Visual walkthrough — Velocity and acceleration in SHM — v = ω√(A² − x²)

1.6.4 · D2 · Physics › Oscillations & Waves › Velocity and acceleration in SHM — v = ω√(A² − x²)

Hum poore walkthrough mein ek hi character use karte hain: ek dot jo ek line par aage-peeche jaata hai, jaise ek wire par bead. Centre se uski doori ko hum $x$ kehte hain. Bas yahi ek idea hai jisse hum shuru karte hain.

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Step 1 — Motion draw karo aur sign convention fix karo

KYA. Ek horizontal wire imagine karo. Ek red dot left aur right slide karta hai, kabhi do walls se aage nahi jaata. Wire ka middle "home" (equilibrium) hai. Dot jitni door tak pahunch sakta hai usse hum $==A==$ — amplitude (metres mein) — kehte hain. Ab hum ek pakka choice karte hain jo hamesha rahegi: rightward positive hai. Toh $x>0$ matlab dot home ke right mein hai, $x<0$ matlab left mein; velocity $v>0$ matlab rightward ja raha hai, $v<0$ matlab leftward.

KYUN. Kisi bhi algebra se pehle hum yeh fix karna chahte hain ki har letter picture par kya hai, aur kaun sa direction "plus" count hoga. $x$ = dot abhi home se kitna door hai (aur kis taraf). $A$ = $|x|$ ka sabse bada possible value. Sign convention woh cheez hai jo baad mein hamare answer ke $\pm$ ko asli meaning degi.

PICTURE. Figure s01 dekho: dot signed position $x$ par baitha hai. $-A$ aur $+A$ par dashed walls hain — ye turning points hain — dot inhe touch karta hai, ek pal ke liye rukta hai, aur palat jaata hai. Green arrow dikhata hai kaun sa direction "positive" hai.

Abhi hum sirf do facts jaante hain, dono picture se seedhe padhne layak hain:

• $x$, $-A$ aur $+A$ ke beech range karta hai;
• walls par dot ruk jaata hai aur palat jaata hai.

Abhi tak koi sine, koi angle nahi — Step 2 mein hum unhe build karenge, assume nahi karenge.

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Step 2 — Spinning dot kyun? Reference circle

KYA. Ek doosra dot lo jo ek circle par steady speed se ghoom raha ho jiska radius $A$ hai. Hum uska angle $==\theta ==$ upar wali vertical se measure karte hain, anticlockwise direction mein. Ab uska shadow horizontally ek vertical ruler par daalo (sideways projection karke up–down axis par). Woh shadow bilkul ek bead ki tarah upar-neeche slide karta hai — aur hum usi up–down shadow ko apna $x$ maanenge, bas sideways rakh ke.

KYUN. Hum circle isliye choose karte hain kyunki uniform circular motion ki shadow smoothly oscillate karti hai (Reference circle (projection of uniform circular motion) trick). $\theta$ ko top se measure karna aur shadow ko usi top axis ki taraf se padhna ek deliberate choice hai: isse shadow, $\theta$ ke sine ke barabar hoti hai. Agar hum side se measure karte, toh cosine milta. Physics same hoti, bookkeeping alag — hum sine convention isliye lete hain taaki parent note ke $x=A\sin (\omega t+\varphi )$ se match ho.

PICTURE. Figure s02 mein circle-dot top se $\theta$ angle par hai. Vertical axis par ek horizontal dotted line daalo: woh axis par height $x$ par milti hai. Right triangle ka hypotenuse $A$ (radius) hai aur jis direction se hum $\theta$ measure karte hain us direction wali side ki length $x$ hai. Kyunki $\theta$ top se measure hota hai, woh side jo $\theta$ ke kholne par badhti hai woh $x=A\sin \theta$ wali hai.

$x=A\sin \theta ⟹\sin \theta =\frac{x}{A}$

• $\sin \theta$ = (angle $\theta$ ke opposite side) $÷$ (hypotenuse) — $\theta$ top se padhne par, woh opposite side exactly shadow $x$ hai.
• $\frac{x}{A}$ = "dot apni poori reach mein kitna aage hai," $-1$ se $+1$ tak ka ek signed number (negative jab circle-dot middle ke neeche ho).
• Steady sweep ka matlab hai $\theta$ time mein linearly badhta hai: $\theta =\omega t+\varphi$, jahan $\omega$ spin rate hai aur $\varphi$ starting angle hai. $\varphi$ ki hum aage zaroorat nahi padhegi — woh bas yeh set karta hai ki $t=0$ par dot kahan tha.

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Step 3 — Velocity matlab shadow ki speed

KYA. Circle-dot, circle par constant speed $\omega A$ (radius $\times$ spin rate) se chalta hai. Uska velocity arrow circle ke saath saath (tangent) point karta hai. Shadow ki velocity sirf us arrow ka hamare measuring axis par projection hai.

KYUN. Velocity ka matlab hai "position ke change ki rate," likha jaata hai $v=\frac{dx}{dt}$. Hum derivative isliye use karte hain kyunki woh exact tool hai jo ek position rule ko speed rule mein convert karta hai. Circle par, $x=A\sin \theta$ ko differentiate karne par ek factor $\omega$ aa jaata hai (dial $\omega$ rate par spin karta hai) aur $\sin$ se $\cos$ ban jaata hai: yahan perpendicular projection saamne aata hai.

PICTURE. Figure s03 mein green tangent arrow ki length $\omega A$ hai. Iska hamare $x$-axis par projection (woh part jo $x$ measure hone ki direction mein hai) dot ki velocity $v$ hai. Dhyan do: jab circle-dot middle ke level par hota hai, tangent arrow poori tarah measuring axis ke saath align hota hai, isliye shadow ki speed sabse zyada hoti hai.

$v=\frac{dx}{dt}=\underset{\text{full circle-speed}}{\underbrace{A\omega }}\underset{\text{kitna axis ke saath hai}}{\underbrace{\cos \theta }}$

• $A\omega$ — circle-dot ki constant speed.
• $\cos \theta$ — us speed ka woh fraction jo hamare $x$-axis ki taraf point karta hai; yeh positive hota hai jab $\theta$ turn ke pehle/chauthei hisse mein ho (dot $+x$ ki taraf ja raha ho) aur wapas aate waqt negative. Yeh sign exactly Step 1 ke convention se $v$ ki sign hai.
• Rearrange karne par: $\cos \theta =\frac{v}{A\omega }$.

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Step 4 — Do shadows, ek circle: important link

KYA. Ab hamare paas usi circle-dot se do readings hain: $\sin \theta =\frac{x}{A},\cos \theta =\frac{v}{A\omega }.$ Ek $\theta$ dono ko control karta hai. Wahi ek angle bridge hai jo hume time hatane deta hai.

KYUN. Hume parwah nahi ki dot kab kahan hai — hum ek aisa rule chahte hain jo kahan hai ($x$) aur kitni tezi se chal raha hai ($v$) ko link kare. Kyunki dono same $\theta$ se aate hain, hum $\theta$ eliminate kar sakte hain us ek law se jo har circle maanta hai: Pythagorean identity sin² + cos² = 1.

PICTURE. Figure s04 circle ke andar right triangle dikhata hai: ek side $=A\sin \theta$, perpendicular side $=A\cos \theta$, hypotenuse $=A$ (radius). Is triangle par Pythagoras hi ${\sin }^2\theta +{\cos }^2\theta =1$ hai.

$\underset{(x/A{)}^2}{\underbrace{{\sin }^2\theta }}+\underset{(v/A\omega {)}^2}{\underbrace{{\cos }^2\theta }}=1$

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Step 5 — Angle hatao, $x$ aur $v$ rakho

KYA. Dono shadow-readings ko identity mein substitute karo aur fractions clear karo.

KYUN. Yeh Step 4 ka payoff hai: identity ek constraint hai jo $x$ aur $v$ ko ek saath baandhti hai, koi $\theta$ nahi bachta. Ek equation, do quantities jo hum actually measure karte hain.

PICTURE. Figure s05 algebra ko triangle ke shrinking ke roop mein track karta hai — har side scale hoti hai — jab tak radius-squared statement sirf $x$ aur $v$ ke baare mein nahi ban jaata.

${\left(\frac{x}{A}\right)}^2+{\left(\frac{v}{A\omega }\right)}^2=1$

Har term ko $A^2{\omega }^2$ se multiply karo:

$\underset{\text{x wale term se}}{\underbrace{{\omega }^2{x}^2}}+\underset{\text{v wale term se}}{\underbrace{v^2}}=\underset{{\text{radius}}^2\text{ scaled}}{\underbrace{A^2{\omega }^2}}$

$v^2$ ke liye solve karo:

$v^2={\omega }^2(A^2-x^2)$

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Step 6 — Formula se do extreme cases padho

KYA. Swing ke dono ends test karo.

KYUN. Ek formula jise tum apne edges par check nahi kar sakte, woh formula nahi jis par tum trust karo. Do special positions plug karo.

PICTURE. Figure s06 wire par speed overlay karta hai: centre par mota arrow, walls par koi arrow nahi.

• Home par, $x=0$: $v=\omega \sqrt{A^2-0}=\omega A$. Yeh $==v_{\max }==$ hai — dot ki sabse tez speed. Circle par dot middle ke level par hai, isliye tangent arrow poori tarah hamare measuring axis ke saath align hai aur uska projection full-length hai.
• Wall par, $x=\pm A$: $v=\omega \sqrt{A^2-A^2}=0$. Dot freeze ho jaata hai palat ne ke liye. Circle par dot bilkul top ya bottom par hai, isliye tangent arrow measuring axis ke seedha across point karta hai (perpendicular); us axis par uska projection isliye zero hai. (Yeh exactly parent note mein "fastest at the edge" wali common galti ka ulta hai.)

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Step 7 — Degenerate cases: koi swing nahi, aur forbidden positions

KYA. Do boundary situations. Pehla, amplitude zero, $A=0$. Doosra, walls se aage position maangna, $|x|>A$.

KYUN. Har honest derivation ko apne extreme inputs par survive karna chahiye aur clearly batana chahiye ki woh kahan use hogi — uska domain.

PICTURE. Figure s07 circle ko origin par ek single point tak shrink dikhata hai (case $A=0$), aur wire par allowed band $-A\le x\le A$ mark karta hai.

Case $A=0$. Circle ka koi radius nahi; dot sirf home par hi ho sakta hai, isliye single legal position $x=0$ hai, aur $v=\omega \sqrt{0-0}=0.$ Koi swing nahi, koi speed nahi — consistent.

Formula ka domain. Dot physically sirf band $-A\le x\le A$ mein rehta hai. Is band ke andar $A^2-x^2\ge 0$ hai, isliye square root ek real, sensible speed deta hai. Bahar ($|x|>A$) hum pooch rahe hote "dot ek aisi jagah par kitna fast hai jahan woh kabhi pahunch nahi sakta?" — aur $A^2-x^2$ negative ho jaata hai. Square root ke neeche negative number algebra ka tarika hai yeh kehne ka ki "impossible position, koi real speed exist nahi karti." Toh formula exactly $-A\le x\le A$ par valid hai, $v=0$ dono ends par aur $v=\pm \omega A$ beech mein.

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Step 8 — Usi picture mein acceleration kahan chhupi hai

KYA. Circle-dot, constant speed par chalne ke bawajood, constantly direction change karta hai, isliye uski acceleration hoti hai. Uniform circular motion ke liye yeh acceleration seedha centre ki taraf point karti hai aur uski magnitude ${\omega }^{2}A$ hoti hai. Iska hamare $x$-axis par projection sliding dot ki acceleration hai.

KYUN ($a={\omega }^{2}R$ derive karna). Centre ki taraf pointing acceleration ${\omega }^{2}R$ kyun hai, $R$ radius ke saath? Velocity arrow ki fixed length $v_c=\omega R$ hai lekin uski direction position ki tarah hi rate $\omega$ par ghoomti hai. Acceleration velocity arrow ke change ki rate hai; length $\omega R$ wala ek vector jiska direction rate $\omega$ par rotate kare, rate $(\text{length})\times (\text{turn rate})=\omega R\times \omega ={\omega }^{2}R$ par change karta hai, aur change inward point karta hai (centre ki taraf). $R=A$ ke saath yeh ${\omega }^{2}A$ hai. (Yahi reasoning hai jisse Step 3 mein length $A$ ki position rate $\omega$ par ghoom kar length $\omega A$ ki velocity deti hai — ab ise velocity arrow par ek baar aur apply karo.)

PICTURE. Figure s08 length ${\omega }^{2}A$ ka inward arrow draw karta hai; hamare $x$-axis par uska projection $-{\omega }^{2}x$ hai — $x$ ke opposite direction mein (hamesha home ki taraf).

$a=-A{\omega }^2\sin \theta =-{\omega }^2\underset{=x}{\underbrace{(A\sin \theta )}}=-{\omega }^{2}x$

• Minus sign = inward arrow ki shadow hamesha home ki taraf point karti hai, displacement ke opposite — SHM ki defining feature.
• Door hone par bada, centre par zero — exactly speed pattern ka ulta.

Yahi relation energy book-keeping hai: $v^2+{\omega }^2{x}^2={\omega }^2{A}^2$ matlab $KE+PE=$ constant hai jisme $\frac{1}{2}m$ cancel ho gaya.

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Ek-picture summary

Figure s09 poori story ek frame mein fold karta hai: reference circle, sliding shadow $x$, tangent velocity jiska projection $v=\omega \sqrt{A^2-x^2}$ hai, aur inward pull jiska projection $a=-{\omega }^{2}x$ hai. Beech mein fast, walls par frozen, hamesha home ki taraf khicha jaata.

Recall Poore walkthrough ki Feynman retelling

Ek wire par dot rakho; woh do walls ke beech slide karta hai jo $A$ door hain, aur hum agree karte hain ki rightward positive count hoga. Ab ek dost ko ek circle par ghoomne do jiska radius bhi $A$ hai aur uski height top se $\theta$ angle maanke padho — unki shadow hi tumhara sliding dot hai, aur woh shadow $A\sin \theta$ ke barabar hai. Tumhara dost steady pace se chalta hai, isliye unka speed arrow (circle ke saath pointing) ki fixed length $\omega A$ hai, aur unka "middle ki taraf kheencha jaana" wala arrow ki fixed length ${\omega }^{2}A$ hai — dono ek length hain jo rate $\omega$ par ghoomti hai, phir ek baar aur ghoomti hai. Speed arrow ka woh part jo tumhare axis ke saath hai woh hai jitna tez tumhara dot chal raha hai ($v$); pull arrow ka woh part jo tumhare axis ke saath hai tumhare dot ka push-back hai ($a$). Jab tumhara dost middle ke level par hota hai, speed arrow poori tarah tumhare axis ke saath lie karta hai — shadow fastest chalti hai — aur yeh exactly tab hota hai jab shadow home par ho. Jab dost bilkul top ya bottom par hota hai, speed arrow seedha tumhare axis ke across point karta hai — us par koi motion nahi — isliye dot wall par freeze ho jaata hai. Circle ke right triangle par Pythagoras (${\sin }^2+{\cos }^2=1$) position aur speed ko ek saath baandhta hai, aur nikalta hai $v=\omega \sqrt{A^2-x^2}$: $\sqrt{A^2-x^2}$ bas "wall se pehle kitna room bacha hai" hai, aur $\pm$ record karta hai dot kaun si taraf ja raha hai. Pull arrow hamesha centre ki taraf aim karta hai, isliye $a=-{\omega }^{2}x$ mein minus hai — woh hamesha dot ko ghar waapas kheeench raha hai.

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Connections

• Velocity and acceleration in SHM — v = ω√(A² − x²) (parent)
• Reference circle (projection of uniform circular motion)
• Pythagorean identity sin² + cos² = 1
• Angular frequency ω and time period T
• Phase and phase difference
• Simple Harmonic Motion — definition a = −ω²x
• Energy in SHM — kinetic and potential

Concept Map