2.3.42. A weight of mass $m$ suspended on a spring of stiffness $k$ is placed on a stand. The spring is not deformed. The stand is quickly removed. Determine the maximum spring extension and maximum load speed.
Solution
After the stand is removed, the object will undergo simple harmonic motion with angular frequency $\omega=\sqrt{k/m}$ with its initial position as the highest position, because the object must have zero speed at this point. The equilibrium point will be at the distance $A$ below the initial position so that $kA=mg$, i.e. $A=mg/k$. This is the amplitude of the motion, and so the maximum elongation of the spring is $x_{\mbox{max}}=2A=2mg/k$. The maximum speed $v_{\mbox{max}}$ is attained at the equilibrium point, and $v_{\mbox{max}}=\omega A=g\sqrt{m/k}$.
