Velocity and acceleration in SHM — v = ω√(A² − x²): Worked examples

1.6.4 · D3 · Physics › Oscillations & Waves › Velocity and acceleration in SHM — v = ω√(A² − x²)

Shuru karne se pehle, ek reminder ki har symbol ka kya matlab hai, taaki kuch bhi unclear na rahe:

Definition Symbols ki cast (sab

$A$ = amplitude — particle kabhi bhi middle se jitna door ja sakta hai, metres mein. Socho do walls ke beech bounce ho raha hai.
$x$ = displacement — abhi is waqt kahan hai, middle se measure karke. Positive = ek side, negative = doosri.
$ω$ (omega) = angular frequency in rad/s — oscillation kitni tezi se "spin" karti hai (dekho Angular frequency ω and time period T). Bada $ω$ = jaldi wobble.
$v$ = velocity (speed sign ke saath: $+$ ek taraf move karna, $-$ doosri taraf).
$v_{m a x}$ = maximum speed — particle sabse tez kabhi jaata hai, jo center par hota hai; hum dikhayenge ki yeh $ω A$ ke barabar hai.
$a$ = acceleration — velocity khud kitni tezi se change ho rahi hai.
$∣ a ∣_{m a x}$ = maximum acceleration magnitude — sabse strong push, jo extremes par hota hai; hum dikhayenge ki yeh $ω^{2} A$ ke barabar hai.
$T = \frac{2 π}{ω}$ = time period, ek poora aane-jaane ka time seconds mein.

Scenario matrix

$v$ aur $a$ ke baare mein SHM ka har question inhi cells mein se ek hota hai. Hum sab cover karenge.

Cell	Case class	Isme kya tricky hai	Example
C1	Interior point, outbound ( $0 < x < A$ , $v > 0$ )	$+$ root choose karna	Ex 1
C2	Interior point, inbound ( $0 < x < A$ , $v < 0$ )	same $x$ , opposite sign	Ex 2
C3	Center ( $x = 0$ )	speed max, accel zero	Ex 3
C4	Extreme ( $x = \pm A$ )	speed zero, accel max	Ex 3
C5	Negative side ( $x < 0$ )	$a$ ka sign $+$ ho jaata hai	Ex 4
C6	Degenerate ( $∣ x ∣ = A$ exactly, ya $A = 0$ )	square root of zero / no motion	Ex 5
C7	Limiting / ratio (max speed ka fraction)	algebra, squaring	Ex 6
C8	Real-world word problem (units, period)	words ko symbols mein translate karna	Ex 7
C9	Exam twist ( $v$ aur $a$ diye hain, $A, ω$ nikalo)	do equations, back-solve	Ex 8

Example 1 — Interior point, outbound (Cell C1)

Worked example Statement

Ek particle ka $A = 0.10 m$ , $ω = 5 rad/s$ hai. Yeh $x = 0.06 m$ par hai, center se door move kar raha hai. Velocity (sign ke saath) aur acceleration nikalo.

Forecast: pehle guess karo — kya speed max ( $ω A$ ) ke closer hai ya zero ke? Kya $a$ yahan positive hai ya negative?

Step 1 — "Leftover room" $A^{2} - x^{2}$ compute karo. $A^{2} - x^{2} = 0.1 0^{2} - 0.0 6^{2} = 0.01 - 0.0036 = 0.0064 = 0.08 m$ Yeh step kyun? $v = ω A^{2} - x^{2}$ : root measure karta hai ki abhi aage kitna amplitude bacha hai. $x = 0.06$ par walls ke andar $0.08 m$ ki "room" bachi hai.

Step 2 — Root ka sign choose karo. Positive side par center se door move karna matlab $+ x$ direction mein move karna, isliye hum $+$ root lete hain. $v = + ω A^{2} - x^{2} = 5 \times 0.08 = + 0.40 m/s$ Yeh step kyun? $\pm$ koi formula quirk nahi hai — yeh encode karta hai ki particle kis taraf ja raha hai (dekho outbound/inbound phase idea).

Step 3 — $a = - ω^{2} x$ se acceleration. $a = - ω^{2} x = - (5)^{2} (0.06) = - 25 \times 0.06 = - 1.5 m/s^{2}$ Yeh step kyun? SHM mein acceleration sirf position par depend karta hai, speed par nahi — yeh defining law $a = - ω^{2} x$ hai Simple Harmonic Motion — definition a = −ω²x se.

Verify: $v_{m a x} = ω A = 5 (0.10) = 0.5 m/s$ , aur $0.40 < 0.5$ ✓ (off-center hone par max se slow, expected). $a$ ka sign negative hai jabki $x$ positive hai → yeh center ki taraf wapas point karta hai, outbound motion ko brake karta hai. Units: $(rad/s) \cdot m = m/s$ ✓, $(rad/s)^{2} \cdot m = m/s^{2}$ ✓.

Neeche ki figure us straight track ka bird's-eye view hai jis par particle slide karta hai. Orange dot dekho — woh particle hai $x = 0.06 m$ par. Upar orange arrow uski velocity hai (outward pointing, $+ 0.40 m/s$ ); neeche teal arrow uska acceleration hai (center ki taraf wapas pointing, $- 1.5 m/s^{2}$ ). Daayein plum bracket "leftover room" $A^{2} - x^{2} = 0.08 m$ dikhata hai jo velocity formula measure karta hai.

Figure — Velocity and acceleration in SHM — v = ω√(A² − x²)

Figure s01 — Ex 1: particle outward move karta hai (orange) jabki uska acceleration already ghar point kar raha hai (teal); plum bracket root term hai.

Example 2 — Same point, inbound (Cell C2)

Worked example Statement

Same particle ( $A = 0.10$ , $ω = 5$ ), same position $x = 0.06 m$ , lekin ab center ki taraf wapas aa raha hai. $v$ aur $a$ nikalo.

Forecast: kya Ex 1 ke comparison mein speed change hogi? Kya acceleration change hogi?

Step 1 — Speed bilkul same hai. $∣ v ∣ = ω A^{2} - x^{2} = 0.40 m/s$ Yeh step kyun? Magnitude sirf kahan ho us par depend karti hai, kis taraf ja rahe ho us par nahi — particle har point par dono taraf same speed se guzarta hai.

Step 2 — Sign flip karo. Positive side se center ki taraf move karna matlab $- x$ direction mein move karna: $v = - 0.40 m/s$ Yeh step kyun? Yahi parent note wala " $\pm$ bhool jaana" trap hai — har interior point ek cycle mein do baar visit hota hai opposite velocity sign ke saath.

Step 3 — Acceleration unchanged hai. $a = - ω^{2} x = - 1.5 m/s^{2}$ Yeh step kyun? $a$ sirf $x$ par depend karta hai. Position same hai, isliye acceleration same hai, travel direction chahe koi bhi ho.

Verify: Ex 1 aur Ex 2 mein equal $∣ v ∣$ aur equal $a$ hai, sirf $v$ ke sign mein difference hai. Yahan $v < 0$ (center ki taraf) aur $a < 0$ (bhi center ki taraf) — particle ghar jaate waqt speed up kar raha hai ✓. Compare karo Ex 1 se jahan $v > 0$ , $a < 0$ → bahar jaate waqt slow down. Same physics, cycle ka opposite half.

Example 3 — Center aur extremes saath mein (Cells C3 & C4)

Worked example Statement

Same particle ( $A = 0.10$ , $ω = 5$ ). (a) center $x = 0$ par, aur (b) extreme $x = + A = 0.10 m$ par $v$ aur $a$ nikalo.

Forecast: in dono points mein se particle sabse fast kahan hai? Kahan usse sabse zyada "push" milti hai?

Step 1 — Center, velocity. $v = ω A^{2} - 0^{2} = ω A = 5 (0.10) = 0.5 m/s = v_{m a x}$ Yeh step kyun? $x = 0$ par poora amplitude "room left" hai, isliye root apne sabse bade value par hai — yeh maximum speed hai.

Step 2 — Center, acceleration. $a = - ω^{2} (0) = 0$ Yeh step kyun? Koi displacement nahi matlab koi restoring pull nahi. Particle beech se bina accelerate hue coast karta hai (yahan sari energy kinetic hai).

Step 3 — Extreme, velocity. $v = ω A^{2} - A^{2} = ω 0 = 0$ Yeh step kyun? Wall par zero room bacha hai; particle momentarily reverse karne ke liye ruk gaya hai.

Step 4 — Extreme, acceleration. $a = - ω^{2} A = - (25) (0.10) = - 2.5 m/s^{2} = ∣ a ∣_{m a x} (directed inward)$ Yeh step kyun? Ghar se sabse door = strongest restoring pull. Yahan sari energy potential hai.

Verify: Center: $v$ max, $a = 0$ . Extreme: $v = 0$ , $∣ a ∣$ max. Yeh perfect opposites hain — exactly wahi anti-phase relationship jo parent note ki table dikhati hai ( $a$ leads $x$ by $180°$ ). Units check: $ω A = 0.5 m/s$ , $ω^{2} A = 2.5 m/s^{2}$ ✓.

Neeche ki figure poore track mein dono formulas ek saath plot karti hai. Orange curve speed $v = ω A^{2} - x^{2}$ hai (solid = outbound $+$ branch, dashed = inbound $-$ branch): notice karo ki yeh center par peak karti hai aur dono walls par zero ho jaati hai. Teal straight line $a = - ω^{2} x$ hai: yeh center par zero hai aur walls par magnitude mein largest hai, neeche ki taraf slope karta hai kyunki $a$ hamesha $x$ ko oppose karta hai. Do markers is example ke do special points pin karte hain.

Figure s02 — Ex 3: speed (orange) ek downward arch hai jo $x = 0$ par peak karti hai; acceleration (teal) origin se guzarni ek straight line hai, edges par largest. Yeh complementary hain — jahan ek max hai, doosra zero hai.

Example 4 — Negative side, acceleration ka sign (Cell C5)

Worked example Statement

Same particle ( $A = 0.10$ , $ω = 5$ ). Yeh $x = - 0.06 m$ par hai (negative side), center ki taraf move kar raha hai. $v$ aur $a$ nikalo, signs ka dhyan rakhte hue.

Forecast: negative side par, restoring acceleration $+$ direction mein point karega ya $-$ direction mein?

Step 1 — Speed magnitude (pehle jaisa). $∣ v ∣ = ω A^{2} - x^{2} = 5 0.1 0^{2} - (- 0.06)^{2} = 5 0.0064 = 0.40 m/s$ Yeh step kyun? $x$ root ke andar squared hai, isliye $x = - 0.06$ same value deta hai jaise $x = + 0.06$ . Position ka sign speed magnitude ko affect nahi karta.

Step 2 — Velocity sign. Negative side se center ki taraf move karna matlab $+ x$ direction mein move karna: $v = + 0.40 m/s$ Yeh step kyun? $x < 0$ se "center ki taraf" increasing $x$ ke same direction mein point karta hai.

Step 3 — Acceleration ka sign flip hota hai. $a = - ω^{2} x = - (25) (- 0.06) = + 1.5 m/s^{2}$ Yeh step kyun? Kyunki $x$ negative hai, $- ω^{2} x$ positive ho jaata hai — pull center ki taraf hai, jo ab $+$ direction mein hai. Yahi $a = - ω^{2} x$ mein minus sign ka poora point hai: yeh hamesha ghar ki taraf aim karta hai, chahe aap kisi bhi side par ho.

Verify: Positive side par (Ex 1) $a$ tha $- 1.5$ ; negative side par yeh $+ 1.5$ hai. Dono $x = 0$ ki taraf point karte hain ✓. Restoring nature dono sides par hold karta hai, confirming Simple Harmonic Motion — definition a = −ω²x.

Example 5 — Degenerate inputs (Cell C6)

Worked example Statement

(a) $A = 0.10$ , $ω = 5$ wala particle reportedly $x = 0.10 m$ par hai. $v$ kya hai? (b) Ek "particle" $A = 0$ ke saath set up hai. Uski motion describe karo.

Forecast: kya particle kabhi $x > A$ par ho sakta hai? $A = 0$ physically kya matlab rakhta hai?

Step 1 — Exactly amplitude par. $v = ω A^{2} - A^{2} = ω 0 = 0$ Yeh step kyun? $0 = 0$ : particle turning point par frozen hai. Yeh boundary case hai — $x$ equal ho sakta hai $A$ ke, lekin exceed nahi kar sakta, kyunki tab $A^{2} - x^{2} < 0$ ho jaata hai aur negative number ka square root real nahi hota. Formula khud $∣ x ∣ > A$ ko forbid karta hai.

Step 2 — Forbidden zone. Agar koi claim kare $x = 0.12 > A$ : $A^{2} - x^{2} = 0.01 - 0.0144 = - 0.0044 < 0$ . Koi real speed exist nahi karti → situation is amplitude ke liye impossible hai. Maths aapko bata raha hai ki walls $\pm A$ par hain.

Step 3 — Zero amplitude. $A = 0$ ke saath: sab $x$ ke liye, $x = 0$ hona zaroori hai (single allowed point), aur $v = ω 0 - 0 = 0$ , $a = - ω^{2} (0) = 0$ . Particle hamesha ke liye center par still baith jaata hai — koi oscillating nahi, degenerate "oscillation."

Verify: (a) $v = 0$ at $∣ x ∣ = A$ agrees with Ex 3. (b) $A = 0 \Rightarrow$ no motion physically sensible hai: koi cheez usse displace nahi karti, isliye koi cheez restore nahi karti. Dono edge cases $v = ω A^{2} - x^{2}$ se consistent hain, real value sirf tab jab $∣ x ∣ \leq A$ ✓.

Example 6 — Maximum speed ka fraction (Cell C7)

Worked example Statement

Kis displacement $x$ par (as a fraction of $A$ ) speed exactly maximum ki ek tihaayi hoti hai? Phir $A = 0.10$ , $ω = 5$ ke liye evaluate karo: $x$ aur accompanying acceleration nikalo.

Forecast: kya $x$ center ke closer hoga ya edge ke? (Yaad karo half-max already $0.866 A$ par hua tha.)

Step 1 — Ratio set up karo. $ω A^{2} - x^{2} = \frac{1}{3} ω A$ Yeh step kyun? $v_{m a x} = ω A$ ; "max ka ek tihaayi" matlab left side $\frac{1}{3} ω A$ ke barabar hai. $ω$ s cancel ho jayenge — speed ka fraction is par depend nahi karta ki oscillation kitni tezi se hai.

Step 2 — $ω$ cancel karo aur dono sides square karo. $A^{2} - x^{2} = \frac{1}{3} A \Rightarrow A^{2} - x^{2} = \frac{1}{9} A^{2}$ Yeh step kyun? Squaring square root hata deta hai taaki hum $x^{2}$ solve kar sakein. (Squaring safe hai yahan: dono sides positive lengths hain.)

Step 3 — $x$ solve karo. $x^{2} = A^{2} - \frac{1}{9} A^{2} = \frac{8}{9} A^{2} \Rightarrow x = \frac{8}{9} A = \frac{2 2}{3} A \approx 0.943 A$ Yeh step kyun? Speed ko poori tarah ek tihaayi tak drop karne ke liye, wall ke bahut paas hona padega — speed edge ke paas steeply collapse hoti hai (yeh parent ke half-speed-at- $0.866 A$ result se match karta hai).

Step 4 — Numbers. $x \approx 0.943 \times 0.10 = 0.0943 m, a = - ω^{2} x = - 25 (0.0943) \approx - 2.36 m/s^{2}$ Yeh step kyun? Jab $x$ pata chal gaya, acceleration seedha $a = - ω^{2} x$ se milta hai.

Verify: Speed check karo: $v = 5 0.1 0^{2} - 0.094 3^{2} = 5 0.01 - 0.008889 = 5 0.001111 \approx 5 (0.03333) = 0.1667 m/s$ , aur $\frac{1}{3} v_{m a x} = \frac{1}{3} (0.5) = 0.1667 m/s$ ✓. $ω$ cancel hua, isliye fraction $0.943 A$ kisi bhi SHM ke liye universal hai.

Example 7 — Real-world word problem (Cell C8)

Worked example Statement

Ek loudspeaker cone SHM mein oscillate karta hai period $T = 0.004 s$ ke saath aur total peak-to-peak travel $2.0 mm$ hai. (a) uski maximum speed, aur (b) center se $0.5 mm$ door hone par uski speed nikalo.

Forecast: speakers fast vibrate karte hain — kya aap max speed centimetres/s ya metres/s mein expect karte ho?

Step 1 — Words ko symbols mein translate karo. Peak-to-peak $= 2 A$ , isliye amplitude $A = \frac{1}{2} (2.0 mm) = 1.0 mm = 1.0 \times 1 0^{- 3} m$ . Period se angular frequency: $ω = \frac{2 π}{T} = \frac{2 π}{0.004} = 500 π \approx 1570.8 rad/s$ . Yeh step kyun? "Peak-to-peak" full swing hai = amplitude ka double — ek classic word-problem trap. Aur $ω = 2 π / T$ ek measured time period se hamare formulas tak ka bridge hai (dekho Angular frequency ω and time period T).

Step 2 — Maximum speed. $v_{m a x} = ω A = 1570.8 \times 1.0 \times 1 0^{- 3} \approx 1.571 m/s$ Yeh step kyun? Max speed center par hoti hai, $v = ω A$ . Tiny amplitude ke bawajood, huge $ω$ isse fast banata hai — isliye speakers air ko audibly move karte hain.

Step 3 — $x = 0.5 mm$ par speed. $v = ω A^{2} - x^{2} = 1570.8 (1.0 \times 1 0^{- 3})^{2} - (0.5 \times 1 0^{- 3})^{2}$ $= 1570.8 (1.0 - 0.25) \times 1 0^{- 6} = 1570.8 0.75 \times 1 0^{- 3} \approx 1570.8 (0.8660) (1 0^{- 3}) \approx 1.360 m/s$ Yeh step kyun? $x = 0.5 mm = \frac{1}{2} A$ , isliye position form mein plug karte hain. Koi time nahi chahiye.

Verify: $x = \frac{1}{2} A$ par, $v / v_{m a x} = 1 - (1/2)^{2} = 3 /2 \approx 0.866$ , aur indeed $1.360/1.571 \approx 0.866$ ✓. Units: $rad/s \times m = m/s$ ✓. Answer metres/second mein hai, jaise ek fast small oscillation ke liye forecast tha.

Example 8 — Exam twist: $A$ aur $ω$ back-solve karo (Cell C9)

Worked example Statement

Examiner batata hai: ek SHM particle ka kisi instant par speed $v = 3 m/s$ aur acceleration magnitude $∣ a ∣ = 16 m/s^{2}$ displacement $x = 0.04 m$ par hai. $ω$ nikalo, phir amplitude $A$ .

Forecast: aapke paas do measured quantities hain aur do unknowns — kaun sa formula $ω$ seedha $A$ ki zaroorat ke bina deta hai?

Step 1 — Acceleration se $ω$ nikalo (isme $A$ nahi hai). $∣ a ∣ = ω^{2} ∣ x ∣ \Rightarrow ω^{2} = \frac{∣ a ∣}{∣ x ∣} = \frac{16}{0.04} = 400 \Rightarrow ω = 20 rad/s$ Yeh step kyun? Law $a = - ω^{2} x$ sirf $a$ , $ω$ , $x$ ko link karta hai — teen quantities jo ya to known hain ya wanted hain, koi $A$ nahi. Magnitudes lena sign hata deta hai, aur divide karna $ω^{2}$ isolate kar deta hai. Pehle $ω$ solve karo taaki baaki sab unlock ho.

Step 2 — $A$ nikalne ke liye velocity use karo. $v = ω A^{2} - x^{2} \Rightarrow A^{2} - x^{2} = \frac{v}{ω} = \frac{3}{20} = 0.15$ Yeh step kyun? Ab $ω$ known hai, velocity formula mein ek hi unknown hai, $A$ . Dono sides ko $ω$ se divide karo taaki root se strip ho jaye.

Step 3 — Square karo aur $A$ solve karo. $A^{2} - x^{2} = 0.1 5^{2} = 0.0225 \Rightarrow A^{2} = 0.0225 + 0.0 4^{2} = 0.0225 + 0.0016 = 0.0241$ $A = 0.0241 \approx 0.1552 m$ Yeh step kyun? Squaring square root undo karta hai (safe: dono sides positive), phir $x^{2}$ cross karke add karte hain aur final root lete hain amplitude nikalne ke liye.

Verify: Answers ko seedha dono master formulas mein plug karo. Acceleration: $∣ a ∣ = ω^{2} x = 400 (0.04) = 16 m/s^{2}$ ✓. Velocity: $v = 20 0.0241 - 0.0016 = 20 0.0225 = 20 (0.15) = 3 m/s$ ✓. Sanity: $∣ x ∣ = 0.04 < A = 0.155$ , isliye reported point legitimately range ke andar hai ✓. Bonus ke roop mein, time period $T = 2 π / ω \approx 0.314 s$ hai.

Recall Matrix par quick self-test

"Same $x$ , opposite travel direction" kaun si cell hai? ::: C2 — speed identical, velocity sign flipped. Negative side par, jab $x < 0$ ho to acceleration ka sign kya hoga? ::: Positive, kyunki $a = - ω^{2} x$ aur $x$ negative hai — phir bhi center ki taraf point karta hai. Agar aap formula mein $∣ x ∣ > A$ feed karo to kya hoga? ::: Yeh negative number ka square root demand karta hai → physically impossible; walls $\pm A$ par hain. Amplitude ke kis fraction par speed maximum ki ek tihaayi hoti hai? ::: $x = \frac{2 2}{3} A \approx 0.943 A$ . Ek word problem mein, "peak-to-peak = 2 mm" se amplitude kya milta hai? ::: $A = 1 mm$ (peak-to-peak ka aadha).

Connections

Velocity and acceleration in SHM — v = ω√(A² − x²) (parent)
Simple Harmonic Motion — definition a = −ω²x
Energy in SHM — kinetic and potential
Phase and phase difference
Reference circle (projection of uniform circular motion)
Angular frequency ω and time period T
Pythagorean identity sin² + cos² = 1