Simple Harmonic Motion and Uniform Circular Motion — Question 4

Question 4

Two same-frequency oscillations combine as fixed vectors

Two displacements along one line add: $x_1=.030\cos\omega t$ and $x_2=.040\cos(\omega t+\pi/2)$, in meters. The frequency is the same for both.

1. Express their sum as $R\cos(\omega t+\phi)$ with $R>0$.

2. Find the allowed resultant amplitude range if the relative phase can vary. Explain why adding the amplitudes directly fails for the specified phase.

Question 4 - Solution

Two same-frequency oscillations combine as fixed vectors

(a) Derivation and calculation

$$x=.030\cos\omega t-.040\sin\omega t=R\cos(\omega t+\phi).$$$$R\cos\phi=.030,\qquad R\sin\phi=.040.$$$$\boxed{R=.0500\,\mathrm m},\qquad\boxed{\phi=\operatorname{atan2}(.040,.030)=.92730\,\mathrm{rad}}.$$

(b) Result and interpretation

For arbitrary relative phase $\delta$,$$R^2=A_1^2+A_2^2+2A_1A_2\cos\delta.$$$$|A_1-A_2|\le R\le A_1+A_2\Rightarrow\boxed{.0100\le R\le.0700\,\mathrm m}.$$The upper bound requires equal phase; the specified quadrature gives the vector magnitude instead.

Physical check

At $t=0$, $x=.0300\,\mathrm m$ and $v=-.0400\omega$, both reproduced by the resultant phasor.