Hooke's law — spring force F = −kx: Visual walkthrough

1.3.11 · D2 · Physics › Work, Energy & Power › Hooke's law — spring force F = −kx

Step 1 — Spring draw karo aur uska "ghar" name karo

KYA. Ek asli spring, flat rakhi hui, koi chhoo nahi raha. Ek length hai jo use pasand hai — uski natural length. Us resting position ko number line ka zero mark karo.

KYUN. Neeche ke saare measurements "ghar se kitna door" hain. "Kitna door" bolne se pehle, yeh agree karna zaroori hai ki ghar kahan hai. Ghar $= 0$ .

PICTURE. Figure mein, grey block $x = 0$ wale tick par baitha hai. Use right kheencho toh pointer positive numbers ki taraf jaata hai; left push karo toh negative numbers ki taraf jaata hai.

Figure — Hooke's law — spring force F = −kx

Step 2 — Energy ek valley mein rehti hai

KYA. Forces ki jagah, energy se shuru karo. Spring mein stored energy ko $U$ kaho. Plot karo ki spring har position $x$ ke liye kitni energy hold karti hai. Woh plot ek smooth U-shaped valley hai jiska lowest point $x = 0$ par hai.

KYUN. Ghar par spring kuch store nahi karti; use stretch KARO YA compress karo toh woh aur store karti hai. "Dono sides par zyada energy, beech mein sabse kam" bilkul valley ki shape hai. Energy se shuru karna (force se nahi) hamare liye prove karna possible banata hai ki koi bhi spring linear hogi — hum ise Step 5 mein use karenge.

PICTURE. Green curve $x = 0$ par minimum tak dip karti hai. Is curve par rakha ek marble downhill roll karta hai — bottom ki taraf, ghar ki taraf. Woh marble yaad rakho; woh hi restoring force hai.

Step 3 — Force valley ki steepness hai

KYA. Koi bhi formula aane se pehle, us cheez ka naam rakho jo hum dhundh rahe hain. Phir "marble downhill roll karta hai" ko ek formula mein badlo: force energy curve ki minus steepness hai.

$F = - \frac{d U}{d x}$

YEH TOOL KYUN — derivative $\frac{d U}{d x}$ . Hume ek aisa number chahiye jo bataye ki valley kitni tilted hai bilkul wahan jahan block baitha hai. Ek simple ratio "rise over run" sirf straight line describe karta hai — lekin haari valley curve karti hai. Derivative $\frac{d U}{d x}$ bilkul isi kaam ke liye bana tool hai: yeh ek point par curve ko touch karne wali tangent line ki slope hai — steepness bilkul yahan. Koi aur tool curve ki local steepness measure nahi karta; isliye yeh yahan aata hai.

Term by term:

$F$ — block par spring ki force, abhi define ki gayi.
$\frac{d U}{d x}$ — position $x$ par energy curve ki slope (jab tum right move karte ho toh energy kitni tezi se badhti hai).
minus sign — force ko downhill point karwata hai: jahan curve right ki taraf upar jaati hai, marble left push hota hai.

PICTURE. Valley ki right wall par tangent upar tilt karti hai (positive slope), toh $- slope$ left ki taraf point karta hai — ghar ki taraf wapas. Left wall par tangent neeche tilt karti hai (negative slope), toh $- slope$ right ki taraf point karta hai — phir se ghar ki taraf. Bilkul bottom par tangent flat hai: slope $= 0$ , force $= 0$ .

Step 4 — Bottom mein zoom in karo: har valley parabola jaisi dikhti hai

KYA. Koi bhi smooth valley lo aur uske lowest point par zoom in karo. Kareeb se, woh special dikhna band ho jaati hai — woh sabse simple possible valley jaisi dikhti hai: ek parabola, curve $U = \frac{1}{2} k x^{2}$ .

YEH TOOL KYUN — Taylor expansion. Hum bottom ke paas shape jaanna chahte hain, door nahi. Taylor expansion woh tool hai jo ek smooth curve ko ek point ke paas simple powers of $x$ ki sum se rebuild karta hai: ek constant piece, ek straight-line piece ( $x$ ), ek bowl piece ( $x^{2}$ ), ek gentle-bending piece ( $x^{3}$ ), aur aage bhi: $U (x) = height at bottom U (0) + tilt at bottom U^{'} (0) x + curviness \frac{1}{2} U^{''} (0) x^{2} + higher powers \frac{1}{6} U^{'''} (0) x^{3} + \dots$

Term by term, aur zyaadatar terms kyun gayab ho jaate hain ya fade karte hain:

$U (0)$ — bottom par energy. Ek constant. Saari energy ko upar ya neeche shift karne se forces par koi asar nahi, toh hum ise drop karte hain (bottom ko $0$ set karte hain).
$U^{'} (0) x$ — bottom par tilt. Lekin valley ka bottom flat hota hai: iska slope $0$ hai. Toh $U^{'} (0) = 0$ aur yeh poora term gayab ho jaata hai. (Yahi "equilibrium" ka matlab hai: koi force nahi, toh koi slope nahi.)
$\frac{1}{2} U^{''} (0) x^{2}$ — curviness. Yeh pehla term hai jo bachta hai, aur yeh $x^{2}$ ke proportional hai: ek parabola.
$\frac{1}{6} U^{'''} (0) x^{3} + \dots$ — higher powers. Yahan key intuition hai: chhote $x$ ke liye, bada power chhota number hota hai. Agar $x = 0.1$ , toh $x^{2} = 0.01$ lekin $x^{3} = 0.001$ — das guna chhota, aur $x^{4}$ aur bhi chhota. Toh jis pal tum ghar ke kaafi kareeb ho, $x^{2}$ bowl har higher term ke upar tower karta hai. ("Smooth" precisely woh promise hai ki yeh higher pieces exist karte hain aur aise shrink karte hain — koi sudden kinks nahi.) Isliye sirf $x^{2}$ tak rakhna aalas nahi hai; bottom ke paas yeh poori kahani hai.

PICTURE. Figure ek bumpy real energy curve (blue) aur parabola (yellow) ko overlay karta hai jo $x = 0$ ke paas use hug karti hai. $x = 0$ ke kareeb shaded band ke andar woh indistinguishable hain; sirf bahar, jahan $x^{3}$ aur aage wake up karte hain, woh alag hote hain. Isliye har chhoti oscillation ek spring hai.

Step 5 — Parabola se slide down karo aur $F = - k x$ paao

KYA. Ab hamare paas energy $U = \frac{1}{2} k x^{2}$ hai aur rule $F = - \frac{d U}{d x}$ hai. Inhe combine karo.

KYUN. Step 3 ne hamare liye machine di (force = −steepness). Step 4 ne valley di (ek parabola). Ek ko doosre mein daalo toh force law nikalna chahiye. Yahi payoff hai.

$\frac{1}{2} k x^{2}$ differentiate karo — position $x$ par parabola ki slope $k x$ hai: $F = - \frac{d U}{d x} = - \frac{d}{d x} (\frac{1}{2} k x^{2}) = - k x$

Term by term:

$\frac{d}{d x} (\frac{1}{2} k x^{2}) = k x$ — parabola ki steepness $x$ ke saath straight-line badhti hai: do guna door, do guna steep.
minus — Step 3 se inherited: force wall neeche ki taraf point karti hai.
result $F = - k x$ — origin se guzarne wali ek straight line jiska slope $- k$ hai.

PICTURE. Figure $F$ ko $x$ ke against plot karta hai: ek neeche ki taraf slope karne wali straight line. Iska slope $- k$ hai. Vertical axis cross karo aur force ka sign exactly flip hota hai jaise $x$ ka sign flip hota hai — "restoring" ka fingerprint.

Step 6 — Ek line par charo sign-cases (kuch skip nahi)

KYA. $x$ ke har sign ko line $F = - k x$ ke against check karo — ek bada stretch, ek chhota stretch, degenerate rest point, aur ek bada compression.

KYUN. Ek law jise tumne sirf "right stretch kiya" ke liye test kiya woh law hai jis par tum trust nahi karte. Har sign scenario walk karo taaki reader kabhi koi unshown case na mile. Far-out cases ( $∣ x ∣$ bada) bhi "farther = harder" magnitude dikhate hain, lekin unka asli kaam yahan confirm karna hai ki sign ab bhi behave karta hai.

Case	$x$	$- k x$ deta hai	$F$ ki direction	Matlab
Big stretch	$x ≫ 0$	negative (bada)	left ki taraf point karta hai	ghar wapas strong pull
Small stretch	$x > 0$	negative	left ki taraf point karta hai	block ko ghar wapas pull karta hai
At rest	$x = 0$	$0$	koi force nahi	equilibrium, valley ka bottom
Big compression	$x ≪ 0$	positive (bada)	right ki taraf point karta hai	ghar wapas strong push

PICTURE. Ek number line par chaar chhote blocks, har ek ke saath force arrow. Notice karo: har arrow zero mark ki taraf point karta hai, aur far-out arrows sabse lambe hain. Yeh, drawn, "restoring" ka matlab hai.

Step 7 — Energy triangle hai, rectangle nahi

KYA. Ghar ( $x^{'} = 0$ ) se kisi positive stretch $x > 0$ tak stretch karne par kitni energy store hoti hai? Yeh stretching-force line ke neeche ka area hai — ek triangle, jis se $U = \frac{1}{2} k x^{2}$ milta hai.

YEH TOOL KYUN — integral (curve ke neeche area). Stretching force $0$ se $k x$ tak grow karti hai jab tum bahar jaate ho; yeh kabhi constant nahi hoti, toh plain (force $\times$ distance) galat hai — woh maximum force ka rectangle hoga. Har position par badalti force ko add karne ke liye, tum $F_{ext} d x^{'}$ accumulate karte ho poori trip mein: woh accumulation hi integral hai, aur geometrically yeh graph ke neeche ka area hai. $U = \int_{0}^{x} F_{ext} > 0 k x^{'} d x^{'} = \frac{1}{2} k x^{2}$

Term by term:

$k x^{'}$ — har intermediate position $x^{'}$ par tumhara haath jo outward (positive) force lagaata hai, with $0 \leq x^{'} \leq x$ .
$\int_{0}^{x} \dots d x^{'}$ — ghar se $x$ tak (limits $0$ to $x$ , dono non-negative) (force $\times$ tiny step) ke saare thin slivers ko jodo.
$\frac{1}{2} k x^{2}$ — triangle ka area: base $x$ , height $k x$ , toh $\frac{1}{2} (base) (height)$ .

PICTURE. Blue triangle (sahi, $\frac{1}{2} k x^{2}$ ) dashed red rectangle (galat $k x \cdot x = k x^{2}$ ) ke andar baitha hai. Rectangle exactly triangle ka do guna hai — yahi famous factor of $\frac{1}{2}$ hai.

Ek-picture summary

Upar sab kuch, compressed: valley left par ( $U = \frac{1}{2} k x^{2}$ ) aur line right par ( $F = - k x$ ) ek hi object ke do views hain — line valley ki slope hai, negated. Left-to-right padho: energy ki shape $\to$ uski steepness $\to$ restoring force.

Recall Feynman retelling — poori walk plain words mein

Ek spring energy se bani valley ke bottom mein rehti hai. Bottom mein baithke, woh khush hai — kuch push nahi. Agar tum use kisi bhi wall par drag karo, woh wapas roll karna chahti hai, aur bilkul uske neeche wall ki steepness hai ki use kitna dhakka lagta hai. Kisi bhi valley ke bottom mein zoom in karo aur woh hamesha same smooth bowl jaisi dikhti hai — ek parabola, $\frac{1}{2} k x^{2}$ — kyunki kareeb se tiny higher-power wiggles ( $x^{3}$ , $x^{4}$ , …) $x^{2}$ bowl se tezi se shrink karte hain aur simply matter nahi karte. Us bowl ki steepness position $x$ par $k x$ nikli, aur "roll downhill" sign flip karta hai, toh spring ka push-back hai $- k x$ : do guna door, do guna hard, hamesha ghar ki taraf — chahe tumne use stretch kiya ya squish kiya. Aur energy store karne ke liye tum apne haath ki growing outward push ko poori trip par add karte ho — woh line ke neeche triangle hai, $\frac{1}{2} k x^{2}$ , jo tumhare naive guess ka half hai, kyunki force nothing se shuru hui thi.

F = - k x

mein minus sign pictorially kahan se aata hai?

F = - d U / d x

se — force energy valley mein downhill point karti hai, us direction ke opposite jahan energy badhti hai.

Har stable system rest ke paas spring jaisa kyun dikhta hai?

Taylor-expanding its energy minimum karo, constant aur linear terms drop ho jaate hain aur higher powers ghar ke paas tezi se shrink karti hain,

\frac{1}{2} U^{''} (0) x^{2}

— ek parabola — bachta hai, toh

F = - U^{''} (0) x = - k x

F

–

x

graph par,

k

kya hai?

Straight line

F = - k x

ki (neeche ki taraf) slope ki magnitude.

Stored energy paane ke liye hum

+ k x^{'}

integrate kyun karte hain (

- k x^{'}

nahi)?

Kyunki hum apne haath ka kaam track karte hain, jo outward force

F_{ext} = + k x^{'}

supply karta hai spring ke against; stretch par

x^{'}

0

x > 0

tak run karta hai, toh integrand positive hai.

Connections

Parent topic — poora statement aur examples.
Conservative forces and potential energy — rule $F = - d U / d x$ jo Step 3 mein use hua.
Work done by a variable force — Step 7 mein triangle ke peeche ka integral.
Elastic potential energy — $\frac{1}{2} k x^{2}$ reservoir.
Simple Harmonic Motion — $F = - k x$ ko motion mein daalo aur oscillation milti hai, $ω = k / m$ .
Conservation of mechanical energy — valley height (PE) aur speed (KE) trade karti hai.
Interatomic forces — "generic valley" argument atomic bonds par apply hua.