Energy in Simple Harmonic Motion — Question 8

Question 8

A weak quartic term changes turning-point energy before it changes the center

A conservative oscillator has $U(x)=kx^2/2+\lambda x^4/4$, with $k,\lambda>0$, and turns at $x=\pm A$. Its mass is $m$.

1. Find total energy and speed at equilibrium. Compare the result with the harmonic prediction for the same amplitude.

2. Find the condition on $A$ for the quartic contribution to turning-point energy to be at most $5\%$ of the quadratic contribution. Does the potential describe exact SHM?

Question 8 - Solution

A weak quartic term changes turning-point energy before it changes the center

(a) Derivation and calculation

$$E=U(A)=\tfrac12kA^2+\tfrac14\lambda A^4.$$$$\tfrac12mv_{\max}^2=E-U(0)\Rightarrow\boxed{v_{\max}=\sqrt{\frac{kA^2}{m}+\frac{\lambda A^4}{2m}}}.$$$$\frac{v_{\max}}{A\sqrt{k/m}}=\sqrt{1+\frac{\lambda A^2}{2k}}>1.$$$$F=-\frac{dU}{dx}=-kx-\lambda x^3.$$

(b) Result and interpretation

At amplitude $A$, the relative quartic contribution is$$\frac{U_4}{U_2}=\frac{\lambda A^4/4}{kA^2/2}=\frac{\lambda A^2}{2k}.$$$$\frac{\lambda A^2}{2k}\le.05\Rightarrow\boxed{A\le\sqrt{.10k/\lambda}}.$$Finite excursions are not exact SHM because $F/x=-(k+\lambda x^2)$ varies with position.

Physical check

As $\lambda A^2/k\to0$, the usual $v_{\max}=A\sqrt{k/m}$ is recovered.