The Force Law for Simple Harmonic Motion — Question 6

Question 6

A tension-only elastic cord can invalidate the sinusoidal solution

A mass hangs from a vertical elastic cord with stiffness $k$ when stretched, but the cord goes slack under compression. Its static extension is $y_e=mg/k$. A proposed SHM solution about equilibrium has amplitude $A$, with downward extension $y=y_e+x$.

1. Find the condition under which the cord remains taut throughout the motion.

2. For $m=.20\,\mathrm{kg}$, $k=100\,\mathrm{N/m}$, $g=10\,\mathrm{m/s^2}$ and $A=.030\,\mathrm{m}$, decide whether the SHM solution is valid. What is the force law once the cord goes slack?

Question 6 - Solution

A tension-only elastic cord can invalidate the sinusoidal solution

(a) Derivation and calculation

For the proposed sinusoid,$$y=y_e+A\cos\psi,\qquad y_e=mg/k,\qquad y_{\min}=y_e-A.$$$$\mathcal T=ky>0\ \text{throughout}\iff\boxed{A<mg/k}.$$At equality, tension just touches zero at the upper turning point; for $A>y_e$ the assumed spring law would demand compression.

(b) Result and interpretation

$$y_e=\frac{0.20(10)}{100}=0.0200\,\mathrm m,\quad y_{\min}=0.0200-0.0300=-0.0100\,\mathrm m.$$The proposed global SHM is invalid. The actual equation is piecewise:$$\boxed{m\ddot y=\begin{cases}mg-ky,&y>0,\\mg,&y\le0.\end{cases}}$$A slack interval is free fall until the cord becomes taut again.

Physical check

The admissibility constraint must be checked before using a sinusoidal solution; an elastic cord cannot exert a compressive force.