1.3.11 · D3Hooke's law — spring force F = −kx
Intuition Yeh page kis liye haiParent note ne tumhe law F = − k x U = 2 1 k x 2 problem chahe jo bhi scenario laaye — stretched, compressed, hanging, oscillating, series, parallel, zero-displacement — tumne woh pattern pehle se dekha hua hai. Hum pehle ek scenario matrix banate hain, phir har cell ke liye ek example solve karte hain.
Un do tools ki reminders jinpe hum rely karte hain (dono fully built hain parent note mein ):
Kuch bhi solve karne se pehle, chaliye har tarah ke situations list karte hain jo ek Hooke's-law problem mein aa sakti hain. Har row ek "cell" hai; neeche ke worked examples ko us cell ke saath tag kiya gaya hai jo woh fill karte hain.
Cell
Ise kya alag banata hai
Example
A. Stretch (x>0) positive displacement, force wapas kheenchti hai
Ex 1
B. Compression (x<0) negative displacement, force bahar dhakkelti hai — sign trap
Ex 2
C. Zero displacement (x=0) degenerate case: force aur energy dono zero
Ex 3
D. Energy vs force scaling x Ex 4
E. Vertical hanging mass gravity spring ko balance karti hai, k
Ex 5
F. Energy → speed (motion) spring PE kinetic energy mein convert hoti hai
Ex 6
G. Elastic limit exceeded limiting/degenerate: law break ho jaati hai
Ex 7
H. Combined springs (twist) series & parallel effective k
Ex 8
I. Work over an interval do non-zero positions ke beech work
Ex 9
Hum x
Worked example Stretched springk = 150 N/m x = 0.08 m
Forecast: Compute karne se pehle guess karo — kya force positive hai ya negative? 10 N se badi ya choti?
Step 1 — Force. F = − k x F = − ( 150 ) ( 0.08 ) = − 12 N Yeh step kyun? Direct substitution; minus sign law mein hi built-in hai.
Step 2 — Sign padho. x > 0 F < 0 12 N wapas home ki taraf kheenchti hai.
Yeh step kyun? Number direction-encoded hai; hum sign ko ek physical arrow mein translate karte hain.
Step 3 — Energy. U = 2 1 k x 2 U = 2 1 ( 150 ) ( 0.08 ) 2 = 2 1 ( 150 ) ( 0.0064 ) = 0.48 J Yeh step kyun? Energy F x area hai, F ⋅ x
Verify: Units: ( N/m ) ( m ) = N ( N/m ) ( m 2 ) = N⋅m = J F ⋅ x = 0.96 aadha hai — correct, kyunki force 0 se badhi thi. ✓
Worked example Compressed springWohi spring (k = 150 N/m 4 cm compress ki gayi hai, yaani x = − 0.04 m
Forecast: Kya force negative hogi (jaise Ex 1 mein) ya positive? Kya energy negative hogi?
Step 1 — Force. Negative x F = − k x = − ( 150 ) ( − 0.04 ) = + 6 N Yeh step kyun? Do minus signs multiply hokar plus bante hain. Yahi Cell B ka poora point hai.
Step 2 — Interpret karo. x < 0 F > 0 baahaar ki taraf, wapas home ki taraf dhakkelti hai — Ex 1 se opposite direction, bilkul jaisa "restoring" demand karta hai.
Yeh step kyun? Minus sign ka matlab "force hamesha negative hai" NAHI hai; iska matlab hai "force displacement ka oppose karta hai."
Step 3 — Energy. Square sign ko khatam kar deta hai:
U = 2 1 ( 150 ) ( − 0.04 ) 2 = 2 1 ( 150 ) ( 0.0016 ) = 0.12 J Yeh step kyun? x 2 ≥ 0
Neeche wali picture applied force k x Red vertical lines ko follow karo jo x = + 0.04 x = − 0.04 2 1 k x 2 dono triangles ka area same (0.12 J) hai — same distance se stretch aur compress karne mein bilkul same energy lagti hai, chahe forces opposite directions mein point karein.
Verify: Force sign Ex 1 se flip hua (jaisa opposite displacement ke liye hona chahiye) ✓. Energy positive ✓. Figure mein, left aur right red triangles equal area ke mirror images hain ✓.
Worked example Spring apni natural length parEk spring apni natural length par baithi hai: x = 0 F U
Forecast: Dono zero? Ek zero? Trick question?
Step 1 — Force. F = − k ( 0 ) = 0 Yeh step kyun? Natural length par restore karne ke liye kuch nahi hai — equilibrium ka matlab zero net spring force hai.
Step 2 — Energy. U = 2 1 k ( 0 ) 2 = 0 Yeh step kyun? Displacement nahi matlab koi stored work nahi.
Step 3 — Yeh cell kyun matter karta hai. Ex 2 ki figure dekho: straight force line bilkul origin se guzarti hai. x = 0 x = 0 Simple Harmonic Motion mein.
Verify: Dono zero chahe k F x
Worked example Stretch ko triple karox U 1 F 1 3 x
Forecast: Dono ×3? Force ×3 aur energy ×9? Ya dono ×9?
Step 1 — Force scaling. F ∝ x F 1 F 2 = x 3 x = 3. Yeh step kyun? Force x linear hai — graph ek straight line hai.
Step 2 — Energy scaling. U ∝ x 2 U 1 U 2 = x 2 ( 3 x ) 2 = 9. Yeh step kyun? Energy ek triangle ka area hai; base aur height ko triple karne se 3 × 3 = 9
Verify: k = 150 x = 0.02 U 1 = 2 1 ( 150 ) ( 0.0004 ) = 0.03 x = 0.06 U 2 = 2 1 ( 150 ) ( 0.0036 ) = 0.27 0.27/0.03 = 9
Worked example Vertical spring par massm = 2 kg x = 0.16 m g = 9.8 m/s 2 k
Forecast: Lagbhag 100 N/m? Zyada? Kam?
Step 1 — Forces balance karo. Rest par net force zero hai, isliye upar ki taraf spring force neeche ki taraf weight ke barabar hai:
k x = m g . Yeh step kyun? "Rest par" ka matlab acceleration zero hai, isliye forces cancel hote hain (Newton's first law).
Step 2 — k
k = x m g = 0.16 ( 2 ) ( 9.8 ) = 0.16 19.6 = 122.5 N/m . Yeh step kyun? Balance equation rearrange karne se stiffness isolate hoti hai.
Verify: Units ( kg ⋅ m/s 2 ) / m = N/m k x = 122.5 × 0.16 = 19.6 = m g
Worked example Spring ek block launch karti haik = 800 N/m x = 0.05 m m = 0.2 kg
Forecast: Kuch m/s? Tens of m/s?
Step 1 — Stored energy. Compressed spring hold karti hai
U = 2 1 k x 2 = 2 1 ( 800 ) ( 0.05 ) 2 = 2 1 ( 800 ) ( 0.0025 ) = 1.0 J . Yeh step kyun? Yahi energy reservoir (Elastic potential energy ) hai jo work karne ke liye available hai.
Step 2 — Kinetic energy mein convert karo. Frictionless hai, isliye saari PE, KE ban jaati hai (Conservation of mechanical energy ):
2 1 m v 2 = U . Yeh step kyun? Koi energy leak nahi hoti, isliye lost PE = gained KE.
Step 3 — v
v = m 2 U = 0.2 2 ( 1.0 ) = 10 ≈ 3.16 m/s . Yeh step kyun? KE formula rearrange karne se speed isolate hoti hai.
Verify: Units J / kg = m 2 / s 2 = m/s 2 1 ( 0.2 ) ( 3.16 ) 2 ≈ 1.0 = U
Worked example Jab Hooke's law fail hota haik = 500 N/m F = − k x ∣ x ∣ elastic limit x m a x = 0.20 m x = 0.35 m F = − k x U = 2 1 k x 2
Forecast: Haan, bas plug in karo? Ya nahi?
Step 1 — Limit check karo. x = 0.35 m > x m a x = 0.20 m Yeh step kyun? Is page par har formula assume karta hai ki spring abhi bhi elastic hai — ek straight F x
Step 2 — Conclusion. x m a x permanently deform ho jaati hai; F x F = − k x U = 2 1 k x 2 apply nahi hote . Tum x = 0.35 Yeh step kyun? Yeh model ki boundary hai. Ise recognize karna hi "answer" hai.
Step 3 — Kya abhi bhi sach hai. x m a x = 0.20 m tak formulas hold karte hain. Is spring ki maximum energy jo woh elastic rehte hue store kar sakti hai woh hai
U lim = 2 1 k x m a x 2 = 2 1 ( 500 ) ( 0.20 ) 2 = 2 1 ( 500 ) ( 0.04 ) = 10 J . Yeh step kyun? Hum valid model ko uski edge tak le jaate hain, usse aage nahi, aur ek concrete number paate hain.
Verify: 0.35 > 0.20 U lim = 10
Worked example Do springs, series aur parallelDo identical springs, har ek ka k = 100 N/m series mein (end to end) join kiye hain aur (b) jab parallel mein (side by side) rakhe hain.
Forecast: Kya series ek spring se zyada stiff hogi ya floppy? Parallel?
Step 1 — Series. Series mein same force dono ko stretch karti hai, isliye displacements add hote hain:
k series 1 = k 1 + k 1 = 100 2 ⇒ k series = 50 N/m . Yeh step kyun? Softer chain: ek lambi spring har newton mein zyada stretch hoti hai, isliye effective k girti hai.
Step 2 — Parallel. Parallel mein dono same displacement share karte hain, isliye forces add hoti hain:
k parallel = k + k = 200 N/m . Yeh step kyun? Do springs ek saath kheenchti hain toh zyada stiffly resist karti hain, isliye k badhti hai.
Verify: Series 50 < 100 < 200 1/50 = 1/100 + 1/100
Worked example Ek point se doosre point tak stretch karnak = 400 N/m x 1 = 0.02 m x 2 = 0.05 m
Forecast: Kya yeh 2 1 k ( x 2 − x 1 ) 2
Step 1 — Work ek energy difference hai. Tumhara kiya gaya work stored energy ke change ke barabar hai (Work done by a variable force ):
W = U ( x 2 ) − U ( x 1 ) = 2 1 k x 2 2 − 2 1 k x 1 2 . Yeh step kyun? Force constant nahi hai, isliye hum force × distance use nahi kar sakte; balki hum triangular areas ka difference lete hain.
Step 2 — Compute karo.
W = 2 1 ( 400 ) [ ( 0.05 ) 2 − ( 0.02 ) 2 ] = 200 ( 0.0025 − 0.0004 ) = 200 ( 0.0021 ) = 0.42 J . Yeh step kyun? Dono endpoints ko energy formula mein plug karo aur subtract karo.
Step 3 — Trap check. Naive 2 1 k ( x 2 − x 1 ) 2 = 200 ( 0.03 ) 2 = 0.18 galat hai — spring pehle se partly stretched hai, isliye force k x 1 Yeh step kyun? Areas subtract karo, interval ko zero se fresh stretch treat mat karo.
Verify: U ( 0.05 ) = 200 ( 0.0025 ) = 0.5 U ( 0.02 ) = 200 ( 0.0004 ) = 0.08 = 0.42
Recall Scenario checklist — kya tum har ek ko ek cell mein rakh sakte ho?Ek spring ek puck ko push karti hai: kaun sa cell? ::: Cell F (energy → speed)
Ek spring 3 cm compress ki gayi: kaun sa cell, aur kya F F > 0 k k x = m g x 1 x 2 2 1 k ( x 2 2 − x 1 2 ) 2 1 k ( x 2 − x 1 ) 2
Mnemonic Yaad rakhne wali ek lineSign − k x 2 1 k x 2