Simple Harmonic Motion — Question 5

Question 5

A camera can mistake fast oscillation for slow oscillation

A camera samples displacement every $\Delta t=0.10\,\mathrm{s}$. One possible motion is $x_1(t)=A\cos(2\pi ft)$ with $f=2.0\,\mathrm{Hz}$. A second is $x_2(t)=A\cos(2\pi f_2t)$ with $f_2=8.0\,\mathrm{Hz}$. Both start at $+A$.

1. Show that their recorded positions agree at every camera sample.

2. Would a signed velocity measurement at $t=0.10\,\mathrm{s}$ distinguish them? Compare the two velocities in units of $A\,\mathrm{s^{-1}}$.

Question 5 - Solution

A camera can mistake fast oscillation for slow oscillation

(a) Derivation and calculation

At frame time $t_n=0.10n\,\mathrm s$,$$\psi_{1n}=2\pi(2)(0.10n)=0.4\pi n,$$$$\psi_{2n}=2\pi(8)(0.10n)=1.6\pi n=2\pi n-\psi_{1n}.$$$$x_2(t_n)=A\cos(2\pi n-\psi_{1n})=A\cos\psi_{1n}=\boxed{x_1(t_n)}.$$

(b) Result and interpretation

Differentiate the continuous fields before evaluating the time:$$\frac{v_1(0.10)}A=-4\pi\sin(0.4\pi)=\boxed{-11.9513\,\mathrm{s^{-1}}},$$$$\frac{v_2(0.10)}A=-16\pi\sin(1.6\pi)=\boxed{+47.8053\,\mathrm{s^{-1}}}.$$A signed velocity measurement distinguishes both direction and magnitude, although the sampled positions coincide.

Physical check

The frame rate is $10\,\mathrm{Hz}$ and $f_2=10-f_1$, giving the cosine sampling alias shown explicitly.