Phasors — Question 3

Question 3

Amplitude limits decide whether a cancellation design is possible

Three coherent scalar waves have adjustable phases but fixed amplitudes $2$, $3$ and $6\,\mathrm{mm}$.

1. Find the smallest and largest possible resultant amplitudes.

2. If the largest amplitude is reduced to $4\,\mathrm{mm}$, can complete cancellation be achieved? Derive the required relative angle between the first two phasors.

Question 3 - Solution

Amplitude limits decide whether a cancellation design is possible

Method and mathematical working

For the minimum, align all shorter vectors against the longest whenever that longest length exceeds their sum. Complete cancellation becomes possible precisely when the lengths can close a polygon.$$\begin{aligned}R_{\max}&=2+3+6=11,\\R_{\min}&=\max(6-2-3,0)=1,\\4^2&=2^2+3^2+2(2)(3)\cos\theta.\\[4pt]\cos\theta&=\frac{4^2-2^2-3^2}{2(2)(3)}=\frac14,\qquad\theta=1.31812\,\mathrm{rad}.\end{aligned}$$

(a) Calculation and conclusion

The first two can supply at most $2+3=5\,\mathrm{mm}$ against the largest. The minimum is therefore $6-5=1\,\mathrm{mm}$ and maximum is $2+3+6=11\,\mathrm{mm}$. The minimum occurs when the smaller vectors align opposite the largest.

(b) Interpretation and limits

With the third amplitude $4$, a closed triangle is possible. The first two must have resultant magnitude $4$: $4^2=2^2+3^2+2(2)(3)\cos\delta$. Thus $\cos\delta=1/4$, giving $\delta=1.3181\,\mathrm{rad}$. The third phasor points opposite their resultant.

Physical check

The replacement length $4$ lies between $|3-2|$ and $3+2$, so it can oppose the resultant of the other two. The computed cosine $1/4$ lies in the required interval $[-1,1]$.