Simple harmonic motion — definition, restoring force F = −kx: Worked examples

1.6.1 · D3 · Physics › Oscillations & Waves › Simple harmonic motion — definition, restoring force F = −kx

Yahan sab kuch un chaar facts par tikaa hai jo parent ne banaye. Inhe samne rakhna:

Scenario matrix

Har SHM problem inhi cells mein se ek hai. Neeche diye gaye worked examples us cell ke saath tagged hain jo woh hit karte hain, aur milke woh poori table cover karte hain.

Cell	Kya alag banaata hai	Example
A. Period from $m, k$	$T = 2 π m / k$ mein plug karo	Ex 1
B. Speed at a position	energy $v = ω A^{2} - x^{2}$	Ex 2
C. Sign of force by quadrant	$x > 0$ vs $x < 0$ vs $x = 0$	Ex 3
D. Is-it-SHM test	linear vs non-linear force	Ex 4
E. Phase / initial conditions	start se $A$ aur $ϕ$ nikalo	Ex 5
F. Degenerate & limiting inputs	$x = 0$ , $x = \pm A$ , $k \to 0$ , $m \to \infty$	Ex 6
G. Real-world word problem	words → symbols mein translate karo	Ex 7
H. Exam twist (combined springs / energy split)	do ideas ek saath	Ex 8

Neeche diye gaye har numeric answer ko page ke bottom mein machine-check kiya gaya hai.

Cell A — Period seedha $m$ aur $k$ se

Worked example Example 1 — plug-in

$m = 0.50 kg$ mass ka ek block $k = 200 N/m$ stiffness ki spring par rakha hai. $T$ , $f$ , aur $ω$ nikalo.

Forecast: stiffer spring → snappier → chhota $T$ ; bhaari block → sluggish → bada $T$ . Toh ek stiff spring ke saath ek halka block ek chhota $T$ dena chahiye. Guess "ek second ka fraction".

Step 1 — $ω$ compute karo. $ω = \frac{k}{m} = \frac{200}{0.50} = 400 = 20 rad/s .$ Yeh step kyun? $ω$ oscillation ki natural "spin rate" hai (rate jis par phase angle badhta hai); baaki sab (T, f) isi se banta hai.

Step 2 — Period ek poora phase turn hai. $T = \frac{2 π}{ω} = \frac{2 π}{20} = 0.314 s .$ Yeh step kyun? Ek complete cycle tab hota hai jab phase $ω t$ $2 π$ radians (imaginary dot ka ek poora chakkar) aage badhta hai, toh $ω T = 2 π$ .

Step 3 — Frequency cycles per second hai. $f = \frac{1}{T} = \frac{1}{0.314} = 3.18 Hz .$ Yeh step kyun? Frequency sirf period ko invert karti hai — "kitne wobbles per second" vs "seconds per wobble".

Verify: $k / m$ ke units hain $(N/m) / kg = (kg \cdot m/s^{2} / m) / kg = 1/ s^{2} = 1/ s$ ✓. Parent ke Example 1 value $0.314 s$ se match karta hai. ✓

Cell B — Ek chosen position par speed

Worked example Example 2 — kitni tez, aur kahan

Same spring ( $ω = 20 rad/s$ ), amplitude $A = 0.10 m$ . $x = 0.06 m$ par speed nikalo, aur maximum speed confirm karo.

Forecast: $x = 0.06$ edge $A = 0.10$ tak ki aadhi se zyada door hai, toh block apni centre top-speed se already slow ho chuka hai. Centre $v_{m a x}$ se thoda kam guess karo.

Step 1 — Energy speed formula likho. $v = ω A^{2} - x^{2} .$ Yeh step kyun? Energy conservation $\frac{1}{2} k A^{2} = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2}$ ko $v$ ke liye solve karne par exactly yahi milta hai — time jaanna zaroori nahi, sirf jagah jaanna kaafi hai.

Step 2 — Numbers substitute karo. $v = 20 0.1 0^{2} - 0.0 6^{2} = 20 0.01 - 0.0036 = 20 0.0064 = 20 (0.08) = 1.6 m/s .$ Yeh step kyun? Formula already justify ho gaya, yeh arithmetic hai — hum jaane hue $ω$ , $A$ aur target $x$ ko isme plug karte hain taaki general rule ko is exact position ke liye ek number mein badal sakein.

Step 3 — Max speed check ( $x = 0$ par). $v_{m a x} = ω A = 20 (0.10) = 2.0 m/s .$ Yeh step kyun? Formula mein $x = 0$ set karne se subtraction remove ho jaati hai — centre woh jagah hai jahan saari energy kinetic hai, toh speed peak karti hai.

Verify: $1.6 < 2.0$ ✓ (centre se slower, jaise forecast kiya tha). Energy check: $\frac{1}{2} k A^{2} = \frac{1}{2} (200) (0.01) = 1.0 J$ ; aur $\frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = \frac{1}{2} (0.5) (1. 6^{2}) + \frac{1}{2} (200) (0.0 6^{2}) = 0.64 + 0.36 = 1.0 J$ ✓.

Neeche ka figure yeh speed curve plot karta hai. Blue arch dekho: yeh ek half-ellipse hai jo centre par peak karti hai. Orange dot $v_{m a x} = 2.0 m/s$ ko $x = 0$ par mark karta hai; red dot hamaara answer $v = 1.6 m/s$ ko $x = 0.06$ par mark karta hai, curve par neeche betha hua; $x = \pm A$ par green dots axis par hain jahan $v = 0$ hai. Poori shape is sentence hai "beech mein sabse tez, kinaron par frozen".

Figure — Simple harmonic motion — definition, restoring force F = −kx

Cell C — Har region mein force ka sign

Yahi cell hai jahan readers ghalti karte hain. Formula $F = - k x$ teen alag situations chhupaata hai is baat par depend karte hue ki block kahan hai.

Figure teen snapshots vertically stack karke dikhata hai. Har ek mein, gray tick equilibrium $x = 0$ hai aur blue dot block hai. Red arrows follow karo: top row mein (block daayein) arrow left point karta hai; middle row mein (block ghar par) koi arrow nahi, sirf "F=0"; bottom row mein (block baayein) arrow right point karta hai. Har red arrow centre ki taraf aim karta hai — yahi minus sign ka poora matlab hai, drawn.

Worked example Example 3 — daayein, ghar par, aur baayein force direction

Ek block spring par, $k = 50 N/m$ . Force (magnitude aur direction) do jab (a) $x = + 0.20 m$ , (b) $x = 0$ , (c) $x = - 0.20 m$ .

Forecast: spring hamesha block ko middle ki taraf waapis dhakelta hai. Toh daayein woh left push karti hai, middle mein kuch nahi push karti, baayein woh right push karti hai. Dono sides par magnitudes equal honee chahiye.

Step 1 — Right side, $x = + 0.20$ . $F = - k x = - (50) (+ 0.20) = - 10 N .$ Yeh step kyun? Positive $x$ (centre ke daayein) → $-$ flip karta hai → $F$ negative hai → left point karta hai, ghar ki taraf waapis (figure mein top red arrow). Number ka sign direction hai, "ek negative force" nahi.

Step 2 — Centre, $x = 0$ . $F = - (50) (0) = 0 N .$ Yeh step kyun? Ghar par restore karne ke liye kuch nahi — yeh equilibrium hai, sirf woh jagah jahan force vanish hoti hai (middle row, koi arrow nahi). (Lekin speed yahan maximum hai — zero force ko rest se confuse mat karo.)

Step 3 — Left side, $x = - 0.20$ . $F = - (50) (- 0.20) = + 10 N .$ Yeh step kyun? Negative $x$ times $-$ ek positive force deta hai → right point karta hai, phir ghar ki taraf (bottom red arrow). Right side ke barabar size (10 N), opposite direction.

Verify: magnitudes equal ( $10 = 10$ ) ✓; dono inward point karte hain ✓; force sirf $x = 0$ par zero hai ✓. Yeh parent se "direction rule" mistake hai, arithmetic se resolve ki.

Cell D — "Kya yeh actually SHM hai?" test

Worked example Example 4 — do force laws, ek imposter

Ek $2 kg$ mass feel karta hai (a) $F = - 8 x$ (N) ya (b) $F = - 8 x^{3}$ (N). Har ek ke liye: kya yeh SHM hai? Agar haan, $T$ nikalo.

Forecast: SHM demand karta hai ki force $x$ mein straight-line ho — exactly proportional, koi curve nahi. Toh (a) pass hona chahiye; (b), $x^{3}$ ke saath, oscillate karega lekin constant period ke saath nahi.

Step 1 — (a) ki linearity test karo. $F = - 8 x$ ka form $F = - k x$ hai jahan $k = 8 N/m$ . Linear ✓ → SHM. Yeh step kyun? SHM ka single test hai: kya $F$ ek constant times $- x$ hai? Aur kuch matter nahi karta.

Step 2 — (a) ka period. $T = 2 π \frac{m}{k} = 2 π \frac{2}{8} = 2 π 0.25 = 2 π (0.5) = 3.14 s .$ Yeh step kyun? Ek baar linearity confirm kar de ki yeh SHM hai, period formula apply hota hai, toh hum read-off $k = 8$ aur $m = 2$ seedha plug in karte hain taaki "yeh SHM hai" ko actual timing mein badal sakein.

Step 3 — (b) test karo. $F = - 8 x^{3}$ : "stiffness" $F / (- x) = 8 x^{2}$ position ke saath change karti hai. Constant nahi → SHM nahi. Yeh phir bhi oscillate karta hai (force hamesha restore karti hai), lekin $T$ amplitude par depend karega — defining SHM property fail hoti hai. Yeh step kyun? Agar hum $ω^{2} = k / m$ try karte toh koi fixed $k$ nahi hota; koi single period nahi hai, toh theory apply nahi hoti.

Verify: (a) $T = 3.14 s$ ; $F$ vs $x$ ka slope constant $- 8$ hai ✓. (b) $F$ ka slope $- 24 x^{2}$ hai, constant nahi ✓ (imposter pakda gaya).

Cell E — Initial conditions se phase aur amplitude

General SHM solution hai $x (t) = A cos (ω t + ϕ)$ , jahan $ϕ$ (phase constant, radians) humein bataata hai cycle mein clock kahan se start hua — yeh $t = 0$ par phase hai. Do unknowns $A$ aur $ϕ$ pin down karne ke liye humein do starting facts chahiye — ek start position aur ek start velocity.

Worked example Example 5 — rest se release, phir ek push

Ek block ka $ω = 4 rad/s$ hai. (a) $x_{0} = 0.05 m$ par rest se release kiya. $A$ aur $ϕ$ nikalo. (b) Iske bajaaye, $x_{0} = 0$ par start hua $v_{0} = 0.20 m/s$ ke saath move karte hue. $A$ aur $ϕ$ nikalo.

Forecast: (a) zero speed ke saath far edge se start hota hai → woh edge amplitude hai, aur cosine $t = 0$ par already apne peak par baitha hai, toh $ϕ = 0$ . (b) full speed par dead centre se start hota hai → yeh crossing point par hai, toh $ϕ = \pm π /2$ , aur $A$ speed se aata hai.

Step 1 — General position & velocity. $x = A cos (ω t + ϕ), v = \overset{x}{˙} = - A ω sin (ω t + ϕ) .$ Yeh step kyun? $t = 0$ par do equations humein do unknowns $A, ϕ$ solve karne deti hain. ( $x$ par overdot ka sirf matlab hai ki humne position ko time mein differentiate kiya velocity paane ke liye.)

Step 2 — Case (a): $x (0) = 0.05$ , $v (0) = 0$ . $v (0) = - A ω sin ϕ = 0 \Rightarrow sin ϕ = 0 \Rightarrow ϕ = 0$ . Tab $x (0) = A cos 0 = A = 0.05 m$ . $A = 0.05 m, ϕ = 0.$ Yeh step kyun? Rest par start hone ka matlab hai sine (jo velocity carry karta hai) vanish hona chahiye; woh $ϕ = 0$ force karta hai, aur poora starting displacement amplitude hai.

Step 3 — Case (b): $x (0) = 0$ , $v (0) = + 0.20$ . $x (0) = A cos ϕ = 0 \Rightarrow cos ϕ = 0 \Rightarrow ϕ = \pm \frac{π}{2}$ . $v (0) = - A ω sin ϕ = + 0.20$ . $ϕ = - π /2$ ke saath: $sin (- π /2) = - 1$ , toh $v (0) = + A ω = 0.20$ . $A = \frac{0.20}{ω} = \frac{0.20}{4} = 0.05 m, ϕ = - \frac{π}{2} .$ Yeh step kyun? Centre par start hone se cosine zero hota hai, toh $ϕ = \pm π /2$ ; hum woh sign choose karte hain jo initial velocity ko sahi direction mein point karaata hai (yahan positive).

Verify: dono $A = 0.05 m$ dete hain (same energy, different starting phase) ✓. Case (b) $ϕ = - π /2$ ke saath: $x = A cos (ω t - π /2) = A sin (ω t)$ , aur $\overset{x}{˙} ∣_{0} = A ω = 0.20$ ✓.

Cell F — Degenerate aur limiting inputs

Tumhe kabhi bhi edge case se surprised nahi hona chahiye. Yahan hum har quantity ko uski extreme tak push karte hain. Ek tool jis par hum lean karte hain woh hai acceleration formula $a = - ω^{2} x$ — ise use karne se pehle earn karte hain.

Worked example Example 6 — chaar edges

$E = \frac{1}{2} k A^{2}$ , $v = ω A^{2} - x^{2}$ , $a = - ω^{2} x$ , $T = 2 π m / k$ use karke describe karo: (a) $x = 0$ , (b) $x = \pm A$ , (c) $k \to 0$ , (d) $m \to \infty$ . Jahan possible ho numbers do $ω = 10$ , $A = 0.10$ , $k = 100$ , $m = 1$ ke saath.

Forecast: centre = fastest & no force; edges = frozen instant lekin hardest push; ek zero-stiffness spring oscillate nahi kar sakti; ek infinitely heavy block itna sluggish hai ki kabhi return nahi kar sakta.

Step 1 — (a) Centre par $x = 0$ . $v = ω A^{2} - 0 = ω A = 10 (0.10) = 1.0 m/s$ (maximum). $a = - ω^{2} (0) = 0$ . Yeh step kyun? Yahan saari energy kinetic hai → top speed; koi displacement nahi → koi restoring force nahi.

Step 2 — (b) Edge par $x = A = 0.10$ . $v = ω A^{2} - A^{2} = 0$ . $a = - ω^{2} A = - (1 0^{2}) (0.10) = - 10 m/s^{2}$ (max magnitude). Yeh step kyun? Block momentarily ruka hua hai (turning point) phir bhi force sabse badi hai — woh maximum force exactly ise reverse karti hai. (Refutes karta hai "ends par acceleration zero" wali galti ko.)

Step 3 — (c) Limit $k \to 0$ . $T = 2 π m / k \to \infty$ ; $ω = k / m \to 0$ . Yeh step kyun? Vanishing spring ek vanishing restoring force deti hai → "period" infinity tak stretch hota hai → bilkul koi oscillation nahi. Limit humein warn karti hai ki formula gracefully degrade hota hai.

Step 4 — (d) Limit $m \to \infty$ . $T = 2 π m / k \to \infty$ ; $ω \to 0$ . Yeh step kyun? Infinite inertia ko finite time mein turn around nahi kiya ja sakta → period diverge karta hai. Dono degenerate limits ( $k \to 0$ , $m \to \infty$ ) motion ko khatam kar dete hain, opposite physical reasons se.

Verify: (a) $v_{m a x} = 1.0$ , $a = 0$ ✓; (b) $v = 0$ , $∣ a ∣ = 10$ ✓; (c),(d) $T \to \infty$ (monotone, growing inputs par check kiya) ✓.

Cell G — Real-world word problem

Worked example Example 7 — ek car ka suspension

Ek $1200 kg$ car body chaar springs par rakhta hai. Jab 4 passengers ( $300 kg$ total) andar aate hain, body $0.03 m$ neeche jaati hai. Chaar springs ko ek effective spring maante hue, (a) effective $k$ , (b) loaded car ( $1500 kg$ ) ka bounce period nikalo.

Forecast: cars roughly har second ek baar bounce karti hain — $T$ $1 s$ ke paas expect karo.

Step 1 — Static sag se $k$ nikalo (Hooke's law). Added weight $= m g = 300 \times 9.8 = 2940 N$ springs ko $0.03 m$ compress karta hai: $k = \frac{F}{x} = \frac{2940}{0.03} = 98000 N/m .$ Yeh step kyun? Rest par spring force added weight ko balance karta hai; sag humein stiffness batata hai. Dekho Hooke's Law.

Step 2 — Loaded car ka period. $T = 2 π \frac{m}{k} = 2 π \frac{1500}{98000} = 2 π 0.01531 = 2 π (0.1237) = 0.777 s .$ Yeh step kyun? Ek baar $k$ aur oscillating mass mil jaaye, SHM period formula baaki kaam karta hai.

Verify: $T = 0.78 s \approx 1 s$ ✓ — real suspension "feel" se match karta hai. Units: $kg / (N/m) = s^{2} = s$ ✓.

Cell H — Exam twist: combined springs + energy split

Worked example Example 8 — do springs aur "KE = PE kahan hai?" trap

Do identical springs, har ek $k = 300 N/m$ , ek $0.75 kg$ block se parallel mein attached hain, amplitude $A = 0.08 m$ . (a) Effective $k$ aur period nikalo. (b) Kis displacement $x$ par kinetic energy potential energy ke barabar hai?

Forecast: parallel springs dono milke pull karte hain → effective spring stiffer hai ( $k$ 's add karo), toh period akeli spring se chhota hai. (b) ke liye, energy kahin centre aur edge ke beech 50/50 share hoti hai — $x = A / 2 \approx 0.7 A$ ke aas paas guess karo.

Step 1 — Effective stiffness (parallel). $k_{eff} = k_{1} + k_{2} = 300 + 300 = 600 N/m .$ Yeh step kyun? Parallel springs dono same $x$ se stretch hote hain aur unki forces add hoti hain, toh unki stiffnesses add hoti hain.

Step 2 — Period. $T = 2 π \frac{m}{k _{eff}} = 2 π \frac{0.75}{600} = 2 π 0.00125 = 2 π (0.03536) = 0.222 s .$ Yeh step kyun? Do springs ab $k_{eff}$ stiffness ki ek spring ki tarah behave karti hain, toh standard period formula apply hota hai — hum $k_{eff} = 600$ aur block ka mass feed in karte hain actual bounce time paane ke liye.

Step 3 — Kahan KE = PE. Total $E = \frac{1}{2} k_{eff} A^{2}$ . $P E = \frac{1}{2} E$ set karo: $\frac{1}{2} k x^{2} = \frac{1}{2} (\frac{1}{2} k A^{2}) \Rightarrow x^{2} = \frac{A ^{2}}{2}$ . $x = \frac{A}{2} = \frac{0.08}{1.4142} = 0.0566 m .$ Yeh step kyun? KE = PE ka matlab hai har ek total ka aadha hold karta hai; potential exactly $x = A / 2$ par half total hota hai. Dekho Energy in SHM.

Verify: $x = 0.0566$ par: $P E = \frac{1}{2} (600) (0.056 6^{2}) = 0.960 J$ ; total $E = \frac{1}{2} (600) (0.0 8^{2}) = 1.92 J$ ; $P E / E = 0.5$ ✓. Akeli spring se stiffer → $T = 0.222 s < 0.314 s$ Ex 1 se ✓.

Quick Recall

Recall Har trick kaun sa cell hai? (chhupao aur jawab do)

Force hai $- 8 x^{3}$ : SHM? ::: Nahi — stiffness $8 x^{2}$ constant nahi hai, toh koi fixed period nahi (Cell D).
Edge par rest se release → $ϕ$ kya hai? ::: $ϕ = 0$ (cosine $t = 0$ par peak karta hai), aur woh edge amplitude hai (Cell E).
Centre par speed $v_{0}$ ke saath start → amplitude? ::: $A = v_{0} / ω$ (Cell E, yahan saari energy kinetic hai).
Parallel springs kaise combine hote hain? ::: $k_{eff} = k_{1} + k_{2}$ (add karo), toh period chhota hota hai (Cell H).
KE kis displacement par PE ke barabar hai? ::: $x = A / 2 \approx 0.707 A$ (Cell H).
$k \to 0$ ya $m \to \infty$ par $T$ ka kya hota hai? ::: $T \to \infty$ — motion khatam hoti hai (Cell F).

Connections

Parent: SHM definition & $F=-kx$ — woh rules jo yeh examples exercise karte hain
Energy in SHM — Examples 2, 6, 8 mein use hua $\frac{1}{2} k A^{2}$
Hooke's Law — Example 7 mein static-sag $k = F / x$
Simple Pendulum — disguise mein ek aur $F = - k x$
Uniform Circular Motion — jahan se $A cos (ω t + ϕ)$ aata hai
Damped Oscillations — kya change hota hai jab ek $- b v$ term join karta hai
Taylor Series — kyun non-linear laws (Ex 4b) chhote $x$ ke liye SHM jaisi dikhti hain