Phasors — Question 9

Question 9

Different frequencies do not form a rigid rotating phasor diagram

Two oscillations are $A\cos\omega t$ and $B\cos(2\omega t+\phi)$, with $A,B\ne0$. A student adds two fixed arrows and rotates the result at rate $\omega$.

1. Explain why that construction fails.

2. For $A=B$ and $\phi=0$, compare the true displacement at $t=0$ and $t=\pi/\omega$ with the proposed single-frequency result.

Question 9 - Solution

Different frequencies do not form a rigid rotating phasor diagram

Method and mathematical working

Factor out one frequency and inspect the remaining coefficient. If it still depends on time, it is not a constant single-frequency phasor.$$\begin{aligned}Ae^{i\omega t}+Be^{i(2\omega t+\phi)}&=e^{i\omega t}(A+Be^{i(\omega t+\phi)}),\\y(\pi/\omega)&=-A+B\cos\phi.\\[4pt]A=B,\ \phi=0&:\quad y(0)=2A,\\y(\pi/\omega)&=A\cos\pi+A\cos2\pi=-A+A=0.\end{aligned}$$

(a) Calculation and conclusion

The arrows rotate at different rates. Factoring out $e^{i\omega t}$ leaves $A+Be^{i(\omega t+\phi)}$, a time-dependent complex coefficient. A fixed resultant phasor represents only one common frequency.

(b) Interpretation and limits

At $t=0$ the true value is $2A$, but at $t=\pi/\omega$ it is $-A+A=0$. The fixed-arrow construction would predict $2A\cos\omega t=-2A$ at the second time. Harmonic components require separate frequency phasors before reconstructing the time signal.

Physical check

For equal amplitudes and $\phi=0$, the actual displacement at this time is zero, whereas treating the two initial phasors as a permanent sum predicts $-2A$. Agreement at one instant is not sufficient.