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

1.6.4 · Physics › Oscillations & Waves

KYA describe kiya ja raha hai

Simple Harmonic Motion mein ek particle yeh obey karta hai: $x(t)=A\sin (\omega t+\varphi )$

• $A$ = amplitude (max displacement, units m)
• $\omega$ = angular frequency (rad/s)
• $x$ = equilibrium se displacement time $t$ par

Humein velocity $v$ aur acceleration $a$ chahiye — dono time ke function ke roop mein aur (zyada useful) position $x$ ke function ke roop mein.

Velocity KAISE derive karein (scratch se)

Step 1 — Position ko differentiate karo. $x=A\sin (\omega t+\varphi )$ $v=\frac{dx}{dt}=A\omega \cos (\omega t+\varphi )$ Yeh step kyun? Velocity literally position ki time-derivative hai; chain rule ek $\omega$ neeche kheench laata hai.

Step 2 — Time eliminate karo taaki $v$ milo $x$ ke terms mein. Hum identity ${\sin }^2\theta +{\cos }^2\theta =1$ use karte hain jahan $\theta =\omega t+\varphi$ hai.

Dono equations se: $\sin (\omega t+\varphi )=\frac{x}{A},\cos (\omega t+\varphi )=\frac{v}{A\omega }$

Yeh step kyun? Hamare paas do unknown trig values hain lekin ek constraint hai jo unhe link karti hai — toh hum time variable ko poori tarah khatam kar sakte hain.

Step 3 — Pythagorean identity apply karo. ${\left(\frac{x}{A}\right)}^2+{\left(\frac{v}{A\omega }\right)}^2=1$

$A^2{\omega }^2$ se multiply karo: $\frac{x^2{\omega }^2}{1}+\frac{v^2}{1}=A^2{\omega }^2\Rightarrow v^2={\omega }^2(A^2-x^2)$

Acceleration KAISE derive karein

Step 1 — Velocity ko differentiate karo. $a=\frac{dv}{dt}=\frac{d}{dt}(A\omega \cos (\omega t+\varphi ))=-A{\omega }^2\sin (\omega t+\varphi )$

Step 2 — Pehchano ki $A\sin (\omega t+\varphi )=x$. $\boxed{a=-{\omega }^{2}x}$

Minus sign kyun matter karta hai: yahi SHM ki definition hai — acceleration displacement ke proportional aur opposite direction mein. Minus nahi, toh oscillation nahi.

WHY yeh actually energy conservation hai

Total energy $E=\frac{1}{2}k{A}^2$ jahan $k=m{\omega }^2$. $\underset{KE}{\underbrace{\frac{1}{2}m{v}^2}}+\underset{PE}{\underbrace{\frac{1}{2}m{\omega }^2{x}^2}}=\frac{1}{2}m{\omega }^2{A}^2$ $\frac{1}{2}m$ cancel karo: $v^2+{\omega }^2{x}^2={\omega }^2{A}^2\Rightarrow v=\omega \sqrt{A^2-x^2}$ Same formula — proof hai ki yeh energy book-keeping hai.

Phase relationships (Dual coding)

Worked Examples

Common Mistakes (Steel-manned)

Flashcards

Connections

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

Concept Map

Deep Dive

• D1 · Foundations — har symbol zero se
• D2 · Visual walkthrough — derivation pictures mein
• D3 · Worked examples — har scenario
• D4 · Exercises — graded, full solutions
• D5 · Question bank — concept traps