Spring potential energy — derivation — Hinglish notes

1.3.12 · Physics › Work, Energy & Power

WHY a single $F \times d$ fails

HOW to derive it — from scratch

Hum woh kaam compute karte hain jo tum spring ko dheere-dheere $0$ se $x$ tak stretch karne mein karte ho. Tumhari applied force spring ko balance karni chahiye, isliye $F_{applied} = + k x$ .

Step 1 — motion ko chhote-chhote tukdon mein kaato. Ek tiny displacement $d x^{'}$ par, position barely change hoti hai, isliye force $k x^{'}$ effectively constant hoti hai.

Yeh step kyun? Work-as-force-times-distance tab hi kaam karta hai jab force constant ho. Slice ko infinitesimal banana yeh sach kar deta hai.

Step 2 — ek slice par kiya gaya kaam. $d W = F_{applied} d x^{'} = k x^{'} d x^{'}$

Yeh step kyun? Yeh sirf $W = F d$ hai jo ek aise tukde par apply kiya gaya hai jo itna chhota hai ki $F$ us par change nahi hoti.

Step 3 — saare slices add karo (integrate karo) $0$ se $x$ tak. $W = \int_{0}^{x} k x^{'} d x^{'} = k \int_{0}^{x} x^{'} d x^{'} = k [\frac{x ^{'2}}{2}]_{0}^{x} = \frac{1}{2} k x^{2}$

Yeh step kyun? Total kaam har infinitesimal $d W$ ka sum hai — woh sum exactly ek integral hai.

Step 4 — kiye gaye kaam ko stored energy ke barabar rako. Kyunki tumne yeh kaam ek conservative force ke against kiya aur spring ab ise hold kar rahi hai, $U = \frac{1}{2} k x^{2}$

Yeh step kyun? Ek conservative force ke liye, us ke against kiya gaya kaam = potential energy gain. Ek ideal spring mein koi energy heat mein nahi jaati.

Figure — Spring potential energy — derivation

Steel-man your mistakes

Recall Feynman: ek 12-saal ke bacche ko samjhao

Socho ek rubber band kheench rahe ho. Pehle aasaan lagta hai, lekin jitna zyada kheencho, utna zyada woh wapas kheenchta hai. Toh tumhara "kheenchne ka kaam" shuru mein chhota hota hai aur end mein bada. Total kaam nikaalने ke liye, tum poore raaste ke liye sabse mushkil pull use nahi kar sakte — yeh cheating hogi. Balki shuru (zero) aur end ke beech average lo. Stiffness times stretch times stretch ka aadha — bas itni hi energy rubber band ab hold kar raha hai, kuch fling karne ke liye wapas snap karne ki intezaar mein.

Active Recall

Spring force constant kyun nahi hoti?

Hooke's law

F = - k x

ke anusaar, force equilibrium se displacement ke saath linearly badhti hai.

Spring potential energy formula kya hai?

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

F = - k x

U

derive karo.

W = \int_{0}^{x} k x^{'} d x^{'} = \frac{1}{2} k x^{2}

; yeh conservative force ke liye stored PE ke barabar hai.

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

mein ½ ka factor kyun hai?

Average force

\frac{k x}{2}

hai (linearly 0 se

k x

tak); equivalently

F

–

x

ke neeche triangle ka area.

Compression positive ya negative energy store karta hai?

Positive —

x

squared hai, isliye

U \geq 0

stretch aur compression dono ke liye.

Agar stretch double ho, stored energy kaise change hoti hai?

Woh quadruple ho jaati hai, kyunki

U \propto x^{2}

x

kahan se measure kiya jaata hai?

Spring ki natural (relaxed/equilibrium) length se.

Force-displacement graph ke neeche ka area kya represent karta hai?

Kiya gaya kaam = stored energy.

Connections

Hooke's Law — woh force law jise hum integrate karte hain.
Work done by a variable force — general $W = \int F d x$ method.
Conservative forces and potential energy — isliye force ke against kaam = stored $U$ .
Conservation of mechanical energy — launching example mein use kiya gaya.
Simple Harmonic Motion — $U = \frac{1}{2} k x^{2}$ oscillation drive karta hai.
Gravitational potential energy — contrast: wahan $U = m g h$ (constant force, height mein linear).

Spring potential energy — derivation

WHAT we are finding