The Force Law for Simple Harmonic Motion — Question 3

Question 3

Turning on a constant force launches motion about a new center

A horizontal mass $m=0.20\,\mathrm{kg}$ on an ideal spring $k=80\,\mathrm{N/m}$ is initially at rest at $x=0$, its original equilibrium. At $t=0$, a constant force $F_0=4.0\,\mathrm{N}$ in the positive direction is switched on and retained. There is no impulse at switch-on.

1. Find the new equilibrium and the complete motion $x(t)$.

2. Find the first maximum displacement and maximum speed. Why does the particle not simply stop at the new equilibrium?

Question 3 - Solution

Turning on a constant force launches motion about a new center

(a) Derivation and calculation

$$m\ddot x+kx=F_0,\qquad x_e=F_0/k=4/80=0.0500\,\mathrm m.$$$$q=x-x_e:\quad\ddot q+\omega^2q=0,\quad\omega=\sqrt{80/0.20}=20\,\mathrm{rad/s}.$$$$q(0)=-x_e,\quad\dot q(0)=0\Rightarrow\boxed{x(t)=0.0500[1-\cos20t]\,\mathrm m}.$$

(b) Result and interpretation

$$v(t)=20(0.0500)\sin20t=1.00\sin20t\,\mathrm{m/s}.$$$$t_{\max}=\frac\pi{20}=0.15708\,\mathrm s,\qquad\boxed{x_{\max}=2x_e=0.100\,\mathrm m}.$$$$\boxed{v_{\max}=1.00\,\mathrm{m/s}},\qquad x=x_e:\ F_{\mathrm{net}}=0,\ a=0,\ |v|=v_{\max}.$$The equilibrium crossing has nonzero velocity, so the undamped mass overshoots.

Physical check

The force step changes acceleration but supplies no impulse: position and velocity remain continuous at $t=0$.