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

1.6.2 · D4 · Physics › Oscillations & Waves › SHM differential equation — solution - x = A cos(ωt + φ)

Shuru karne se pehle, ek baar toolkit yaad kar lo — har answer neeche sirf inhi se bana hai:

L1 — Recognition

Goal: formula padhna aur ek number nikalna. Abhi koi manipulation nahi.

Exercise 1.1. Ek particle ka equation hai $x (t) = 0.04 cos (30 t + 0.5)$ (SI units). Iska amplitude $A$ , angular frequency $ω$ , aur phase constant $ϕ$ likho.

Recall Solution 1.1

Template $x = A cos (ω t + ϕ)$ se term by term compare karo. Yeh sirf pattern-matching hai — cosine ko multiply karne wala number $A$ hai; andar $t$ ko multiply karne wala number $ω$ hai; bacha hua added angle $ϕ$ hai.

$A = 0.04$ m
$ω = 30$ rad/s
$ϕ = 0.5$ rad

Yeh kaisa dikhta hai: $A$ batata hai ki wiggle centre se kitni door tak jaati hai; $ω$ batata hai ki internal clock kitni tezi se tick karta hai; $ϕ$ batata hai ki $t = 0$ par cosine wave par motion kahan hai.

Exercise 1.2. Usi particle ke liye, period $T$ aur frequency $f$ nikalo.

Recall Solution 1.2

Cosine repeat hoti hai jab iska andar wala part $2 π$ se badh jaata hai. $2 π$ kyun? Kyunki cosine $2 π$ radians ki circumference wale circle par define hoti hai — reference circle ka ek full chakkar ek oscillation hai. $T = \frac{2 π}{ω} = \frac{2 π}{30} \approx 0.2094 s, f = \frac{1}{T} = \frac{ω}{2 π} = \frac{30}{2 π} \approx 4.775 Hz .$

L2 — Application

Goal: derived formulas mein initial conditions ya physical constants daalna.

Exercise 2.1. $m = 0.2$ kg ka ek block $k = 50$ N/m ki spring par $x_{0} = 0.06$ m tak kheencha jaata hai aur rest se release hota hai. $ω$ , $A$ , $ϕ$ nikalo, aur $x (t)$ likho.

Recall Solution 2.1

Step 1 — $ω$ . Newton + Hooke se, $ω = k / m$ . $ω = 50/0.2 = 250 \approx 15.81 rad/s .$ Step 2 — $A$ . Rest se release matlab $v_{0} = 0$ , isliye $A = x_{0}^{2} + \frac{v _{0}^{2}}{ω ^{2}} = 0.0 6^{2} + 0 = 0.06 m .$ Kyun: rest se shuru karna matlab turning point se shuru karna, aur turning point hi amplitude hai. Step 3 — $ϕ$ . $tan ϕ = - v_{0} / (ω x_{0}) = 0$ , aur $x_{0} > 0$ se $cos ϕ > 0$ zaroori hai, isliye $ϕ = 0$ . Answer: $x (t) = 0.06 cos (15.81 t)$ m.

Exercise 2.2. Usi block ke liye. Maximum speed aur maximum acceleration nikalo.

Recall Solution 2.2

$v_{m a x} = A ω = 0.06 \times 15.81 \approx 0.9487 m/s (centre par x = 0) .$ $a_{m a x} = A ω^{2} = 0.06 \times 250 = 15 m/s^{2} (ends par x = \pm A) .$ Alag-alag jagah kyun: speed wahan greatest hoti hai jahan pull zero ho (centre); acceleration wahan greatest hoti hai jahan pull strongest ho (extremes). Isi fact ka energy view Energy in SHM mein dekho.

Exercise 2.3. Ek pendulum-like SHM ka $T = 1.5$ s aur amplitude $A = 0.08$ m hai. $ω$ aur $v_{m a x}$ nikalo.

Recall Solution 2.3

Period relation ko invert karo: $ω = 2 π / T = 2 π /1.5 \approx 4.189$ rad/s. $v_{m a x} = A ω = 0.08 \times 4.189 \approx 0.3351 m/s .$ (Ek swinging pendulum ke $ω$ ke physics ke liye, Simple pendulum dekho.)

L3 — Analysis

Goal: $ϕ$ ke liye correct quadrant choose karna, ya aisi relations ko invert karna jo soch maangti hain.

Exercise 3.1. Ek particle ( $ω = 10$ rad/s) $x_{0} = + 0.03$ m par hai aur $v_{0} = - 0.4$ m/s ke saath $- x$ direction mein move kar raha hai. $A$ aur $ϕ$ nikalo, aur $x (t)$ likho.

Recall Solution 3.1

Amplitude: $A = x_{0}^{2} + \frac{v _{0}^{2}}{ω ^{2}} = 0.0 3^{2} + \frac{( - 0.4 ) ^{2}}{1 0 ^{2}} = 0.0009 + 0.0016 = 0.0025 = 0.05 m .$ Phase — nazuk hissa. $tan ϕ = - v_{0} / (ω x_{0}) = - (- 0.4) / (10 \times 0.03) = + 1.333$ . Calculator seedha $arctan (1.333) = 0.9273$ rad deta hai. Lekin $tan$ har $π$ par repeat hoti hai, isliye $ϕ$ ya toh $0.9273$ ya $0.9273 - π = - 2.214$ ho sakta hai. Hume $x_{0}$ aur $v_{0}$ ki actual signs use karke choose karna hoga — neeche figure dekho.

$x_{0} = A cos ϕ > 0 \Rightarrow cos ϕ > 0$ .
$v_{0} = - A ω sin ϕ < 0 \Rightarrow sin ϕ > 0$ . Dono positive $\Rightarrow ϕ$ quadrant I mein hai, isliye $ϕ = + 0.9273$ rad. Answer: $x (t) = 0.05 cos (10 t + 0.9273)$ m.

Figure — SHM differential equation — solution - x = A cos(ωt + φ)

Exercise 3.2. Same $ω = 10$ , same starting position $x_{0} = + 0.03$ , lekin ab $v_{0} = + 0.4$ m/s ke saath $+ x$ direction mein move kar raha hai. $ϕ$ nikalo.

Recall Solution 3.2

$A$ unchanged hai: $A = 0.05$ m. $tan ϕ = - v_{0} / (ω x_{0}) = - (+ 0.4) / (0.3) = - 1.333$ , raw $arctan = - 0.9273$ milta hai. Sign check: $cos ϕ > 0$ (kyunki $x_{0} > 0$ ) aur $v_{0} > 0 \Rightarrow sin ϕ < 0$ . Dono quadrant IV point karte hain, isliye $ϕ = - 0.9273$ rad. 3.1 jaisi hi magnitude hai lekin negative — motion ki direction ne phase ka sign flip kar diya.

L4 — Synthesis

Goal: ek problem mein kai tools combine karna — energy, position–velocity, timing.

Exercise 4.1. SHM mein ek particle ka $A = 0.10$ m aur $ω = 8$ rad/s hai. Jab yeh $x = 0.06$ m par ho tab uski speed nikalo. Phir jab $x = A$ ho tab speed nikalo.

Recall Solution 4.1

Time-free speed relation use karo (parent derivation ke Step 2 se): $v = \pm ω A^{2} - x^{2} = \pm 8 0.1 0^{2} - 0.0 6^{2} = \pm 8 0.01 - 0.0036 = \pm 8 0.0064 = \pm 0.64 m/s .$ $x = A = 0.10$ par: $v = \pm 8 0.1 0^{2} - 0.1 0^{2} = 0$ . Kyun: extreme par particle momentarily rukta hai phir wapas muda ta hai — turning point. Yeh Energy in SHM se match karta hai: wahan saari energy potential hai, kinetic kuch nahi.

Exercise 4.2. Usi particle ke liye, kis displacement $x$ par kinetic energy potential energy ke barabar hogi?

Recall Solution 4.2

Kinetic $K E = \frac{1}{2} m v^{2} = \frac{1}{2} m ω^{2} (A^{2} - x^{2})$ ; potential $P E = \frac{1}{2} m ω^{2} x^{2}$ ( $k = m ω^{2}$ use karke). Barabar karo: $A^{2} - x^{2} = x^{2} \Rightarrow x^{2} = \frac{A ^{2}}{2} \Rightarrow x = \frac{A}{2} = \frac{0.10}{2} \approx 0.0707 m .$ Yeh neat kyun hai: answer $x = A / 2$ $m$ aur $ω$ se independent hai — har SHM mein energy same fractional displacement par 50/50 split hoti hai.

Exercise 4.3. $t = 0$ par $x = + A$ se start karke, $x = 0$ tak pahunchne mein kitna time lagta hai? ( $ω = 8$ rad/s use karo.)

Recall Solution 4.3

$+ A$ se rest ke saath start karna matlab $ϕ = 0$ : $x (t) = A cos (8 t)$ . Chahiye $cos (8 t) = 0$ , jiska pehla positive solution $8 t = π /2$ hai. $t = \frac{π}{2 ω} = \frac{π}{16} \approx 0.1963 s .$ Yeh kaisa dikhta hai: yeh full period $T = 2 π /8 \approx 0.785$ s ka ek-chauthai hai — ek extreme se centre tak jaana ek-chauthai cycle hai. $0.785/4 \approx 0.196$ ✓.

L5 — Mastery

Goal: multi-step reasoning, do states ke beech timing, aur scratch se phase + physics combine karna.

Exercise 5.1. $m = 0.25$ kg ka ek mass $k = 100$ N/m ki spring par $t = 0$ par $x_{0} = + 0.05$ m par hai aur $v_{0} = - 1.0$ m/s se move kar raha hai. (a) $ω$ , $A$ , $ϕ$ nikalo. (b) $x (t)$ likho. (c) Woh earliest time $t > 0$ nikalo jab mass pehli baar centre $x = 0$ se guzre.

Recall Solution 5.1

(a) $ω$ : $ω = k / m = 100/0.25 = 400 = 20$ rad/s. Amplitude: $A = x_{0}^{2} + v_{0}^{2} / ω^{2} = 0.0 5^{2} + (- 1.0)^{2} /2 0^{2} = 0.0025 + 0.0025 = 0.005 \approx 0.07071$ m. Phase: $tan ϕ = - v_{0} / (ω x_{0}) = - (- 1.0) / (20 \times 0.05) = + 1.0$ , raw $arctan (1) = π /4 = 0.7854$ . Sign check: $x_{0} > 0 \Rightarrow cos ϕ > 0$ ; $v_{0} < 0 \Rightarrow sin ϕ > 0$ . Quadrant I → $ϕ = + π /4 \approx 0.7854$ rad. ✓

(b) $x (t) = 0.07071 cos (20 t + \frac{π}{4})$ m.

(c) Chahiye pehla $t > 0$ jab $x = 0$ ho, yaani $cos (20 t + \frac{π}{4}) = 0$ . Cosine $\pm π /2, \pm 3 π /2, \dots$ par zero hoti hai. Andar wala $t = 0$ par $π /4$ se start hota hai aur badh ta hai, isliye pehla zero $20 t + \frac{π}{4} = \frac{π}{2}$ par hai: $20 t = \frac{π}{2} - \frac{π}{4} = \frac{π}{4} \Rightarrow t = \frac{π}{80} \approx 0.03927 s .$

Exercise 5.2. Do particles same $ω = 6$ rad/s aur same amplitude ke saath oscillate karte hain. Particle A ka $x_{A} = A cos (6 t)$ hai; particle B, A se $π /3$ phase lead karta hai. $x_{B} (t)$ likho aur woh time interval nikalo jitna B apne $+ A$ extreme par A se pehle pahunchta hai.

Recall Solution 5.2

" $π /3$ lead karna" matlab B ka internal clock $π /3$ radians aage hai, isliye andar add karo: $x_{B} = A cos (6 t + \frac{π}{3})$ . (Phase and phase difference dekho.) A $+ A$ tab pahunchta hai jab uska argument $= 0$ ho ( $t = 0$ par). B $+ A$ tab pahunchta hai jab uska argument $6 t + \frac{π}{3} = 0$ ho, yaani $t = - \frac{π}{18}$ — B apne extreme par pehle pahuncha tha, $Δ t = \frac{Δ ϕ}{ω} = \frac{π /3}{6} = \frac{π}{18} \approx 0.1745 s .$ Phase ko $ω$ se divide kyun: $ω$ "radians of phase" ko "seconds of time" mein convert karta hai, kyunki phase $ω$ rad per second ki rate se advance hoti hai.

Exercise 5.3. $A = 0.12$ m, $ω = 5$ rad/s, $ϕ = 0$ wala particle $+ A$ se start karta hai. Pehli baar $x = + A$ se $x = + A /2$ tak jaane mein kitna time lagta hai?

Recall Solution 5.3

$x (t) = 0.12 cos (5 t)$ . Set karo $0.12 cos (5 t) = 0.06 \Rightarrow cos (5 t) = 0.5 \Rightarrow 5 t = arccos (0.5) = \frac{π}{3}$ (pehla positive value). $t = \frac{π}{15} \approx 0.2094 s .$ Yeh kaisa dikhta hai: reference-circle point ka projection $A$ se $A /2$ tak drop karne ke liye $6 0^{\circ}$ sweep karta hai — woh $6 0^{\circ} = π /3$ exactly wahi hai jo $arccos (0.5)$ return karta hai.

Recall Self-test checklist (sabhi try karne ke baad hi open karo)

Formula se $A, ω, ϕ$ extract karo ::: term-by-term match karo; units report karo (m, rad/s, rad). Spring se $ω$ nikalo ::: $ω = k / m$ . $x_{0}, v_{0}$ se Amplitude ::: $A = x_{0}^{2} + v_{0}^{2} / ω^{2}$ (pehle velocity ko $ω$ se divide karo). Sahi $ϕ$ quadrant choose karo ::: $cos ϕ$ ka sign (= sign of $x_{0}$ ) AND $sin ϕ$ ka sign ( $v = - A ω sin ϕ$ se) check karo. Centre se door speed ::: $v = \pm ω A^{2} - x^{2}$ , kabhi $A ω$ nahi. Do states ke beech time ::: argument ke liye solve karo, phir $t = (argument) / ω$ ; phase-to-time hai $Δ t = Δ ϕ / ω$ .