Simple Harmonic Motion and Uniform Circular Motion — Question 9

Question 9

A rotating observation axis changes the measured frequency

A point moves around a circle of radius $A$ at constant laboratory angular speed $\omega>0$. Its initial angle is zero. A sensor measures projection on an axis rotating counterclockwise at constant rate $\Omega$, also starting at angle zero.

1. Find the sensor coordinate and its period when $\omega\ne\Omega$.

2. For $\Omega=\omega$, explain the constant sensor reading despite continuing laboratory acceleration. Can its second time derivative be identified with the laboratory acceleration projected on the axis?

Question 9 - Solution

A rotating observation axis changes the measured frequency

(a) Derivation and calculation

$$\mathbf r=A(\cos\omega t,\sin\omega t),\quad\mathbf e=(\cos\Omega t,\sin\Omega t).$$$$\boxed{u=\mathbf r\cdot\mathbf e=A\cos[(\omega-\Omega)t]}.$$$$\boxed{\omega_{\mathrm{measured}}=|\omega-\Omega|},\quad\boxed{T=\frac{2\pi}{|\omega-\Omega|}}.$$

(b) Result and interpretation

$$\ddot u=\ddot{\mathbf r}\cdot\mathbf e+2\dot{\mathbf r}\cdot\dot{\mathbf e}+\mathbf r\cdot\ddot{\mathbf e}.$$$$\Omega=\omega:\quad u=A,\quad\ddot u=0,\quad\ddot{\mathbf r}\cdot\mathbf e=-\omega^2A.$$$$-\omega^2A+2\omega^2A-\omega^2A=0.$$The moving basis contributes the extra derivative terms; the second coordinate derivative is not simply projected laboratory acceleration.

Physical check

When $\Omega=0$, the fixed-axis result is recovered.