The Force Law for Simple Harmonic Motion — Question 1

Question 1

Gravity shifts the spring equilibrium without changing its frequency

A mass $m=0.50\,\mathrm{kg}$ hangs from an ideal spring with $k=200\,\mathrm{N/m}$. Let downward extension from the natural length be $y$ and take $g=10\,\mathrm{m/s^2}$. The mass is released from rest at the natural length. The spring can exert both tension and compression.

1. Derive the equilibrium extension and the motion equation using displacement from equilibrium.

2. Find the amplitude about equilibrium, angular frequency and greatest extension.

Question 1 - Solution

Gravity shifts the spring equilibrium without changing its frequency

(a) Derivation and calculation

Balance forces with downward positive:$$m\ddot y=mg-ky,\qquad0=mg-ky_e.$$$$\boxed{y_e=\frac{mg}{k}=\frac{0.50(10)}{200}=0.0250\,\mathrm m}.$$Shift to $x=y-y_e$:$$m\ddot x=mg-k(x+y_e)=-kx,\qquad\boxed{\ddot x+\frac{k}{m}x=0}.$$

(b) Result and interpretation

$$\omega=\sqrt{k/m}=\sqrt{200/0.50}=\boxed{20\,\mathrm{rad/s}}.$$$$x(0)=-y_e,\quad\dot x(0)=0\Rightarrow x(t)=-0.0250\cos20t.$$$$\boxed{A=0.0250\,\mathrm m},\qquad y(t)=0.0250(1-\cos20t),\quad\boxed{y_{\max}=0.0500\,\mathrm m}.$$

Physical check

Gravity contributes a constant force that cancels at the shifted equilibrium; the linear force slope remains $-k$.