Energy in Simple Harmonic Motion — Question 10

Question 10

A one-sided auxiliary spring makes unequal turning distances

A mass moves on a frictionless line with a main spring of constant $k$. At $x>0$ it also compresses an auxiliary spring of constant $3k$, initially just touching at $x=0$. At $x<0$ the auxiliary spring loses contact. Both natural equilibria are at $x=0$, and contacts are ideal and elastic. Release the mass from rest at $x=-A$.

1. Find the positive turning position and the speed at $x=0$.

2. Find the full cycle period and explain why the motion is periodic but not one global sinusoid.

Question 10 - Solution

A one-sided auxiliary spring makes unequal turning distances

(a) Derivation and calculation

$$U(x)=\begin{cases}\tfrac12kx^2,&x\le0,\\\tfrac12(4k)x^2,&x\ge0.\end{cases}\qquad E=\tfrac12kA^2.$$$$2kx_+^2=\tfrac12kA^2\Rightarrow\boxed{x_+=A/2}.$$$$\tfrac12mv_0^2=E\Rightarrow\boxed{|v_0|=A\sqrt{k/m}}.$$

(b) Result and interpretation

Let $\omega=\sqrt{k/m}$. The negative and positive excursions each occupy a half-cycle of their own oscillator:$$t_- =\frac\pi\omega,\qquad t_+=\frac\pi{2\omega}.$$$$\boxed{T=t_-+t_+=\frac{3\pi}{2}\sqrt{m/k}}.$$Position and velocity join continuously at zero; the two half-cycles have different frequencies.

Physical check

The period is independent of amplitude within the ideal piecewise-linear model, although the complete waveform is not sinusoidal.