SHM differential equation — solution - x = A cos(ωt + φ): Worked examples

1.6.2 · D3 · Physics › Oscillations & Waves › SHM differential equation — solution - x = A cos(ωt + φ)

Yeh page $x=A\cos(\omega t+\phi)$ solution ke liye ek worked-example machine hai. Equation ek baar solve hoti hai; asli art yeh hai ki do constants $A$ (amplitude) aur $ϕ$ (phase) ko jo bhi starting state diya jaye usके liye sahi choose kiya jaye. Har possible starting state — position ka sign, velocity ka sign, wall pe start karna, centre pe start karna, sab zero — ka apna worked cell neeche hai.

Symbols kaam karne se pehle, do master formulas yaad karo (dono parent note mein derive hue hain, kabhi guess nahi hue):

Do sign checks (inhe memorise karo, har jagah use karo)

$t = 0$ par hamaara solution aur uski derivative yeh hai: $x_{0} = A cos ϕ, v_{0} = - A ω sin ϕ .$ Kyunki $A > 0$ aur $ω > 0$ :

Upar wali picture phase compass hai: $cos ϕ$ ka sign (left/right) aur $sin ϕ$ ka sign (up/down) pin karo aur arrow ek quadrant mein land karega. Har example mein ise refer karo.

Scenario matrix

Har SHM "find $x (t)$ " problem in cells mein se ek hai. Har cell ko cover karne wala Example named hai. $(x_{0}, v_{0})$ ke chaar nonzero sign combinations sab present hain (Cells E, F, G, K).

Cell	Starting position $x_{0}$	Starting velocity $v_{0}$	$ϕ$ kaun sa quadrant hai?	Covered by
A	$+ A$ (wall pe)	$0$ (rest se release)	$ϕ = 0$	Ex 1
B	$- A$ (far wall pe)	$0$	$ϕ = π$	Ex 2
C	$0$ (centre)	$+$ (right move)	$ϕ = - π /2$	Ex 3
D	$0$ (centre)	$-$ (left move)	$ϕ = + π /2$	Ex 4
E	$+$	$+$	Quadrant IV: $- π /2 < ϕ < 0$	Ex 5
F	$+$	$-$	Quadrant I: $0 < ϕ < π /2$	Ex 6
G	$-$	$-$	Quadrant II: $π /2 < ϕ < π$	Ex 7
K	$-$	$+$	Quadrant III: $- π < ϕ < - π /2$	Ex 8
H	$0$ aur $0$	$0$	degenerate: koi motion nahi	Ex 9
I	word problem (pendulum)	given period + push	full pipeline	Ex 10
J	exam twist (pehli baar position pe time dhundhna)	—	cosine ko invert karna	Ex 11

Har example mein hum same spring use karte hain jab tak aur na bataya jaye: $m = 0.5 kg$ , $k = 200 N/m$ , isliye $ω = k / m = 200/0.5 = 400 = 20 rad/s .$

Example 1 — Cell A: $+$ wall se rest pe release

Worked example Statement

Spring mass ko $x_{0} = + 0.1 m$ tak kheencha aur rest se release kiya ( $v_{0} = 0$ ). $x (t)$ aur max speed dhundho.

Forecast: cosine cycle mein "far right pe stationary start" kahan baithta hai? ( $ϕ$ pehle se guess karo padhe bina.)

1. Amplitude. $A = x_{0}^{2} + v_{0}^{2} / ω^{2} = 0. 1^{2} + 0 = 0.1 m$ . Yeh step kyun? $v_{0} = 0$ ke saath poori energy stretch-energy hai, isliye start point hi turning point hai — yeh literally amplitude ki definition hai.

2. Phase. $cos ϕ$ ka sign $x_{0} > 0$ (positive) ka sign hai aur $sin ϕ$ ka sign $v_{0} = 0$ (zero) ka ulta hai. Positive cosine, zero sine ⇒ $ϕ = 0$ . Yeh step kyun? $tan ϕ = - v_{0} / (ω x_{0}) = 0$ se $ϕ = 0$ ya $π$ milta hai; positive-cosine check $π$ ko eliminate karta hai.

3. Assemble. $x (t) = 0.1 cos (20 t) m$ ; $v_{m a x} = A ω = 0.1 \times 20 = 2 m/s$ . Yeh step kyun? Do constants ko master solution $x = A cos (ω t + ϕ)$ mein daalna hi poora point hai — constants hi sirf unknowns bache the.

Verify: $t = 0$ par: $x = 0.1 cos 0 = 0.1$ ✓, $v = - 0.1 \cdot 20 sin 0 = 0$ ✓. $A ω$ ki units: $m \cdot s^{- 1} = m/s$ ✓.

Example 2 — Cell B: $-$ wall se rest pe release

Worked example Statement

Same spring, $x_{0} = - 0.1 m$ tak kheencha, rest se release ( $v_{0} = 0$ ). $x (t)$ dhundho.

Forecast: left wall se start karna Example 1 se half-turn different hona chahiye. Kaun sa $ϕ$ ?

1. Amplitude. $A = (- 0.1)^{2} + 0 = 0.1 m$ . Yeh step kyun? Amplitude ek distance hai, isliye $x_{0}$ ka sign hat jaata hai — squaring se woh mit jaata hai.

2. Phase. $cos ϕ$ ka sign $x_{0} < 0$ (negative) ka sign leta hai; $sin ϕ = 0$ . Negative cosine, zero sine ⇒ $ϕ = π$ . Yeh step kyun? $tan ϕ = 0$ phir $0$ ya $π$ offer karta hai; ab negative-cosine requirement $π$ force karti hai. Yahi reason hai ki "hamesha $ϕ = 0$ " wali galti fail karti hai.

3. Assemble. $x (t) = 0.1 cos (20 t + π) = - 0.1 cos (20 t) m$ . Yeh step kyun? Hum $A = 0.1$ aur $ϕ = π$ ko $x = A cos (ω t + ϕ)$ mein substitute karte hain; identity $cos (θ + π) = - cos θ$ isse cleaner mirror form mein rewrite karti hai.

Verify: $x (0) = 0.1 cos π = - 0.1$ ✓. $v (0) = - 0.1 \cdot 20 sin π = 0$ ✓.

Example 3 — Cell C: centre, right move

Worked example Statement

Mass $x_{0} = 0$ se guzar raha hai $+ x$ direction mein $v_{0} = + 2 m/s$ ke saath. $x (t)$ dhundho.

Forecast: centre par woh sabse tez hai, isliye yeh $v_{m a x}$ ka moment hai. Yeh cosine ko sine mein badal dena chahiye. Sine ka kaun sa sign?

1. Amplitude. $A = 0 + v_{0}^{2} / ω^{2} = (2/20)^{2} = 0.1 m$ . Yeh step kyun? Centre par poori energy kinetic hai, isliye $A = v_{0} / ω$ — reach poori tarah speed se milti hai.

2. Phase. Yahan $x_{0} = 0$ hai, isliye formula $tan ϕ = - v_{0} / (ω x_{0})$ undefined hai (zero se division) — hum ise drop karte hain aur direct rule use karte hain. $x_{0} = 0$ ⇒ $cos ϕ = 0$ ⇒ $ϕ = \pm π /2$ . Phir $v_{0} > 0$ ko $sin ϕ < 0$ chahiye (kyunki $v_{0} = - A ω sin ϕ$ ). Negative sine ⇒ $ϕ = - π /2$ . Yeh step kyun? $tan ϕ \to - \infty$ $+ π /2$ aur $- π /2$ mein distinguish nahi kar sakta; sirf velocity ka sign kar sakta hai.

3. Assemble. $x (t) = 0.1 cos (20 t - π /2) = 0.1 sin (20 t) m$ . Yeh step kyun? Constants substitute karne ke baad $cos (θ - π /2) = sin θ$ use karne se pure sine milta hai — yeh centre-crossing start ki pehchaan hai.

Verify: $x (0) = 0.1 cos (- π /2) = 0$ ✓. $v (0) = - 0.1 \cdot 20 sin (- π /2) = + 2$ ✓.

Example 4 — Cell D: centre, left move

Example 5 — Cell E: positive position, positive velocity ( $ϕ$ ka Quadrant IV)

Worked example Statement

Mass $x_{0} = + 0.06 m$ par hai aur $v_{0} = + 1.6 m/s$ ke saath right ki taraf outward move kar raha hai. $A$ aur $ϕ$ dhundho.

Forecast: + side par start karna aur abhi bhi outward speed karna — wall tak nahi pahuncha, isliye cycle mein turning point se pehle hai. Kya $ϕ$ positive hai ya negative?

1. Amplitude. $A = 0.0 6^{2} + (1.6/20)^{2} = 0.0036 + 0.0064 = 0.01 = 0.1 m$ . Yeh step kyun? Position kuch reach deta hai, bachi hui speed baaki deti hai; Pythagoras unhe add karta hai kyunki energy (kinetic + potential) add hoti hai.

2. Phase. $cos ϕ$ sign = $x_{0} > 0$ ka sign (+); $sin ϕ$ sign = $v_{0} > 0$ ka ulta ⇒ negative. Positive cosine + negative sine = Quadrant IV, isliye $- π /2 < ϕ < 0$ . $tan ϕ = - v_{0} / (ω x_{0}) = - 1.6/ (20 \cdot 0.06) = - 1.3 \overline{3}$ ⇒ $ϕ = arctan (- 1.333...) = - 0.9273 rad$ (yeh Quadrant IV mein land karta hai, isliye seedha sahi answer hai). Yeh step kyun? Kyunki raw $arctan$ already ek Quadrant IV angle return karta hai, koi $\pm π$ correction nahi chahiye — compass confirm karta hai.

3. Assemble. $x (t) = 0.1 cos (20 t - 0.9273) m$ . Yeh step kyun? Found constants master solution mein daalo; kuch aur simplify nahi hota, isliye $ϕ$ ko numeric radian value ki tarah rakhte hain.

Verify: $A cos ϕ = 0.1 cos (- 0.9273) = 0.06$ ✓ ( $= x_{0}$ ). $- A ω sin ϕ = - 0.1 \cdot 20 \cdot sin (- 0.9273) = 1.6$ ✓ ( $= v_{0}$ ).

Example 6 — Cell F: positive position, negative velocity ( $ϕ$ ka Quadrant I)

Worked example Statement

Example 5 jaisi hi numbers lekin ab mass $x_{0} = + 0.06 m$ par inward (wapas) move kar raha hai, $v_{0} = - 1.6 m/s$ . $ϕ$ dhundho.

Forecast: wall ke baad wapas aa raha hai — outgoing swing ka baad wala hissa. Kya $ϕ$ ab positive hona chahiye?

1. Amplitude. Same: $A = 0.0 6^{2} + (- 1.6/20)^{2} = 0.1 m$ (velocity squared).

2. Phase. $cos ϕ$ sign = $x_{0} > 0$ (+); $sin ϕ$ sign = $v_{0} < 0$ ka ulta ⇒ positive. Positive cosine + positive sine = Quadrant I, isliye $0 < ϕ < π /2$ . Ab $tan ϕ = - v_{0} / (ω x_{0}) = - (- 1.6) / (20 \cdot 0.06) = + 1.3 \overline{3}$ aur $arctan (1.333...) = + 0.9273 rad$ — already Quadrant I mein hai. ✓ Yeh step kyun? $v_{0}$ ka sign flip hua, jisne $tan ϕ$ ka sign flip kiya, same-magnitude angle ko Quadrant IV (Ex 5) se Quadrant I mein bhej diya. Same position, ulti travel ⇒ mirror-image phase.

3. Assemble. $x (t) = 0.1 cos (20 t + 0.9273) m$ . Yeh step kyun? Constants substitute karna solution complete karta hai; positive $ϕ$ curve ko aise shift karta hai ki mass $t = 0$ par wapas aa raha ho.

Verify: $A cos ϕ = 0.1 cos (0.9273) = 0.06$ ✓. $- A ω sin ϕ = - 2 sin (0.9273) = - 1.6$ ✓.

Example 7 — Cell G: negative position, negative velocity (arctan galat quadrant mein land karta hai)

Worked example Statement

Mass $x_{0} = - 0.06 m$ par hai aur $v_{0} = - 1.6 m/s$ ke saath left move kar raha hai (far wall ki taraf ja raha hai). $ϕ$ dhundho. Arctan trap dekho.

Forecast: − side par aur aur bhi − direction mein ja raha hai. Compass kehta hai cosine negative aur sine positive → Quadrant II. Kya raw calculator woh dega?

1. Amplitude. $A = (- 0.06)^{2} + (- 1.6/20)^{2} = 0.1 m$ .

2. Phase (the trap). $cos ϕ$ sign = $x_{0} < 0$ (−); $sin ϕ$ sign = $v_{0} < 0$ ka ulta ⇒ positive. Negative cosine + positive sine = Quadrant II, isliye $π /2 < ϕ < π$ . Lekin $tan ϕ = - v_{0} / (ω x_{0}) = - (- 1.6) / (20 \cdot (- 0.06)) = 1.6/ (- 1.2) = - 1.3 \overline{3}$ , aur calculator $arctan (- 1.333...) = - 0.9273 rad$ return karta hai — woh Quadrant IV hai, galat jagah! $π$ add karo: $ϕ = - 0.9273 + π = 2.2143 rad$ (Quadrant II ✓). Yeh step kyun? $tan$ har $π$ pe repeat hota hai, isliye $arctan$ exactly $π$ se off ho sakta hai. Sign compass batata hai kab add karna hai — yahan cosine negative honi chahiye, isliye angle ko Quadrant II mein push karte hain.

3. Assemble. $x (t) = 0.1 cos (20 t + 2.2143) m$ . Yeh step kyun? Corrected constants substitute karna solution finish karta hai; bada positive $ϕ$ mass ko $t = 0$ par negative side pe outward rakhta hai.

Verify: $A cos ϕ = 0.1 cos (2.2143) = - 0.06$ ✓. $- A ω sin ϕ = - 2 sin (2.2143) = - 1.6$ ✓.

Figure Cells E, F aur G ke liye same magnitude problem ko ek time axis par plot karta hai: identical amplitude, teen alag phases curve ko left/right slide karte hain. Woh sliding hi $ϕ$ hai.

Example 8 — Cell K: negative position, positive velocity ( $ϕ$ ka Quadrant III)

Worked example Statement

Mass $x_{0} = - 0.06 m$ par hai lekin right (centre ki taraf wapas) move kar raha hai $v_{0} = + 1.6 m/s$ ke saath. $ϕ$ dhundho. Yeh last missing sign combination hai.

Forecast: − side par lekin ghar wapas ja raha hai. Compass kehta hai cosine negative, sine negative → Quadrant III. Kya calculator wahan land karega, ya correct karna padega?

1. Amplitude. $A = (- 0.06)^{2} + (1.6/20)^{2} = 0.0036 + 0.0064 = 0.1 m$ (signs squared away).

2. Phase (the trap again). $cos ϕ$ sign = $x_{0} < 0$ (−); $sin ϕ$ sign = $v_{0} > 0$ ka ulta ⇒ negative. Negative cosine + negative sine = Quadrant III, isliye $- π < ϕ < - π /2$ . Lekin $tan ϕ = - v_{0} / (ω x_{0}) = - (1.6) / (20 \cdot (- 0.06)) = - 1.6/ (- 1.2) = + 1.3 \overline{3}$ , aur calculator $arctan (1.333...) = + 0.9273 rad$ return karta hai — woh Quadrant I hai, galat! $π$ subtract karo: $ϕ = 0.9273 - π = - 2.2143 rad$ (Quadrant III ✓). Yeh step kyun? $arctan$ ne do cosine-signs ko ek mein compress kar diya; compass negative cosine demand karta hai, isliye hum $- π$ se Quadrant III mein shift karte hain. Yeh Example 7 ka mirror hai lekin velocity reversed ke saath.

3. Assemble. $x (t) = 0.1 cos (20 t - 2.2143) m$ . Yeh step kyun? Corrected constants substitute karna solution finish karta hai; negative $ϕ$ mass ko $t = 0$ par negative side pe inward rakhta hai.

Verify: $A cos ϕ = 0.1 cos (- 2.2143) = - 0.06$ ✓. $- A ω sin ϕ = - 2 sin (- 2.2143) = + 1.6$ ✓.

Example 9 — Cell H: degenerate case (zero, zero)

Worked example Statement

Mass $x_{0} = 0$ par baith gaya hai $v_{0} = 0$ ke saath. $x (t)$ kya hai?

Forecast: na position, na push — kya yeh kabhi hilega?

1. Amplitude. $A = 0 + 0/ ω^{2} = 0$ . Yeh step kyun? Amplitude total reach hai; koi displacement nahi aur koi speed nahi toh kisi bhi reach ko fund karne ke liye zero energy hai.

2. Phase. $A = 0$ ke saath, $ϕ$ koi mayne nahi rakhta — $0 \cdot cos (anything) = 0$ . Phase meaningless ho jaata hai (undefined lekin irrelevant). Yeh step kyun? Yeh honest edge case hai: do constants ek trivial solution mein collapse ho jaate hain. Aisa quadrant mat dhundho jo exist hi nahi karta.

3. Assemble. $x (t) = 0$ sab $t$ ke liye — mass hamesha equilibrium par rehta hai. Yeh step kyun? $A = 0$ ko $x = A cos (ω t + ϕ)$ mein substitute karna zero function chhod deta hai — physically obvious "kuch nahi hota" state.

Verify: $\overset{x}{¨} + ω^{2} x = 0 + ω^{2} \cdot 0 = 0$ ✓ — zero function trivially ODE solve karta hai. Physically: equilibrium ek valid (boring) SHM state hai.

Example 10 — Cell I: word problem, period + push se ek pendulum

Worked example Statement

Ek simple pendulum small angles se swing karta hai, period $T = 2 s$ hai. $t = 0$ par uska bob bottom pe hai ( $x_{0} = 0$ ) aur use arc ke along $v_{0} = + 0.15 m/s$ wali push di jaati hai. $ω$ , $A$ , $ϕ$ , aur $x (t)$ dhundho.

Forecast: swing ka bottom = motion ka centre. Example 3 jaisi hi shape, sirf $ω$ period se aata hai $k / m$ se nahi.

0. Legality check. Ek pendulum sirf small-angle approximation mein $\overset{x}{¨} = - ω^{2} x$ follow karta hai, jahan $sin θ \approx θ$ restoring force ko displacement mein linear banata hai. Kyunki push chota hai, swing chhota rehta hai aur yeh linear SHM solution valid hai — warna period amplitude pe depend karta aur $x = A cos (ω t + ϕ)$ apply nahi hota. Yeh step kyun? Poori matrix true (linear) SHM assume karti hai; formulas use karne se pehle confirm karna hoga ki system qualify karta hai.

1. Angular frequency. $ω = 2 π / T = 2 π /2 = π \approx 3.1416 rad/s$ . Yeh step kyun? Cosine tab repeat hota hai jab uska argument $2 π$ badhe, isliye ek full period $ω T = 2 π$ correspond karta hai. Hum ise invert karte hain.

2. Amplitude. $A = 0 + v_{0}^{2} / ω^{2} = v_{0} / ω = 0.15/ π \approx 0.04775 m$ . Yeh step kyun? Centre pe start ⇒ sab kinetic ⇒ reach $= v_{0} / ω$ — exactly Cell C ki logic.

3. Phase. $x_{0} = 0$ $tan ϕ$ ko undefined karta hai, isliye direct rule use karo: $cos ϕ = 0$ aur $v_{0} > 0$ ko $sin ϕ < 0$ chahiye ⇒ $ϕ = - π /2$ .

4. Assemble. $x (t) = \frac{0.15}{π} cos (π t - \frac{π}{2}) = \frac{0.15}{π} sin (π t) m$ . Yeh step kyun? Teen found constants substitute karne ke baad $cos (θ - π /2) = sin θ$ use karne se centre-start swing ka clean sine milta hai.

Verify: $x (0) = 0$ ✓. $v (0) = - A ω sin (- π /2) = A ω = (0.15/ π) \cdot π = 0.15$ ✓. $2 π / T$ ki units: $rad / s$ ✓.

Example 11 — Cell J: exam twist (cosine invert karke time dhundhna)

Worked example Statement

Example 1 ki motion $x (t) = 0.1 cos (20 t)$ use karo: pehli baar $t_{1} > 0$ par mass $x = 0.05 m$ par kab pahunchta hai? Wahan speed bhi dhundho.

Forecast: $0.05$ wall tak ka aadha raasta hai. Kyunki cosine top ke paas steep hota hai aur baad mein flat, kya $t_{1}$ ek quarter period se zyada hoga ya kam?

1. Equation setup karo. $0.05 = 0.1 cos (20 t) \Rightarrow cos (20 t) = 0.5$ . Yeh step kyun? Hum position jaante hain aur time chahiye, isliye cosine undo karna padega — $arccos$ yehi karta hai: "kaun sa angle is cosine wala hai?"

2. Invert karo. $20 t = arccos (0.5) = π /3 rad$ (sabse chhota positive solution). Isliye $t_{1} = \frac{π /3}{20} = \frac{π}{60} \approx 0.05236 s$ . Yeh step kyun? $arccos$ $[0, π]$ mein pehla angle return karta hai; $ω = 20$ se divide karna angle ko seconds mein convert karta hai. (Baad ke times $2 π - π /3$ wagera honge — hume pehla chahiye.)

3. Wahan speed — time-free relation use karo. $v = \pm ω A^{2} - x^{2} = - 20 0. 1^{2} - 0.0 5^{2} = - 20 0.0075 \approx - 1.732 m/s$ . Yeh step kyun? Mass $+ A$ se inward start hua, isliye $x = + 0.05$ par neeche jaate waqt woh $- x$ direction mein ja raha hai ⇒ negative sign. $v = \pm ω A^{2} - x^{2}$ use karne se dobara differentiate karna avoid hota hai.

Verify: Period $T = 2 π /20 \approx 0.314 s$ ; ek quarter period (centre tak pahunchne ke liye) $0.0785 s$ hai, aur $t_{1} \approx 0.0524 s < 0.0785$ ✓ (halfway point centre se pehle — cosine top ke paas tezi se girta hai, forecast se match karta hai). Speed magnitude $1.732 < v_{m a x} = 2$ m/s ✓.

Recall Kisi bhi starting snapshot ke liye one-line recipe

$A = x_{0}^{2} + v_{0}^{2} / ω^{2}$ compute karo; $ϕ$ ki magnitude $tan ϕ = - v_{0} / (ω x_{0})$ se nikalo (jab tak $x_{0} = 0$ na ho, wahan $ϕ = \pm π /2$ directly use karo); phir quadrant set karo: $cos ϕ$ ka sign $x_{0}$ ka share karta hai, aur $sin ϕ$ $v_{0}$ ka ulta hota hai.

Kaun se cell mein

ϕ = 0

hai?

+ A

par rest se release (positive position, zero velocity).

Kaun se cell mein

ϕ = π

hai?

- A

par rest se release (negative position, zero velocity).

Centre, right move —

ϕ

kya hai?

- π /2

(toh

x = A sin ω t

Centre, left move —

ϕ

kya hai?

+ π /2

(toh

x = - A sin ω t

Negative position, positive velocity —

ϕ

kaun se quadrant mein hai?

Quadrant III (

- π < ϕ < - π /2

); raw arctan se

π

subtract karo.

Raw

arctan

ϕ

ke liye

π

kab add karna padta hai?

Jab required

cos ϕ

ka sign (yaani

x_{0}

ka)

arctan

ke return se disagree kare — usually

x_{0} < 0

par.

tan ϕ = - v_{0} / (ω x_{0})

kab unusable hai?

Jab

x_{0} = 0

ho (zero se division); tab

ϕ = \pm π /2

use karo

v_{0}

ke sign se choose karo.

Zero-zero degenerate case mein

ϕ

kya hai?

Undefined aur irrelevant —

A = 0

hai isliye phase kuch bhi multiply nahi karta.