Simple Harmonic Motion and Uniform Circular Motion — Question 7

Question 7

A quarter-period sample supplies the missing quadrature

A harmonic oscillator of known period $T$ is sampled at $t$ and $t+T/4$. The readings are $p=.030\,\mathrm{m}$ and $q=-.040\,\mathrm{m}$. Let $\omega=2\pi/T$.

1. Find its amplitude and velocity at the first sample without fitting a full sinusoid.

2. Derive the corresponding amplitude formula for samples separated by phase $\Delta$ instead. Why does the reconstruction fail at an integer half-period separation?

Question 7 - Solution

A quarter-period sample supplies the missing quadrature

(a) Derivation and calculation

$$p=A\cos\psi,\quad q=A\cos(\psi+\pi/2)=-A\sin\psi.$$$$\boxed{A=\sqrt{p^2+q^2}=\sqrt{.030^2+.040^2}=.0500\,\mathrm m}.$$$$\boxed{v(t)=-A\omega\sin\psi=\omega q=-.0400\omega\,\mathrm{m/s}}.$$

(b) Result and interpretation

For a general sample phase separation $\Delta$,$$q=p\cos\Delta-A\sin\psi\sin\Delta.$$$$A\sin\psi=\frac{p\cos\Delta-q}{\sin\Delta}\Rightarrow\boxed{A^2=p^2+\left(\frac{p\cos\Delta-q}{\sin\Delta}\right)^2}.$$$$\Delta=n\pi\Rightarrow q=(-1)^np.$$At these separations the second sample supplies no independent quadrature; small $|\sin\Delta|$ amplifies errors.

Physical check

At $\Delta=\pi/2$, the general result reduces exactly to $A^2=p^2+q^2$.