Energy in Simple Harmonic Motion — Question 6

Question 6

Completing the square identifies the oscillation energy vertically

A mass $m$ hangs from a vertical spring of constant $k$. Let $y$ be downward extension from natural length. Choose total potential $U(y)=ky^2/2-mgy$.

1. Rewrite $U$ in terms of displacement $x=y-mg/k$ and identify the constant offset.

2. The mass is released from rest at $y=0$. Find the speed at equilibrium and the lower turning extension using energy. Why is using spring energy alone insufficient?

Question 6 - Solution

Completing the square identifies the oscillation energy vertically

(a) Derivation and calculation

With downward extension $y$,$$U(y)=\tfrac12ky^2-mgy=\tfrac12k\left(y-\frac{mg}{k}\right)^2-\frac{m^2g^2}{2k}.$$$$x=y-y_e,\quad y_e=mg/k:\qquad\boxed{U=\tfrac12kx^2-\frac{m^2g^2}{2k}}.$$

(b) Result and interpretation

Release at $y=0,v=0$ sets total energy to zero. At equilibrium,$$\tfrac12mv_e^2=\frac{m^2g^2}{2k}\Rightarrow\boxed{v_e=g\sqrt{m/k}}.$$At the lower turning point,$$0=\tfrac12ky^2-mgy=y(ky/2-mg)\Rightarrow\boxed{y_{\max}=2mg/k}.$$

Physical check

The additive potential constant has no effect on force; the excursion from equilibrium is $A=mg/k$.