Phasors — Question 5

Question 5

Differentiation rotates a phasor and depends on the time convention

At a fixed point use the convention $y(t)=\operatorname{Re}[Ze^{i\omega t}]$, with $Z=.003+.004i\,\mathrm{m}$ and $\omega=20\,\mathrm{rad/s}$.

1. Find the velocity and acceleration phasors and the instantaneous velocity at $t=0$.

2. How do the derivative factors change if the convention is instead $\operatorname{Re}[We^{-i\omega t}]$? State how $W$ represents the same real motion.

Question 5 - Solution

Differentiation rotates a phasor and depends on the time convention

Method and mathematical working

Differentiate the complex time dependence before taking the real part. Multiplication by $i$ rotates a phasor by a quarter turn; two differentiations supply a minus sign.$$\begin{aligned}Z_v&=i\omega Z_y,\qquad Z_a=-\omega^2 Z_y,\\Z_v&=20i(0.003+0.004i)=-0.080+0.060i\,\mathrm{m/s}.\\[4pt]Z_a&=-400(0.003+0.004i)=-1.20-1.60i\,\mathrm{m/s^2},\\v(0)&=\operatorname{Re}Z_v=-0.080\,\mathrm{m/s}.\end{aligned}$$

(a) Calculation and conclusion

The velocity phasor is $i\omega Z=-.080+.060i\,\mathrm{m/s}$. Acceleration is $-\omega^2Z=-1.20-1.60i\,\mathrm{m/s^2}$. Thus $\dot y(0)=-.080\,\mathrm{m/s}$.

(b) Interpretation and limits

With negative temporal exponent, differentiation multiplies by $-i\omega$; the second derivative still multiplies by $-\omega^2$. The same real motion uses $W=Z^*$. Mixing the two conventions without conjugating phases reverses velocity signs.

Physical check

The magnitudes are $\omega|Z_y|$ and $\omega^2|Z_y|$. Changing to an $e^{-i\omega t}$ convention requires consistent conjugation and changes the derivative multiplier to $-i\omega$.