Energy in Simple Harmonic Motion — Question 4

Question 4

Changing stiffness transfers energy in a phase-dependent way

An ideal oscillator of mass $m$ and stiffness $k$ has amplitude $A$. An external mechanism instantaneously changes the stiffness to $4k$ without changing the equilibrium or imparting an impulse. Position and velocity are continuous.

1. Find the new energy and amplitude if the switch occurs at a turning point.

2. Find them if it occurs at equilibrium. Account for the different work supplied by the mechanism.

Question 4 - Solution

Changing stiffness transfers energy in a phase-dependent way

(a) Derivation and calculation

With $x$ and $v$ unchanged across the stiffness switch,$$E_+=\tfrac12mv^2+\tfrac12(4k)x^2,\quad\boxed{\Delta E=\tfrac12(4k-k)x^2=\tfrac32kx^2}.$$$$\omega_+=\sqrt{4k/m}=2\omega,\qquad A_+^2=\frac{2E_+}{4k}.$$

(b) Result and interpretation

At a turning point, $x=\pm A,v=0$:$$E_+=2kA^2=4E,\quad\boxed{A_+=A},\quad\boxed{W_{\mathrm{switch}}=3E}.$$At equilibrium, $x=0$:$$E_+=\tfrac12mv^2=E,\quad\boxed{A_+=A/2},\quad W_{\mathrm{switch}}=0.$$

Physical check

Changing a parameter can do work without delivering an impulse; energy conservation across the switch requires including the actuator.