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

WHAT is being described

A particle in Simple Harmonic Motion obeys: $x(t) = A\sin(\omega t + \phi)$

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

We want velocity $v$ and acceleration $a$ — both as functions of time and (more usefully) as functions of position $x$ .

HOW to derive velocity (from scratch)

Step 1 — Differentiate position. $x = A\sin(\omega t + \phi)$ $v = \frac{dx}{dt} = A\omega\cos(\omega t + \phi)$ Why this step? Velocity is literally the time-derivative of position; the chain rule pulls down an $\omega$ .

Step 2 — Eliminate time to get $v$ in terms of $x$ . We use the identity $\sin^2\theta + \cos^2\theta = 1$ with $\theta = \omega t + \phi$ .

From the two equations: $\sin(\omega t + \phi) = \frac{x}{A}, \qquad \cos(\omega t + \phi) = \frac{v}{A\omega}$

Why this step? We have two unknown trig values but one constraint linking them — so we can kill the time variable entirely.

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

Multiply through by $A^2\omega^2$ : $\frac{x^2\omega^2}{1} + \frac{v^2}{1} = A^2\omega^2 \quad\Rightarrow\quad v^2 = \omega^2(A^2 - x^2)$

HOW to derive acceleration

Step 1 — Differentiate velocity. $a = \frac{dv}{dt} = \frac{d}{dt}\big(A\omega\cos(\omega t + \phi)\big) = -A\omega^2\sin(\omega t + \phi)$

Step 2 — Recognize $A\sin(\omega t+\phi) = x$ . $\boxed{\,a = -\omega^2 x\,}$

Why the minus sign matters: it is the definition of SHM — acceleration proportional to displacement and directed oppositely. No minus, no oscillation.

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

WHY it's really energy conservation

Total energy $E = \frac{1}{2}kA^2$ with $k = m\omega^2$ . $\underbrace{\tfrac12 m v^2}_{KE} + \underbrace{\tfrac12 m\omega^2 x^2}_{PE} = \tfrac12 m\omega^2 A^2$ Cancel $\tfrac12 m$ : $v^2 + \omega^2 x^2 = \omega^2 A^2 \;\Rightarrow\; v = \omega\sqrt{A^2 - x^2}$ Same formula — proof it's energy book-keeping.

Phase relationships (Dual coding)

Quantity	Expression	Max value	Phase lead over $x$
$x$	$A\sin(\omega t)$	$A$	—
$v$	$A\omega\cos(\omega t)$	$\omega A$	$90°$ ahead
$a$	$-A\omega^2\sin(\omega t)$	$\omega^2 A$	$180°$ ahead

Worked Examples

Common Mistakes (Steel-manned)

Flashcards

What is velocity as a function of displacement in SHM?

v = \pm\omega\sqrt{A^2 - x^2}

Where in SHM is speed maximum and what is its value?

At the center (

x=0

);

v_{\max} = \omega A

Where is speed zero in SHM?

At the extremes

x = \pm A

What is acceleration in terms of displacement?

a = -\omega^2 x

Why is the acceleration negative?

It is restoring — always directed opposite to displacement, toward equilibrium.

What is the maximum acceleration and where?

|a|_{\max} = \omega^2 A

, at the extremes

x = \pm A

What identity is used to derive

v(x)

from

x(t)

and

v(t)

\sin^2\theta + \cos^2\theta = 1

The velocity-displacement relation is equivalent to which conservation law?

Conservation of total mechanical energy (

KE + PE =

const).

At what fraction of amplitude is the speed half of maximum?

x = \frac{\sqrt3}{2}A \approx 0.866A

By what phase does velocity lead displacement?

90°

(

\pi/2

); acceleration leads by

180°

Recall Feynman: explain to a 12-year-old

Imagine pushing a swing. At the very top of the swing it stops for a tiny moment before coming back — that's where it's slowest (speed zero). As it whooshes through the bottom (the middle), it's moving the fastest. The " $\sqrt{A^2-x^2}$ " is just measuring how much room is left before you hit the top: lots of room = lots of speed. And the push that turns it around is always trying to drag it back to the middle — that's the " $-\omega^2 x$ " acceleration: bigger pull when you're further out.

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

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, SHM mein particle beech (equilibrium) ke aaju-baaju jhoolta rehta hai. Jab particle kinaare par hota hai (yaani $x = A$ , sabse door), tab woh ek pal ke liye ruk jaata hai turn lene ke liye — wahan speed zero hoti hai. Aur jab particle beech mein hota hai ( $x=0$ ), tab woh sabse tez bhaagta hai, $v_{max}=\omega A$ . Isiliye formula aata hai $v = \omega\sqrt{A^2 - x^2}$ — yeh actually energy conservation hi hai roop badal ke: jitni position pe energy lagi, utni kinetic se kam.

Formula nikaalna easy hai: $x = A\sin(\omega t)$ ko differentiate karo to $v = A\omega\cos(\omega t)$ . Phir $\sin^2 + \cos^2 = 1$ identity use karke time ko hata do — bas $v = \omega\sqrt{A^2-x^2}$ ban jaata hai. Acceleration ke liye $v$ ko phir differentiate karo: $a = -\omega^2 x$ . Yeh minus sign bahut important hai — yeh batata hai ki acceleration hamesha center ki taraf kheechta hai (restoring force). Yahi to SHM ki definition hai.

Common galti: log sochte hain edge par speed max hoti hai kyunki "wahan particle sabse zyada chala" — galat! Edge par to ruk ke palatna padta hai, isliye speed zero. Aur dusri galti: acceleration ko constant maan lena projectile ki tarah — nahi bhai, yahan $a$ position ke saath badalta hai. Yaad rakho: "Center = Fast, Edge = Frozen" aur " $a$ always points home." Isse exam mein numericals jaldi nikal jaayenge.

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

WHAT is being described

HOW to derive velocity (from scratch)

HOW to derive acceleration

WHY it's really energy conservation

Phase relationships (Dual coding)

Worked Examples

Common Mistakes (Steel-manned)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Oscillations & Waves

Connections