Exercises — Superposition principle

WHAT: add the two signed displacements. WHY: in 1-D superposition is algebraic addition — the medium does both instructions at once. $y=(+7)+(-4)=+3\text{cm}$ Answer: $+3\text{cm}$. After the instant, each pulse continues untouched.

Use $A_{\text{res}}=2A\cos (\varphi /2)$. (a) $\varphi =0$: $\cos 0=1\Rightarrow A_{\text{res}}=2A$ — constructive. (b) $\varphi =\pi$: $\cos (\pi /2)=0\Rightarrow A_{\text{res}}=0$ — destructive. This is the backbone of Interference of waves.

WHAT: a $90^{\circ }$ phase gap means the two phasors are at right angles. WHY: so they add like the two legs of a right triangle — Pythagoras. $A_{\text{res}}=\sqrt{6^2+8^2}=\sqrt{100}=10$ Answer: $10$ units. (Look at the red and teal arrows in the figure below.)

Equal amplitudes ⇒ use $A_{\text{res}}=2A\cos (\varphi /2)$. $A_{\text{res}}=2(5)\cos (60^{\circ })=10\times 0.5=5\text{cm}$ Answer: $5\text{cm}$.

$y=2+2-3=+1\text{cm}$. Superposition works for any number of waves — just keep adding. Answer: $+1\text{cm}$.

Amplitude — WHY the cosine rule: unequal arrows form a triangle; the closing side is the resultant. $A_{\text{res}}=\sqrt{3^2+4^2+2(3)(4)\cos 60^{\circ }}=\sqrt{9+16+24(0.5)}=\sqrt{37}\approx 6.08$ Phase — WHY components: break each phasor into horizontal (along $y_1$) and vertical parts, add separately, then take the angle. $x=3+4\cos 60^{\circ }=3+2=5,y=0+4\sin 60^{\circ }=2\sqrt{3}\approx 3.464$ $\alpha =\arctan \left(\frac{3.464}{5}\right)\approx 34.7^{\circ }$ Both $x$ and $y$ are positive → first quadrant, so the plain $\arctan$ is correct here (no quadrant fix needed). Answer: $A_{\text{res}}\approx 6.08$, $\alpha \approx 34.7^{\circ }$ ahead of $y_1$.

$A_{\text{res}}=2A\cos (\varphi /2)=0$ requires $\cos (\varphi /2)=0$, i.e. $\varphi /2=\frac{\pi }{2}$. So $\varphi =\pi$ (the only value in $[0,2\pi ]$). More generally $\varphi =(2n+1)\pi$ for integer $n$. Answer: $\varphi =\pi$.

$A_{\text{res}}(\varphi )=2A\cos (\varphi /2)$, so $I\propto 4A^2{\cos }^2(\varphi /2)$; $I_{\text{max}}$ at $\varphi =0$ is $\propto 4A^2$. At $\varphi =\pi /2$: ${\cos }^2(\pi /4)=(1/\sqrt{2}{)}^2=1/2$. $\frac{I_{\pi /2}}{I_{\text{max}}}={\cos }^2(\pi /4)=\frac{1}{2}$ Answer: $1:2$.

WHAT: add three arrows of equal length spaced evenly around the circle. WHY sum to zero: by symmetry they form a closed equilateral triangle — head meets tail. Check with components: $x=A(\cos 0+\cos 120^{\circ }+\cos 240^{\circ })=A(1-0.5-0.5)=0$ $y=A(\sin 0+\sin 120^{\circ }+\sin 240^{\circ })=A(0+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2})=0$ Answer: $A_{\text{res}}=0$. Perfectly balanced three-phase cancellation.

Use the cosine rule and solve for $\cos \varphi$: $7^2=5^2+3^2+2(5)(3)\cos \varphi \Rightarrow 49=34+30\cos \varphi$ $\cos \varphi =\frac{15}{30}=0.5\Rightarrow \varphi =60^{\circ }$ Answer: $\varphi =60^{\circ }$ (and by symmetry $-60^{\circ }$ or $300^{\circ }$ also works — cosine is even).

Use $\sin P+\sin Q=2\sin \frac{P+Q}{2}\cos \frac{P-Q}{2}$ with $P=kx-\omega t$, $Q=kx+\omega t$: $\frac{P+Q}{2}=kx,\frac{P-Q}{2}=-\omega t$ $y=2A\sin (kx)\cos (\omega t)$ The $x$-part and $t$-part separate — every point oscillates in place with amplitude $2A\sin (kx)$. That is a standing wave (see Standing waves). Nodes where $\sin (kx)=0$, i.e. $x=0,\frac{\pi }{k},\frac{2\pi }{k},\dots$ Answer: $y=2A\sin (kx)\cos (\omega t)$ — a standing wave.

Sum: $y=2A\cos \left(2\pi \frac{f_1-f_2}{2}t\right)\sin \left(2\pi \frac{f_1+f_2}{2}t\right)$. The slow $\cos$ envelope modulates loudness. Loudness peaks twice per envelope cycle, so: $f_{\text{beat}}=|f_1-f_2|=|256-260|=4\text{Hz}$ $f_{\text{heard}}=\frac{f_1+f_2}{2}=\frac{256+260}{2}=258\text{Hz}$ Answer: beats at $4\text{Hz}$; pitch at $258\text{Hz}$. See Beats.

Phase difference from path difference: $\varphi =\frac{2\pi }{\lambda }\Delta$. (a) Silence needs $\varphi =\pi$: $\frac{2\pi }{\lambda }\Delta =\pi \Rightarrow \Delta =\frac{\lambda }{2}=0.25\text{m}$. (b) Max loudness needs $\varphi =2\pi$: $\Delta =\lambda =0.5\text{m}$. Answers: (a) $0.25\text{m}$, (b) $0.5\text{m}$. This is the geometry of Interference of waves.

Amplitude adds to $2A$, and $I\propto A^2$, so combined intensity $\propto (2A{)}^2=4A^2=4I_0$. Naively $I_0+I_0=2I_0$, but constructive gives $4I_0$ — twice what "adding intensities" predicts. Reconcile: the extra energy is not created; it is borrowed from the destructive regions elsewhere where the combined intensity drops to $0$ instead of $2I_0$. Averaged over all phases, $⟨4I_0{\cos }^2(\varphi /2)⟩=2I_0$ — exactly conserved. Answer: $4I_0$; energy conserved by redistribution.

Two equal waves, $\varphi =\pi /2$: amplitude factor? ::: $2A\cos (\pi /4)=\sqrt{2}A$. Path difference for first destructive point? ::: $\Delta =\lambda /2$. Beat frequency from $f_1,f_2$? ::: $|f_1-f_2|$. Opposite-direction identical waves give? ::: A standing wave, $2A\sin (kx)\cos (\omega t)$. Constructive intensity of two $I_0$ sources? ::: $4I_0$ locally; $2I_0$ averaged.