Energy in Simple Harmonic Motion — Question 3

Question 3

The same impulse can add or remove oscillator energy

A horizontal oscillator has mass $m$, instantaneous velocity $v$ and position $x$. A very short impulse $J$ acts in the positive direction, leaving position unchanged. Spring force contributes negligible impulse during the event.

1. Derive the energy change caused by the impulse and the new amplitude.

2. For $m=.50\,\mathrm{kg}$, $k=50\,\mathrm{N/m}$, initial amplitude $.10\,\mathrm{m}$ and $J=.20\,\mathrm{N\,s}$, compare applying the impulse at equilibrium while moving right versus while moving left.

Question 3 - Solution

The same impulse can add or remove oscillator energy

(a) Derivation and calculation

An impulse leaves position continuous and changes momentum:$$x_+=x,\qquad m(v_+-v)=J\Rightarrow v_+=v+J/m.$$$$\boxed{\Delta E=\tfrac12m(v+J/m)^2-\tfrac12mv^2=Jv+\frac{J^2}{2m}}.$$$$\boxed{A_+=\sqrt{x^2+\frac m k(v+J/m)^2}}.$$

(b) Result and interpretation

$$\omega=\sqrt{50/.50}=10\,\mathrm{rad/s},\quad v=\pm\omega A=\pm1.00\,\mathrm{m/s},\quad J/m=.400\,\mathrm{m/s}.$$$$v=+1:\ v_+=1.40,\quad\boxed{A_+=.140\,\mathrm m},\quad\Delta E=.240\,\mathrm J.$$$$v=-1:\ v_+=-.600,\quad\boxed{A_+=.0600\,\mathrm m},\quad\Delta E=-.160\,\mathrm J.$$

Physical check

The same positive impulse can add or remove energy; the term $Jv$ retains the initial direction.