Hooke's law — spring force F = −kx: Question bank

1.3.11 · D5 · Physics › Work, Energy & Power › Hooke's law — spring force F = −kx

Yeh page Hooke's law ke liye ek concept gym hai. Koi bhaari arithmetic nahi — sirf woh traps jo unhe pakad lete hain jo "formula jaante hain" lekin physics ko feel nahi kiya. Question padho, answer chhupao, apni reasoning zabar-dast bol ke kaho, phir reveal karo.

Shuru karne se pehle, neeche di gayi shared picture dekho — is page ke har question mein isi ki baat ho rahi hai.

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

Ek horizontal spring table par rakhi hai: ek end wall se pini hua hai, doosre end par ek block hai. Block ki rest position $x = 0$ hai — yeh natural length hai, jahan spring na stretch hui hai na squeeze. Stretching block ko right kheenchti hai, isliye hum ise $x > 0$ kehte hain. Compressing use left dhakelta hai, isliye woh $x < 0$ hai. Picture mein dekho, spring ka force arrow hamesha wapas $x = 0$ ki taraf point karta hai.

Ab, doosri key picture — force–displacement graph — jo neeche ke har "line vs area" trap ko kuch aisa bana deti hai jo tum dekh sako.

True or false — justify karo

Har answer mein ek reason dena hai, sirf "T" ya "F" nahi.

x = 0

par spring zero force lagata hai.

True —

F = - k (0) = 0

; natural length par restore karne ke liye kuch bhi nahi hota, isliye spring "khush" hai aur kuch bhi push/pull nahi karta.

Ek compressed spring negative force lagata hai.

False — compression ka matlab

x < 0

hai, isliye

F = - k x

positive hota hai (right ki taraf point karta hai, wapas

x = 0

ki taraf). Minus sign force ka sign flip karta hai, yeh use permanently negative nahi banata.

Stretch ko double karne se force aur stored energy dono double ho jaate hain.

False — force double hota hai (

F \propto x

) lekin energy chaar guna ho jaati hai (

U \propto x^{2}

). Yeh split (linear force, quadratic energy) spring ke baare mein sabse zyaada miss ki jaane waali fact hai.

Ek stiffer spring usi stretch par zyaada energy store karti hai.

True —

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

k

ke saath badhta hai, isliye equal

x

par bada-

k

spring zyaada energy hold karta hai; uski

F

-vs-

x

line steeper hoti hai, jo area ka ek uncha triangle deti hai (doosri figure).

Agar ek spring ka force stretch hone par

10 N

hai aur utni hi matra se compress hone par bhi

10 N

hai, toh dono forces same hain.

False — magnitudes match karte hain lekin directions opposite hain (stretch wapas left kheenchta hai, compression wapas right dhakelta hai). "Same" ka matlab same vector hona chahiye, aur yeh nahi hain.

Spring constant

k

jitna zyaada stretch karo utna bada hota jaata hai.

False —

k

line ka fixed slope hai, jo coil ki ek property hai (material, thickness, turns). Jab tum stretch karte ho toh sirf

F

badalta hai,

k

nahi — jab tak tum elastic limit ke andar ho.

Spring stretch karne mein jo kaam tum karte ho woh spring ke stored energy ke barabar hota hai.

True (slow, no-friction stretch ke liye) — spring force ke against jo bhi joule tum lagaate ho woh elastic PE

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

ke roop mein deposit hoti hai; isliye hi Elastic potential energy tumhare kiye gaye work ke barabar hoti hai.

Hooke's law sirf metal coils ki khaas property hai.

False — kisi bhi stable energy minimum ko Taylor-expand karne se leading

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

term milta hai, isliye har stable system (bonds, strings, chhoti swings ke liye pendulums) rest ke paas

F = - k x

jaisa dikhta hai.

Galti pakado

Har item ek believable-sounding galat statement hai. Kharabi dhoondo.

"Energy stored

= F \cdot x = k x \cdot x = k x^{2}

Force constant nahi hai — yeh

0

se shuru hokar stretch ke saath

k x

tak badhti hai. Tumhe average force

\frac{1}{2} k x

use karni padegi, jo

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

deti hai; geometrically yeh shaded triangle area hai

F

-vs-

x

line ke neeche, poora rectangle nahi (doosri figure).

F = - k x

humein batata hai spring force hamesha negative hai, isliye hamesha left point karta hai."

Galat —

F

ka sign

x

ke sign ko track karta hai. Jab compress hota hai (

x < 0

) toh force positive hoti hai aur right point karti hai. Minus sign encode karta hai "displacement ke opposite," kabhi "hamesha negative" nahi.

"Kyunki

F = - k x

hai, bade

x

par spring ko eventually infinitely strong pull karni chahiye."

Formula linear hai, lekin real springs ise sirf elastic limit tak follow karti hain. Us se aage coil permanently deform ho jaata hai aur

F

x

ke proportional nahi rehta — law simply apply nahi hota.

"Natural length par spring ki maximum potential energy hoti hai kyunki woh relaxed hai."

Ulta hai — natural length (

x = 0

) energy ka minimum hai,

U = 0

. Isliye hi wahan force zero hai:

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

aur

U

ke slope apne bottom par zero hai.

"Spring force find karne ke liye integrate karo:

F = \int k x d x

Force aur energy mein confusion ho gayi. Force directly

F = - k x

se padhte hain; tum force ko distance par integrate karte ho work/energy paane ke liye (

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

), aur force wapas paane ke liye energy ko differentiate karte ho (

F = - d U / d x

). Yeh dono operations inverses hain. Dekho Work done by a variable force.

"Vertical spring par rest mein latkay huye mass ke liye, spring force zero hai."

Nahi — rest mein spring force ko gravity balance karna padta hai, isliye

k x = m g \neq = 0

. Net force zero hota hai, lekin spring khud stretch hui hai aur weight ke barabar force se upar kheench rahi hai.

Why questions

Reason explain karo, sirf fact nahi.

F = - k x

mein minus sign kyun hai?

Kyunki force hamesha equilibrium ki taraf wapas point karta hai: jab bhi

x

positive hota hai force negative hoti hai aur vice versa, aur sirf minus sign hi

F

aur

x

ko automatically opposite signs deta hai.

Spring energy

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

kyun hai,

k x^{2}

kyun nahi?

Kyunki force

0

se shuru hoti hai aur

k x

tak badhti hai, isliye stretch ke dauran average force sirf

\frac{1}{2} k x

hoti hai; work

=

average force

\times

distance

= \frac{1}{2} k x \cdot x

. Geometrically yeh shaded triangle hai, rectangle nahi (doosri figure).

Har stable system chhote displacements ke liye spring ki tarah kyun behave karta hai?

Ek stable minimum ke paas energy ka first derivative zero ho jaata hai aur pehla surviving Taylor term

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

hota hai; differentiate karne par

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

milta hai. Isliye Interatomic forces chhote springs ki tarah act karte hain.

F = - k x

oscillation kyun produce karta hai, ek single push kyun nahi?

Ghar ki taraf pointing karne wala restoring force mass ko

x = 0

se aage accelerate karta hai, lekin mass inertia se overshoots kar jaata hai, doosri side se wapas kheencha jaata hai, aur repeat karta hai — woh endless "hamesha ghar ki taraf" hi Simple Harmonic Motion hai.

Hum spring force ko conservative kyun keh sakte hain?

Kyunki yeh ek potential se aata hai,

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

with

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

; jo work yeh karta hai woh sirf start aur end positions par depend karta hai, isliye energy KE aur PE ke beech cleanly swap hoti hai — dekho Conservative forces and potential energy.

F

-vs-

x

graph ka straight line hona itna zaroori kyun hai?

Origin se guzarne wali straight line ka matlab hai force displacement ke strictly proportional hai; uska slope constant

k

hai, aur us ke neeche ka saaf triangle hi energy ko exactly

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

banata hai (doosri figure).

Edge cases

Boundaries, zeros, aur limits — jahan sloppy intuition toot jaati hai.

Exactly

x = 0

par force aur energy kya hai?

Dono zero hain:

F = - k (0) = 0

aur

U = \frac{1}{2} k (0)^{2} = 0

. Natural length ek saath force-free point bhi hai aur energy floor bhi.

Kya stored energy kabhi negative hoti hai?

Nahi —

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

mein

x^{2} \geq 0

hota hai, isliye energy rest par zero aur kisi bhi displacement ke liye positive hoti hai, chahe stretch ho ya compress.

Utni hi doori stretch karne aur compress karne se same energy store hoti hai?

Haan — kyunki

U

x^{2}

par depend karta hai,

x

ka sign erase ho jaata hai;

x = + d

aur

x = - d

dono identical

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

store karte hain.

k \to 0

limit mein (fixed displacement

x

par) force ka kya hota hai?

Spring infinitely floppy ho jaati hai:

F = - k x \to 0

aur

U = \frac{1}{2} k x^{2} \to 0

, isliye kisi bhi fixed stretch par yeh koi resistance nahi deti aur koi energy store nahi karti — effectively spring hi nahi hai.

Agar

k \to \infty

ho aur displacement $x$ fixed rakha jaaye toh kya hoga?

Force aur stored energy dono blow up ho jaate hain (

F = - k x \to \infty

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

); same

x

tak pahunchne ke liye zyaada se zyaada work chahiye, isliye ek truly rigid rod bilkul hilne se inkaar karta hai.

Agar

k \to \infty

ho aur uski jagah force $F$ fixed rakha jaaye toh kya hoga?

Displacement collapse ho jaata hai:

x = - F / k \to 0

, aur stored energy

U = \frac{1}{2} \frac{F ^{2}}{k} \to 0

. Toh ek fixed load ke under ek stiffer spring kam move karta hai aur kam energy bank karta hai — yeh fixed-

x

case se bilkul ulta conclusion hai, isliye "kya fixed hai?" waala sawaal zaroor poochha jaana chahiye.

Oscillation ke dauran jab mass

x = 0

se guzarta hai us pal, kya uski energy zero hoti hai?

Nahi — wahan spring PE zero hoti hai, lekin saari energy kinetic ban gayi hai; total mechanical energy poori tarah conserved rehti hai, per Conservation of mechanical energy.

Agar elastic limit se aage stretch karo toh kya hoga?

Hooke's law fail ho jaata hai — material permanently deform ho jaata hai,

F

x

ke proportional nahi rehta, aur

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

ab (ab partially lost) energy describe nahi karta.

Recall Ek-line summary jo saath le jaao

Force line par rehti hai ( $F = - k x$ , $x$ ke saath sign badlta hai), energy area mein rehti hai ( $\frac{1}{2} k x^{2}$ , hamesha positive, square ke saath badhti hai), aur $k$ woh slope hai jo tab tak nahi badlta jab tak spring toot na jaaye.

Connections

Parent: Hooke's law $F=-kx$
Work done by a variable force · Elastic potential energy · Conservative forces and potential energy · Simple Harmonic Motion · Conservation of mechanical energy · Interatomic forces