Energy in Simple Harmonic Motion — Question 5

Question 5

Catching extra mass changes inertia and can dissipate energy

A horizontal spring oscillator has moving mass $m$, stiffness $k$ and amplitude $A$. An equal mass drops vertically onto it and sticks, bringing zero horizontal momentum. The event is instantaneous; ignore any horizontal impulse from the spring and any change of equilibrium.

1. Find the new amplitude and energy if capture occurs at equilibrium.

2. Compare with capture at a turning point. Why does the same added mass not produce the same energy loss in both cases?

Question 5 - Solution

Catching extra mass changes inertia and can dissipate energy

(a) Derivation and calculation

For capture by an initially stationary equal mass,$$mv=(2m)v_+\Rightarrow v_+=v/2,\qquad\omega_+=\sqrt{k/(2m)}=\omega/\sqrt2.$$$$\Delta K=\tfrac12(2m)(v/2)^2-\tfrac12mv^2=-\tfrac14mv^2.$$

(b) Result and interpretation

At equilibrium, $v=\omega A$ and $U=0$:$$E_+=E/2,\qquad\boxed{A_+=\sqrt{2E_+/k}=A/\sqrt2}.$$At a turning point, $v=0$:$$E_+=E,\quad v_+=0,\qquad\boxed{A_+=A}.$$The post-capture frequency is $\omega/\sqrt2$ in both cases.

Physical check

The horizontal kinetic energy lost in the inelastic capture is zero at a turning point and largest at equilibrium.