Spring-mass system — horizontal, vertical: Visual walkthrough

1.6.8 · D2 · Physics › Oscillations & Waves › Spring-mass system — horizontal, vertical

Hum banayenge, order mein:

"Displacement" ka matlab kya hota hai, aur zero kahan hota hai.
Spring wapas kyun push karta hai (restoring force).
Woh force acceleration ke baare mein ek rule kaise ban jaata hai (Newton).
Woh rule ek wiggle ko kyun force karta hai (motion ki shape).
$ω$ number rule ke andar kahan chupta hai.
$ω$ ko period $T$ mein convert karna.
Spring ko latkaana: kya gravity sab kuch tod deti hai? (Nahi — ek aur picture dikhati hai kyun.)
Edge cases: zero displacement, bhaari mass, stiff spring.

Step 1 — Stage set karo aur "zero" ka naam lo

KYA. Mass $m$ ka ek block frictionless table par baitha hai, spring se bandha hua. Jab kuch push nahi karta, spring apni natural length par hoti hai — na stretched, na squashed. Hum wahan ek ruler lagaate hain aur us point ko $x = 0$ kehte hain.

KYUN. Har measurement ko ek home base chahiye. Hum measure karte hain ki block home se kitna door hai aur us number ko $x$ (metres mein) kehte hain. Positive $x$ ka matlab "right taraf push kiya," negative $x$ ka matlab "left taraf push kiya." Zero fix kiye bina, "displacement" ek aisa word hai jiska koi number nahi.

PICTURE. Lavender block mint line par baitha hai. Ruler ka zero (coral tick) exactly wahin hai jahan relaxed spring khatam hoti hai.

Step 2 — Spring hamesha tumhe ghar ki taraf point karti hai ( $F = - k x$ )

KYA. Block ko right taraf kheencho ( $x > 0$ ): spring stretch hoti hai aur use left taraf kheenchti hai. Use left taraf push karo ( $x < 0$ ): spring squash hoti hai aur use right taraf dhakelta hai. Dono taraf se force $x = 0$ ki taraf point karti hai. Yeh hai Hooke's Law: $F = opposite way - stiffness k how far x$

KYUN. Equation ko ek sentence ki tarah padho. $k$ (newtons per metre) kehta hai "stretch ke har metre par kitne newtons ka push" — stiff spring ka $k$ bada hota hai. $x$ kehta hai force utni hi zyada badhti hai jitna door jaao. Minus sign poori personality hai: force aur displacement hamesha opposite direction mein point karte hain, isliye block hamesha zero ki taraf wapas aata rehta hai.

PICTURE. Do snapshots. Right par, red spring force arrow left taraf point karti hai (ghar wapas). Left par, red arrow right taraf point karti hai (ghar wapas). Force arrow kabhi bhi zero se door nahi point karta.

Step 3 — Force ko acceleration ke baare mein ek rule mein badlo (Newton)

KYA. Newton ka second law kehta hai force barabar mass times acceleration hoti hai. Acceleration velocity kitni tezi se change hoti hai; hum ise $\overset{x}{¨}$ likhte hain (do dots ka matlab hai "rate of change, do baar" — velocity pehla change hai, acceleration doosra). $inertia m acceleration \overset{x}{¨} = spring force - k x$

KYUN. Hamare paas ek force hai (Step 2) lekin hum jaanna chahte hain ki block kaise move karta hai. Newton ka law "kaun si force act karti hai" se "motion kaise change hota hai" tak ka bridge hai. Double-dot notation kyun? Kyunki spring force position $x$ par depend karti hai, lekin Newton acceleration ko control karta hai — wohi $x$ dono tarafon par aata hai, aur yeh self-reference exactly wahi hai jo oscillation banata hai.

PICTURE. Left box: spring force $- k x$ (coral). Arrow Newton's law ke through. Right box: acceleration $\overset{x}{¨}$ (lavender), force ke same direction mein lekin $1/ m$ se scaled.

$m$ se divide karo aur theek karo: $\overset{x}{¨} + \frac{k}{m} x = 0$

$\overset{x}{¨}$ — acceleration.
$\frac{k}{m}$ — ek single positive number jo stiffness aur inertia ko combine karta hai. Bada jab spring stiff ho, chhota jab mass bhaari ho. Is bundle ko yaad rakho — yeh Step 5 ka star hai.

Step 4 — Yeh rule ek wiggle ko kyun force karta hai

KYA. Rule $\overset{x}{¨} = - \frac{k}{m} x$ kehta hai: acceleration hamesha position ka flip hoti hai. Dum right par, position badi-positive hai, toh acceleration badi-negative hai (zor se left taraf kheenchi gayi). Block slow hota hai, rukta hai, reverse karta hai, middle se race karta hai (jahan force zero hai lekin speed sabse zyada hai), left taraf overshoot karta hai, wapas kheencha jaata hai — hamesha ke liye. Push-back plus inertia ka overshoot = endless oscillation.

KYUN. Humein dekhna hai ki yeh equation sirf smooth back-and-forth ke siwa kuch describe nahi kar sakta. Clue: woh function jiska doosri rate-of-change apna khud ka negative hai woh cosine hai. Toh hum ek cosine wave guess karte hain aur check karte hain ki yeh fit hota hai.

PICTURE. $x$ ka cosine curve time ke against. Crests par (ghar se door) chhote acceleration arrows steeply zero ki taraf point karte hain; centre crossings par arrows gayab ho jaate hain aur speed arrows sabse lambe hote hain.

Step 5 — $ω$ number kahan chupta hai

KYA. Guess $x (t) = A cos (ω t + ϕ)$ likho aur ise rule mein wapas daalo.

$A$ — amplitude (metres): sabse badi swing kitni door jaati hai.
$ω$ — angular frequency (radians/second): cosine kitni tezi se cycle karta hai.
$ϕ$ — phase: cycle mein hum kahan se start kiye ( $t = 0$ par kaun sa crest).

Cosine ko do baar differentiate karne par $ω$ ke do factors neeche aa jaate hain aur sign flip ho jaata hai: $\overset{x}{¨} = - ω^{2} A cos (ω t + ϕ) = - ω^{2} x$

KYUN. Hum ise apne physics rule $\overset{x}{¨} = - \frac{k}{m} x$ se compare karte hain. Ek hi $\overset{x}{¨}$ ke liye do expressions term-by-term match karne chahiye: $wiggle se ω^{2} = physics se \frac{k}{m}$

Toh $ω$ koi free choice nahi hai — spring aur mass ise dictate karte hain: $ω = \frac{k}{m}$

PICTURE. Bundle $k / m$ (equation se) cosine ki wiggle-rate $ω^{2}$ mein flow karta hua — dono sides ka labelled matching.

Step 6 — $ω$ se period $T$ tak

KYA. Ek full cosine cycle mein $2 π$ radians ka angle lagta hai. Kyunki $ω$ radians per second hai, ek round trip ka time hai $T = \frac{2 π}{ω} = 2 π \frac{m}{k}$

KYUN. $ω$ ka jawab hai "kitne radians per second"; period $T$ ka jawab hai "ek full cycle mein kitne seconds." Yeh reciprocals hain jo ek turn mein $2 π$ radians se scale hote hain. Isliye hum $2 π$ ko $ω$ se divide karte hain.

PICTURE. Ek full cosine cycle mark kiya hua; crest se next crest tak ka horizontal span $T$ label kiya hua hai, aur saath mein algebra $T = 2 π / ω = 2 π m / k$ .

Step 7 — Ise latkaao: kya gravity picture ko tod deti hai?

KYA. Ab spring latki hui hai aur gravity mass ko $m g$ force se neeche kheenchti hai. $y$ ko natural length se neeche ki taraf point karne do. Gravity pehle spring ko ek resting point $y_{0}$ tak stretch karti hai jahan pulls balance ho jaati hain: $k y_{0} = m g \Rightarrow y_{0} = \frac{m g}{k}$

$g$ — gravity ka pull per kilogram ( $\approx 9.8 m/s^{2}$ ).
$y_{0}$ — static extension: gravity akele spring ko kitna stretch karti hai.

General $y$ tak displace karo; net downward force hai spring-up minus... shifted coordinate $x = y - y_{0}$ (naye ghar se distance) ke saath carefully karte hain: $m \overset{x}{¨} = gravity down m g - spring up k (x + y_{0}) = = 0 by balance (m g - k y_{0}) - k x = - k x$

KYUN. Hum yeh isliye poochh rahe hain kyunki gravity lagta hai bounce ko slow karni chahiye. Algebra dikhata hai constant $m g$ poori tarah constant $k y_{0}$ se swallow ho jaata hai — yeh cancel ho jaate hain. Jo bachta hai woh wohi $- k x$ hai Step 2 se. Gravity home base ko $y_{0}$ neeche shift karta hai; yeh restoring term ko kabhi touch nahi karta. Toh $T = 2 π m / k$ unchanged rehta hai.

PICTURE. Left: horizontal spring jiska home natural length par hai. Right: hanging spring jiska home $y_{0}$ neeche drop hua hai (butter arrow), lekin naye home ke baare mein wohi coral restoring arrow. Cancellation $m g - k y_{0} = 0$ dikhaya gaya hai.

Step 8 — Edge cases (koi pathar mat chhodo)

KYA & KYUN & PICTURE, teenon limits ek saath:

Zero displacement ( $x = 0$ ). Force $= - k \cdot 0 = 0$ : koi push nahi. Agar rest par bhi hai, toh yeh ruk jaata hai — ek legal (boring) solution, amplitude $A = 0$ . Agar through move kar raha hai, toh yeh maximum speed ka point hai (Step 4 ka centre crossing). Kuch nahi toot ta.
Bhaari mass ( $m \to$ large). $T = 2 π m / k$ badhta hai: dheema aur dheema. Inertia spring ko resist karta hai. Bahut bade $m$ ki limit mein, bounce glacial ho jaata hai.
Stiffer spring ( $k \to$ large). $T$ ghatta hai: teza aur teza. Limit $k \to \infty$ mein spring rigid hai, $T \to 0$ — koi oscillation nahi, yeh bas hold kar leta hai.
Zero stiffness ( $k \to 0$ ) ya free mass. $T \to \infty$ : koi restoring force nahi, koi oscillation nahi. Block, ek baar push kiya, hamesha drift karta hai. Yeh woh boundary hai jahan SHM exist karna band ho jaata hai — picture straight-line drift mein degenerate ho jaati hai.

Ek-picture summary

Sab kuch ek canvas par: block aur uski restoring force → Newton ka rule → woh cosine jo yeh force karta hai → matched $ω$ → period → aur gravity ki harmless shift.

Recall Puri walkthrough ki Feynman retelling

Ek block ko smooth table par spring se bandh kar rakhoo aur mark karo jahan woh rest karta hai — use ghar kaho. Use door kheencho aur spring, polite lekin firm hote hue, hamesha use ghar ki taraf kheenchti hai, aur jitna zyada kheencho utna zyada kheenchti hai (yahi hai $F = - k x$ ). Newton kehta hai ek force block ko us direction mein speed up karti hai, lekin block mein motion ka weight (inertia) hai, toh woh ghar ko overshoot karta hai, doosri taraf stretch karta hai, wapas kheencha jaata hai — ek kabhi na khatam hone wala back-and-forth. Yeh back-and-forth exactly ek cosine wave hai, aur sirf ek cheez hai jo uski rhythm set karti hai woh hai tug-per-metre ( $k$ ) divided by mass ki laziness ( $m$ ): stiff aur light matlab tez, floppy aur heavy matlab dheema. Us rhythm ko seconds-per-bounce mein badlo aur pao $T = 2 π m / k$ . Aakhir mein puri cheez ko ceiling se latkaao: gravity home base ko thoda neeche kheenchti hai, lekin ek baar woh naya ghar mil jaye, spring bilkul pehle jaisi behave karti hai — bounce wohi rehta hai. Gravity ghar ko move karta hai; yeh heartbeat nahi badalta.

Recall Quick self-check

Kaun si quantity, badhne par, bounce ko slower banati hai? ::: Mass $m$ — yeh $T = 2 π m / k$ mein root ke neeche baitha hai. Block sabse tezi kahan move karta hai? ::: Equilibrium $x = 0$ par, jahan spring force zero hai aur sari energy kinetic hai. Gravity vertical period kyun nahi badlati? ::: Yeh constant hai, toh $m g - k y_{0} = 0$ ise cancel kar deta hai; sirf naye ghar ke baare mein restoring $- k x$ bachta hai. $k$ aur $m$ ke terms mein $ω$ kya hai? ::: $ω = k / m$ .

Connections

Spring-mass System — horizontal, vertical — parent result jo yeh page visually derive karta hai.
Simple Harmonic Motion — Step 4 ki cosine motion.
Hooke's Law — Step 2 ka $F = - k x$ .
Energy in SHM — Steps 4 aur 8 mein fast-at-centre idea.
Springs in Series and Parallel — $k$ badalta hai, isliye Step 8 ki limits.
Simple Pendulum — contrast: iska period do depend karta hai $g$ par.
Damped Oscillations — Step 4 ki endless wiggle friction ke saath kya ban jaati hai.