Spring-mass system — horizontal, vertical: Worked examples

1.6.8 · D3 · Physics › Oscillations & Waves › Spring-mass system — horizontal, vertical

Scenario matrix

Har spring-mass question in cells mein se ek hoti hai. Neeche har worked example us cell ke saath tagged hai jo wo fill karta hai.

#	Case class	Kya special hai isme
A	Horizontal, $T$ nikalo	Pure $T = 2 π m / k$ , gravity nahi
B	Vertical, sirf static stretch pata	$T = 2 π y_{0} / g$ use karo, $m$ aur $k$ unknown
C	Energy: amplitude ↔ max speed	$\frac{1}{2} k A^{2} = \frac{1}{2} m v_{ma x}^{2}$
D	Kisi general instant par position/velocity	Full $x (t) = A cos (ω t + ϕ)$
E	Springs in parallel (stiffer)	$k_{e f f} = k_{1} + k_{2}$
F	Springs in series (softer)	$1/ k_{e f f} = 1/ k_{1} + 1/ k_{2}$
G	Limiting / degenerate inputs	$k \to \infty$ , $k \to 0$ , $m \to 0$ , $A = 0$
H	Real-world word problem	Words ko symbols mein translate karo
I	Exam twist (ideas combine karo)	Vertical + energy + amplitude limit

Worked example Example 2 — Cell B: vertical, sirf static stretch pata

Tum ek unknown mass ko ek unknown spring par latakate ho aur wo natural length se $y_{0} = 2.5$ cm neeche settle hoti hai. Chote vertical bounces ka period nikalo.

Forecast: Na $m$ alag pata hai, na $k$ alag. Phir bhi $T$ nikal sakte hain? (Haan — parent note ka trick.)

Units convert karo. $y_{0} = 2.5 cm = 0.025$ m. Yeh step kyun? Puri SI mein, warna $g$ match nahi karega.
Stretch formula use karo. Kyunki $k y_{0} = m g$ hai, ratio $m / k = y_{0} / g$ hoga, toh $T = 2 π y_{0} / g$ . Yeh step kyun? Gravity dynamics se cancel ho jaati hai lekin measured stretch ke andar $m / k$ chhupa deti hai — yahi hum exploit karte hain.
Substitute karo ( $g = 9.8$ m/s²): $T = 2 π 0.025/9.8 = 2 π 2.551 \times 1 0^{- 3}$ .
Evaluate karo. $2.551 \times 1 0^{- 3} = 0.05051$ , toh $T = 2 π (0.05051) = 0.317$ s.

Verify: $2.5$ cm ka droop chhota hai → fast bounce → sub-second period ✓. Units: $m / (m/s^{2}) =$ s ✓.

Worked example Example 3 — Cell C: energy, amplitude ↔ max speed

Example 1 ke block ( $m = 2.0$ kg, $k = 50$ N/m) ko use karte hue, tum use $A = 0.15$ m kheeencho aur choddo. Maximum speed aur total energy nikalo.

Forecast: Speed sabse zyada kahan hogi — ends par ya middle mein? (Middle mein, jahan spring relaxed hoti hai.)

$ω$ nikalo. $ω = k / m = 50/2.0 = 25 = 5$ rad/s. Yeh step kyun? Max speed $A ω$ hai; pehle $ω$ chahiye.
Max speed. $v_{ma x} = A ω = 0.15 \times 5 = 0.75$ m/s. Yeh step kyun? $x = 0$ par saari energy kinetic hai (PE $= 0$ ), toh speed peak karti hai. Neeche figure dekho.
Total energy. $E = \frac{1}{2} k A^{2} = \frac{1}{2} (50) (0.15)^{2} = 0.5625$ J. Yeh step kyun? Turning point $x = A$ par, KE $= 0$ , toh saari energy spring PE $= \frac{1}{2} k A^{2}$ hai.

Verify: Centre par KE se cross-check: $\frac{1}{2} m v_{ma x}^{2} = \frac{1}{2} (2.0) (0.75)^{2} = 0.5625$ J $= E$ ✓. Figure mein jo trade dikhti hai us mein energy conserved hai.

Worked example Example 4 — Cell D: kisi chosen instant par position & velocity

Examples 1&3 ka block ( $ω = 5$ rad/s, $A = 0.15$ m) $t = 0$ par rest se $x = + A$ se release kiya jaata hai. $t = 0.20$ s par uski position aur velocity nikalo.

Forecast: $0.20$ s baad phase $ω t = 1.0$ rad $\approx 57°$ hai — quarter cycle se kam, toh wo abhi bhi positive hona chahiye lekin centre ki taraf ja raha hoga.

Sahi solution chuno. Far end se rest se release → $x (t) = A cos (ω t)$ (cosine apne max se shuru hoti hai zero slope ke saath). Yeh step kyun? "Rest se $x = + A$ par" ka matlab $x (0) = A$ , $\overset{x}{˙} (0) = 0$ hai; cosine dono ko match karti hai, toh phase $ϕ = 0$ .
Position. $x = 0.15 cos (5 \times 0.20) = 0.15 cos (1.0) = 0.15 (0.5403) = 0.0810$ m. Yeh step kyun? Direct evaluation; angle $1.0$ radians mein hai.
Velocity. $\overset{x}{˙} = - A ω sin (ω t) = - 0.15 \times 5 \times sin (1.0) = - 0.75 (0.8415) = - 0.631$ m/s. Yeh step kyun? Velocity derivative hai; minus sign kehta hai wo centre ki taraf wapas ja raha hai — hamare forecast se match karta hai.

Verify: Energy check: $\frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = \frac{1}{2} (2) (0.631)^{2} + \frac{1}{2} (50) (0.081)^{2} = 0.398 + 0.164 = 0.5625$ J $= E$ ✓.

Worked example Example 7 — Cell G: limiting & degenerate inputs

(Bada arithmetic nahi) Reason karo ki char extreme cases mein $T = 2 π m / k$ ka kya hota hai.

Forecast: Answer padhne se pehle har ek guess karo.

Infinitely stiff spring, $k \to \infty$ . $T = 2 π m / k \to 0$ . Yeh step kyun? Denominator explode hota hai → root → 0. Ek rigid rod oscillate nahi kar sakti; "bounce" infinitely fast hai (kabhi hota hi nahi).
Zero stiffness, $k \to 0$ . $T \to \infty$ . Yeh step kyun? Bilkul koi restoring force nahi → mass kabhi wapas nahi aati → infinite period (koi oscillation nahi).
Massless block, $m \to 0$ . $T \to 0$ . Yeh step kyun? Overshoot karne ke liye koi inertia nahi → spring use instantly ghar bhej deti hai.
Zero amplitude, $A = 0$ . $T$ unchanged hai ( $= 2 π m / k$ ), lekin mass bas baith jaati hai. Yeh step kyun? Period $A$ par depend nahi karta (yahi SHM ki pehchaan hai — isochronism). $A = 0$ trivial "no motion" solution hai; ghadi usi rate par tick karti rehti hai.

Verify: Sab formula ki structure se consistent hain: $T$ , $m$ ke saath badhta hai, $k$ ke saath girta hai, aur $A$ se independent hai ✓.

Worked example Example 8 — Cell H: real-world word problem

$m = 1200$ kg ki ek car char identical suspension springs par rakhti hai. Har corner $1/4$ weight uthata hai aur $y_{0} = 8.0$ cm compress hota hai. Car body ka bounce period estimate karo.

Forecast: Ek badi car soft springs par — bounce period lagbhag ek second? Dekhte hain.

Model karo. Charon springs ko ek effective spring maano jo poori mass support kar rahi hai. Poore load ke neeche measured droop $y_{0}$ pehle se hi $m / k_{e f f}$ encode karta hai. Yeh step kyun? "Jo load uthata hai uske neeche static stretch" exactly wahi hai jo $T = 2 π y_{0} / g$ ko chahiye — $k$ ya mass split karne ki zarurat nahi.
Convert karo. $y_{0} = 8.0 cm = 0.080$ m.
Apply karo. $T = 2 π y_{0} / g = 2 π 0.080/9.8 = 2 π 8.163 \times 1 0^{- 3}$ .
Evaluate karo. $8.163 \times 1 0^{- 3} = 0.09035$ , toh $T = 2 π (0.09035) = 0.568$ s.

Verify: Lagbhag aadha second per bounce — car ke liye realistic hai (real cars dampers use karti hain, Damped Oscillations dekho, isse jaldi kill karne ke liye). Units: s ✓. Note karo humhe mass ki zarurat hi nahi padi — stretch ne saara kaam kar diya.

Worked example Example 9 — Cell I: exam twist (vertical + energy + amplitude limit)

Ek mass spring par latakti hai; static stretch $y_{0} = 0.05$ m hai. Tum ise amplitude $A$ se oscillate karte ho. Spring kabhi slack na ho (yaani mass kabhi natural length se upar na jaaye) uske liye sabse badi amplitude kya hai?

Forecast: Spring tab tak taut rahegi jab tak uska extension positive rahega. Equilibrium extension $y_{0}$ par hai. Toh swing ka highest point natural length ko overshoot nahi karna chahiye. Guess: $A_{ma x} = y_{0}$ ?

Swing ka top locate karo. Equilibrium (extension $y_{0}$ ) ke baare mein amplitude $A$ se oscillation apne highest point par extension $y_{0} - A$ tak pahunchi hai. Yeh step kyun? Amplitude, equilibrium se turning point tak ki distance hai; top turning point equilibrium se $A$ upar hai.
"Spring not slack" impose karo. Spring apni natural length par hoti hai jab extension $= 0$ hoti hai. Taut rehne ke liye: $y_{0} - A \geq 0 \Rightarrow A \leq y_{0}$ . Yeh step kyun? Agar $A > y_{0}$ toh mass natural length se upar uth jaati hai; spring ko push (compress) karna padega — ek latakti spring nahi kar sakti, toh wo slack ho jaati hai aur motion clean SHM nahi rehti.
Limit aur period state karo. $A_{ma x} = y_{0} = 0.05$ m. Period (jab tak taut rahegi) $T = 2 π y_{0} / g = 2 π 0.05/9.8 = 2 π (0.07143) = 0.449$ s hai. Yeh step kyun? Wahi $T = 2 π y_{0} / g$ trick Energy in SHM/parent se — SHM hold karne tak amplitude se independent.

Verify: $A = A_{ma x} = y_{0}$ par swing ka top exactly natural length (extension $0$ ) ko graze karta hai — boundary case ✓. $T$ : units s, value sub-second, $5$ cm stretch ke saath consistent ✓.

Recall Quick self-test

Horizontal, $m = 2$ kg, $k = 50$ N/m — period? ::: $2 π 2/50 = 1.257$ s. Do springs $50$ aur $30$ N/m parallel mein — effective $k$ ? ::: $80$ N/m (wo add hote hain). Same do series mein — effective $k$ ? ::: $18.75$ N/m (reciprocals add hote hain). Jab $k \to \infty$ , period $T \to$ ? ::: $0$ (infinitely stiff spring oscillate nahi kar sakti). Hanging spring ke slack hone se pehle sabse badi amplitude? ::: $A_{ma x} = y_{0}$ , static stretch.

Connections

Parent: Spring-mass system — yahan use ki gayi formulas.
Simple Harmonic Motion — kyun $T$ , amplitude se independent hai (isochronism, Cell G/I).
Hooke's Law — har cell ke peechhey $F = - k x$ .
Energy in SHM — Cells C, D, I mein use hone wala $\frac{1}{2} k A^{2}$ .
Springs in Series and Parallel — Cells E aur F.
Simple Pendulum — contrast: iska $T$ sach mein $g$ par depend karta hai.
Damped Oscillations — real car (Cell H) damped hai, perpetual nahi.