Question

Consider a simple harmonic oscillator consisting of a one kilogram mass m on a spring with spring/force constant k and length L`. If the mass of the spring m_s is 9% of the attached mass, and k = 66 N/m, and if we determined the attached body is displaced 3 cm and given a downward velocity of 0.4 m/s - calculate ,→ the frequency ω of the motion, and the amplitude A of the motion

Answer 1

$\omega=\sqrt{\frac km}=\sqrt{\frac{66}{1}}=8.12~\frac{rad}{s},$

$v=\omega A\implies A=\frac{v}{\omega}=\frac{0.4}{8.12}=5~cm.$