Spring potential energy — derivation: Visual walkthrough

1.3.12 · D2 · Physics › Work, Energy & Power › Spring potential energy — derivation

Step 0 — Spring kya hai, aur $x$ kya hai?

KYA HAI. Ek spring ki ek aisi length hoti hai jahan woh khush rehti hai: uski natural length — na stretch, na squeeze. Ise equilibrium position kaho. Hum sab kuch wahin se measure karte hain.

Is page par sabse important symbol $x$ hai: equilibrium se displacement. Agar tum end ko $3 cm$ bahar khicho, to $x = + 0.03 m$ . Agar $3 cm$ andar dhako, to $x = - 0.03 m$ . Happy length par $x = 0$ .

PEHLE KYUN define karein. Neeche har formula mein $x$ hai. Agar tumhara matlab secretly "spring ki total length" tha, to bakwaas result milega. Isliye ek equation likhne se pehle hum ise ek picture ke saath pakka kar lete hain.

PICTURE. Relaxed spring, phir wohi spring stretched — laal arrow hi $x$ hai, "end abhi kahan hai" aur "woh kahan rest karta hai" ke beech ka gap.

Figure — Spring potential energy — derivation

Step 1 — Spring kitni zyada kheenchti hai? (Hooke's law)

KYA HAI. Experiment kehta hai: spring ko jitna zyada khiincho, woh utni zyada wapas kheenchti hai — aur woh sabse simple tarike se karta hai, seedhi line proportional. Yahi Hooke's Law hai: $F_{spring} = - k x$

Aaiye ise symbol by symbol padhen, theek jahan woh baitha hai:

$F_{spring}$ — woh force jo spring lagaata hai, newtons $(N)$ mein.
$k$ — spring constant, "stiffness." Bada $k$ = zidd spring. Units: newtons per metre $(N/m)$ , yaani har metre stretch par kitne newtons.
$x$ — hamara Step 0 wala displacement.
minus sign $-$ — yahi spring ki poori personality hai: force displacement ke opposite direction mein point karta hai. Daayein khiincho ( $x > 0$ ), woh baayein kheenchta hai. Baayein dhako ( $x < 0$ ), woh daayein push karta hai. Hamesha $x = 0$ ki taraf wapas. Isliye hum ise restoring force kehte hain.

KYUN seedhi line aur curve nahi? Kyunki ek ideal spring ke liye force versus $x$ ka graph bilkul origin se guzarti ek slanted line hoti hai. Yahi linearity hai jiske wajah se final energy itni clean aati hai. (Real springs extreme stretch par is se hat jaati hain — lekin woh alag note hai.)

PICTURE. Force–displacement line. Gaur karo yeh origin se guzarti hai: $x = 0$ par spring zero force se kheenchti hai. Yahi ek fact — shuruat mein zero force — baad mein naive "rectangle" guess ko defeat karta hai.

Step 2 — Woh force jo tumhe lagaani padegi

KYA HAI. Hum stored energy chahte hain, jo us kaam ke barabar hai jo tum ise slowly stretch karte hue karte ho. End ko accelerate kiye bina move karne ke liye, tumhara pull spring ke pull ko exactly cancel karna chahiye: $F_{applied} = + k x$

Sign $+$ ho gaya kyunki tum baahir kheenchte ho, stretch ki same direction mein.
Size same $k x$ , $F_{spring}$ ke opposite direction — har instant par yeh ek balanced tug-of-war hain.

KYUN hum tumhara force track karte hain, spring ka nahi? Kyunki "spring mein stored energy" = "woh kaam jo tumhe uske against invest karna pada." Hum restoring force ke against push karke store banate hain. Yeh conservative-force bookkeeping hai: conservative force ke against kiya gaya kaam potential energy ke roop mein bank ho jaata hai, heat mein kuch nahi jaata.

PICTURE. Wohi slanted line, ab $F_{applied} = k x$ label ke saath (Step 2 ka mirror, upper half mein drawn). Yahi woh line hai jiska area hum aage measure karne wale hain.

Step 3 — Kyun ek $F \times x$ jhooth hai

KYA HAI. Constant force ke liye, kaam simply $W = F \times d$ hota hai — ek rectangle: height $F$ , width $d$ . Likhne ki ichha hoti hai $W = (k x) \cdot x = k x^{2}$ .

KYUN yeh galat hai. $k x$ force sirf bilkul end par hai. Shuruat mein force $0$ tha (line origin se guzarti hai!). Poori trip ke liye sabse bada force use karna double-counting hai. Figure dekho: line ke neeche ka honest region ek triangle hai, lekin $k x \cdot x$ poora rectangle hai — doona bada.

PICTURE. Rectangle (galat, faint) triangle ke upar drawn (sahi, laal). Triangle bilkul rectangle ka aadha hai. Woh $\frac{1}{2}$ factor jo tum baar baar dekhte ho? Woh khaali upar-baayein wala aadha hai.

Step 4 — Motion ko chhote slices mein kato

KYA HAI. Force move karne ke saath change hota hai, isliye hum poori trip par $F \times d$ use nahi kar sakte. Lekin ek tiny step par woh barely change hota hai. Stretch ko thin strips mein slice karo. $x^{'}$ ko $0$ aur $x$ ke beech kahin ek running position maano (prime ka matlab sirf "woh position jahan hum abhi hain," final $x$ se alag rakhne ke liye). $d x^{'}$ ek thin strip ki width hai.

$x^{'}$ — kuch hissa guzara hua position, $0 \leq x^{'} \leq x$ .
$d x^{'}$ — ek slice ki (infinitesimally small) width.

KYUN slice karo? Kyunki " $W = F \times d$ " tab hi legal hai jab $F$ stable rahe. Slice ko infinitesimally thin banana force ko effectively us par constant bana deta hai — phir legal. Yahi Work done by a variable force wali trick hai.

PICTURE. Triangle vertical strips mein kati hui. Ek strip laal mein highlighted: position $x^{'}$ par uski height force $k x^{'}$ hai, uski width $d x^{'}$ hai.

Step 5 — Ek slice par kiya gaya kaam

KYA HAI. Us single laal strip ke liye, force essentially $k x^{'}$ par constant hai, aur move ki gayi distance $d x^{'}$ hai. To kaam ka chhota sa tukda hai: $d W = F_{applied} d x^{'} = k x^{'} d x^{'}$

$d W$ — is ek strip ke liye kaam ka sliver.
$k x^{'}$ — position $x^{'}$ par (is strip ke liye constant) force: strip ki height.
$d x^{'}$ — strip ki width.
to $d W$ literally ek thin rectangle ka area hai — height $\times$ width.

KYUN. Yeh honest $W = F \times d$ hai sirf wahin use kiya jahan allowed hai: ek aaise piece par jo force ke change karne ke liye bahut thin hai. Har strip apna thoda sa rectangular area of work contribute karta hai.

PICTURE. Single strip par zoom: height $k x^{'}$ , width $d x^{'}$ , area $= d W$ laal mein shaded.

Step 6 — Har slice ko jodo (the integral)

KYA HAI. Total work = $x^{'} = 0$ se $x^{'} = x$ tak saare strip-areas ka sum. Infinitely many infinitely thin strips ko add karna hi integral hai: $W = \int_{0}^{x} k x^{'} d x^{'}$

Symbols padhte hain:

$\int_{0}^{x}$ — "add karo, $x^{'}$ ko $0$ (relaxed) se $x$ (final stretch) tak sweep karte hue."
$k x^{'} d x^{'}$ — $x^{'}$ par strip ka area (Step 5 se).

Ab evaluate karo. Constant $k$ bahar nikalo; $x^{'}$ ka integral $\frac{x ^{'2}}{2}$ hai: $W = k \int_{0}^{x} x^{'} d x^{'} = k [\frac{x ^{'2}}{2}]_{0}^{x} = k (\frac{x ^{2}}{2} - \frac{0 ^{2}}{2}) = \frac{1}{2} k x^{2}$

$[\frac{x ^{'2}}{2}]_{0}^{x}$ — top limit plug in karo minus bottom limit.
bottom limit $0$ deta hai (koi stretch nahi, koi stored work nahi) — achha sanity check hai.

KYUN integral triangle ke barabar hai. Saare strip-areas ka sum wohi hai jitna poore triangle ka area hai: base $= x$ , height $= k x$ , area $= \frac{1}{2} \times x \times k x = \frac{1}{2} k x^{2}$ . Calculus aur geometry agree karte hain — jaise zaroori hai.

PICTURE. Saare strips bhare hue, solid laal triangle mein merge hote hue, base $x$ , height $k x$ , area $\frac{1}{2} k x^{2}$ label ke saath.

Step 7 — Kaam stored energy ban jaata hai

KYA HAI. Spring force conservative hai: jo kaam tumne uske against kharch kiya woh heat mein nahi jaata, woh spring ke andar bank ho jaata hai, wapas dene ke liye taiyaar. To stored potential energy us kaam ke barabar hai: $U = \frac{1}{2} k x^{2}$

KYUN conservative matter karta hai. Ek conservative force ke liye, work-in = energy-stored, perfect exchange rate ke saath. Spring release karo aur woh jo bhi push karega usse exactly $\frac{1}{2} k x^{2}$ kaam wapas karta hai — launch example aur Simple Harmonic Motion ki buniyaad yahi hai. Yeh Conservation of mechanical energy ke neeche bhi hai.

PICTURE. Energy bowl: $U = \frac{1}{2} k x^{2}$ ko $x$ ke against plot karne par ek parabola (ek ghati) milti hai. Sabse nichla point $x = 0$ hai (relaxed, zero stored energy); kisi bhi wall par chadhne mein energy lagti hai.

Step 8 — Degenerate aur edge cases

Reader ko kisi undikhaaye scenario mein mat chhodna. Teen cases:

$x = 0$ (relaxed). $U = \frac{1}{2} k (0)^{2} = 0$ . Koi stretch nahi, koi stored energy nahi. Bowl ka bottom.
Compression ( $x < 0$ ). Maano $x = - 0.03 m$ . Tab $x^{2} = (+ 0.0009)$ — minus square hone par gayab ho jaata hai. To $U = \frac{1}{2} k (0.03)^{2}$ , bilkul waisa hi jaise $0.03 m$ stretch karna. Compression positive energy store karta hai, matching stretch ke barabar.
Symmetry. Kyunki $U$ , $x^{2}$ par depend karta hai, bowl ek perfect mirror hai: left wall (compress) right wall (stretch) se match karti hai. Dono walls chadhti hain; koi bhi zero se neeche nahi jaati.

PICTURE. Parabola phir se, do marked points ke saath, $x = + a$ aur $x = - a$ , same height $U$ par baithe hue — visually prove karte hue ki stretch aur compression equal energy store karte hain.

Ek-picture summary

Sab ek saath: slanted force line $F = k x$ , uska triangular area = kaam = number $\frac{1}{2} k x^{2}$ , aur resulting energy parabola jiska valley floor $x = 0$ hai. Force slope-line hai; energy uske neeche ka area hai.

Recall Feynman retelling — poora walkthrough plain words mein

Ek spring aram mein lazy hoti hai — use thoda sa dhako aur woh barely wapas dhakti hai. Lekin har extra millimetre jo tum kheenchte ho, woh thodi aur zyada fight karta hai, straight-line-fair tarike se: dooni stretch, dooni fight. To "fight versus stretch ka graph" origin se chadhti ek slanted line hai. Jo total effort tum kharch karte ho woh us line ke neeche ki jagah hai — aur zero se shuru hoti slanted line ke neeche ki jagah ek triangle hai, ek box ka aadha, poora box nahi. Wahin se yeh chhupaaya hua "half" aata hai. Stretch ko baal-jitni thin steps mein kato, har ek par easy $force \times step$ karo (force ek baal ke across change nahi ho sakta), aur sab ko jodo — yahi adding-up hai jo integral sign ka matlab hai. Yeh total hota hai half-times-stiffness-times-stretch-squared. Kyunki spring release hone par har cheez wapas karti hai (kuch heat mein nahi jaata), woh number hi woh energy hai jo ab woh hoard kar rahi hai. Squeeze karo stretch ki jagah? Stretch number negative ho jaata hai, lekin tum ise square karte ho, to minus gayab ho jaata hai — squeezing bilkul wohi positive energy store karta hai. Ise sab ek ghati ki tarah picture karo: bottom mein baithna free hai; kisi bhi wall par chadhna, stretch-side ya squeeze-side, tumhe exactly $\frac{1}{2} k x^{2}$ kharch karata hai.

Active Recall

Spring ke liye

W = F \times x

use kyun nahi kar sakte?

Force constant nahi hai — woh ek line par

0

k x

tak badhta hai, isliye kaam line ke neeche ka triangle hai,

\frac{1}{2} k x^{2}

, rectangle

k x \cdot x

nahi.

d W = k x^{'} d x^{'}

kya hai?

Position

x^{'}

par ek thin slice par kiya gaya tiny kaam: force

k x^{'}

times slice-width

d x^{'}

= force line ke neeche ek thin rectangle ka area.

Integral

\int_{0}^{x} k x^{'} d x^{'}

triangle area ke barabar kyun hai?

0

x

tak saare strip-areas ka sum poora triangle reconstruct karta hai:

\frac{1}{2} \times base x \times height k x = \frac{1}{2} k x^{2}

Same amount stretch aur compression ke liye

U

same kyun hai?

U = \frac{1}{2} k x^{2}

x^{2}

par depend karta hai;

(- a)^{2} = a^{2}

, to sign drop ho jaata hai aur dono same positive energy store karte hain.

Yahan work-in = energy-stored kyun hai?

Spring force conservative hai — heat mein koi energy nahi jaati, isliye uske against kiya gaya kaam perfectly

U

ke roop mein bank ho jaata hai.

Connections

Hooke's Law — linear force $F = - k x$ jise hum integrate karte hain (Steps 1–2).
Work done by a variable force — slice-and-sum method (Steps 4–6).
Conservative forces and potential energy — kyun work-in = stored $U$ (Step 7).
Conservation of mechanical energy — $U$ ko wapas motion mein convert karta hai.
Simple Harmonic Motion — parabolic bowl oscillation drive karta hai.
Gravitational potential energy — contrast: constant force → rectangle → $U = m g h$ , linear not quadratic.