Exercises — Superposition principle

KYA: dono signed displacements ko add karo. KYU: 1-D superposition mein yahi algebraic addition hoti hai — medium dono instructions ek saath karta hai. $y=(+7)+(-4)=+3\text{cm}$ Answer: $+3\text{cm}$. Us instant ke baad, har pulse bina kisi change ke continue karta hai.

$A_{\text{res}}=2A\cos (\varphi /2)$ use karo. (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. Yahi Interference of waves ki backbone hai.

KYA: $90^{\circ }$ phase gap ka matlab hai dono phasors right angles par hain. KYU: toh woh ek right triangle ki do legs ki tarah add hote hain — Pythagoras. $A_{\text{res}}=\sqrt{6^2+8^2}=\sqrt{100}=10$ Answer: $10$ units. (Neeche figure mein red aur teal arrows dekho.)

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

$y=2+2-3=+1\text{cm}$. Superposition kitni bhi waves ke liye kaam karta hai — bas add karte jao. Answer: $+1\text{cm}$.

Amplitude — cosine rule kyun: unequal arrows ek triangle banate hain; closing side resultant hai. $A_{\text{res}}=\sqrt{3^2+4^2+2(3)(4)\cos 60^{\circ }}=\sqrt{9+16+24(0.5)}=\sqrt{37}\approx 6.08$ Phase — components kyun: har phasor ko horizontal (along $y_1$) aur vertical parts mein todo, alag alag add karo, phir angle lo. $x=3+4\cos 60^{\circ }=3+2=5,y=0+4\sin 60^{\circ }=2\sqrt{3}\approx 3.464$ $\alpha =\arctan \left(\frac{3.464}{5}\right)\approx 34.7^{\circ }$ $x$ aur $y$ dono positive hain → first quadrant, toh plain $\arctan$ yahaan sahi hai (koi quadrant fix nahi chahiye). Answer: $A_{\text{res}}\approx 6.08$, $\alpha \approx 34.7^{\circ }$ $y_1$ se aage.

$A_{\text{res}}=2A\cos (\varphi /2)=0$ ke liye $\cos (\varphi /2)=0$ chahiye, yaani $\varphi /2=\frac{\pi }{2}$. Toh $\varphi =\pi$ ($[0,2\pi ]$ mein yahi ek value hai). Generally $\varphi =(2n+1)\pi$ integer $n$ ke liye. Answer: $\varphi =\pi$.

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

KYA: equal length ke teen arrows ko circle ke around evenly spaced add karo. WHY sum to zero: symmetry se woh ek closed equilateral triangle banate hain — head meets tail. Components se check karo: $x=A(\cos 0+\cos 120^{\circ }+\cos 240^{\circ })=A(1-0.5-0.5)=0$ $y=A(\sin 0+\sin 120^{\circ }+\sin 240^{\circ })=A(0+\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2})=0$ Answer: $A_{\text{res}}=0$. Perfectly balanced three-phase cancellation.

Cosine rule use karo aur $\cos \varphi$ ke liye solve karo: $7^2=5^2+3^2+2(5)(3)\cos \varphi \Rightarrow 49=34+30\cos \varphi$ $\cos \varphi =\frac{15}{30}=0.5\Rightarrow \varphi =60^{\circ }$ Answer: $\varphi =60^{\circ }$ (aur symmetry se $-60^{\circ }$ ya $300^{\circ }$ bhi kaam karta hai — cosine even hai).

$\sin P+\sin Q=2\sin \frac{P+Q}{2}\cos \frac{P-Q}{2}$ use karo jahan $P=kx-\omega t$, $Q=kx+\omega t$: $\frac{P+Q}{2}=kx,\frac{P-Q}{2}=-\omega t$ $y=2A\sin (kx)\cos (\omega t)$ $x$-part aur $t$-part alag ho jaate hain — har point jagah par oscillate karta hai amplitude $2A\sin (kx)$ ke saath. Yahi ek standing wave hai (dekho Standing waves). Nodes jahan $\sin (kx)=0$, yaani $x=0,\frac{\pi }{k},\frac{2\pi }{k},\dots$ Answer: $y=2A\sin (kx)\cos (\omega t)$ — ek standing wave.

Sum: $y=2A\cos \left(2\pi \frac{f_1-f_2}{2}t\right)\sin \left(2\pi \frac{f_1+f_2}{2}t\right)$. Slow $\cos$ envelope loudness ko modulate karta hai. Loudness peaks envelope cycle mein do baar aate hain, toh: $f_{\text{beat}}=|f_1-f_2|=|256-260|=4\text{Hz}$ $f_{\text{heard}}=\frac{f_1+f_2}{2}=\frac{256+260}{2}=258\text{Hz}$ Answer: beats $4\text{Hz}$ par; pitch $258\text{Hz}$ par. Dekho Beats.

Path difference se phase difference: $\varphi =\frac{2\pi }{\lambda }\Delta$. (a) Silence ke liye $\varphi =\pi$ chahiye: $\frac{2\pi }{\lambda }\Delta =\pi \Rightarrow \Delta =\frac{\lambda }{2}=0.25\text{m}$. (b) Max loudness ke liye $\varphi =2\pi$ chahiye: $\Delta =\lambda =0.5\text{m}$. Answers: (a) $0.25\text{m}$, (b) $0.5\text{m}$. Yahi Interference of waves ki geometry hai.

Amplitude $2A$ tak add hota hai, aur $I\propto A^2$, toh combined intensity $\propto (2A{)}^2=4A^2=4I_0$. Naively $I_0+I_0=2I_0$, lekin constructive $4I_0$ deta hai — "intensities add karo" se double. Reconcile: extra energy create nahi hoti; woh un destructive regions se borrow hoti hai jahan combined intensity $2I_0$ ki jagah $0$ ho jaati hai. Saare phases par average karo toh $⟨4I_0{\cos }^2(\varphi /2)⟩=2I_0$ — exactly conserved. Answer: $4I_0$; energy redistribution se conserved.

Do equal waves, $\varphi =\pi /2$: amplitude factor? ::: $2A\cos (\pi /4)=\sqrt{2}A$. First destructive point ke liye path difference? ::: $\Delta =\lambda /2$. $f_1,f_2$ se beat frequency? ::: $|f_1-f_2|$. Opposite-direction identical waves kya dete hain? ::: Ek standing wave, $2A\sin (kx)\cos (\omega t)$. Do $I_0$ sources ki constructive intensity? ::: $4I_0$ locally; $2I_0$ averaged.