Energy in SHM — KE + PE = ½kA² (constant): Worked examples

1.6.5 · D3 · Physics › Oscillations & Waves › Energy in SHM — KE + PE = ½kA² (constant)

Shuru karne se pehle, main toolbox ko seedhe shabdon mein dobara bata deta hoon taaki koi bhi symbol yahaan unearned na lage:

Scenario matrix

Har energy question actually inhi cases mein se ek hota hai. Right column us example ka naam batata hai jo use cover karta hai.

#	Case class	Kya tricky hai	Covered by
C1	Positive $x$ inside the range ( $0 < x < A$ )	standard "yahaan speed / KE / PE find karo"	Ex 1
C2	Negative $x$ ( $x < 0$ )	kya $x$ ka sign $U$ ya $v$ ko change karta hai?	Ex 2
C3	Degenerate: $x = 0$ (centre)	speed maximum hai, PE = 0	Ex 3
C4	Degenerate: $x = \pm A$ (turning point)	speed zero hai, saari energy PE hai	Ex 3
C5	Fraction/ratio question ("kis $x$ par KE = 3·PE hai?")	energy split se position solve karo	Ex 4
C6	Limiting / scaling ( $A$ , $k$ , ya $m$ badlo)	$E$ , $v_{ma x}$ kaise respond karte hain	Ex 5
C7	Real-world word problem (car suspension)	words → symbols → numbers mein translate karo	Ex 6
C8	Exam twist: energy diya hai, backwards work karo	tumhe $E$ aur $v$ diya gaya hai, $A$ find karo	Ex 7
C9	Impossible / boundary check ( $x > A$ ?)	physically forbidden input ko pehchano	Ex 8

Neeche di gayi figure woh map hai jis par har example rehta hai: energy "bowl" $U = \frac{1}{2} k x^{2}$ (black parabola), flat ceiling $E = \frac{1}{2} k A^{2}$ (red), aur — yahi almost har example mein use hone wala key move hai — KE hamesha bowl aur ceiling ke beech ka vertical gap hota hai. Padhte waqt yeh picture open rakho: Ex 1 gap ko $x = 0.08$ par padhta hai, Ex 3 gap ko centre aur edges par padhta hai, Ex 4 woh jagah dhundta hai jahan gap bowl height ka teen guna ho.

Figure — Energy in SHM — KE + PE = ½kA² (constant)

Worked examples

Worked example Example 1 — C1: range ke andar positive

$x$ Ek mass spring par $k = 200$ N/m, $m = 0.50$ kg, amplitude $A = 0.12$ m ke saath hai. $x = + 0.08$ m par speed aur KE/PE split find karo.

Forecast: guess karo — kya mass centre par jitna fast tha, usse zyada fast hai ya kam? (Yeh slower hona chahiye: yeh middle se door hai.)

Total energy. $E = \frac{1}{2} k A^{2} = \frac{1}{2} (200) (0.12)^{2} = 1.44$ J. (figure mein red ceiling.) Yeh step kyun? Ceiling sirf $A$ se fix hoti hai — baaki sab usi se nikalta hai.
PE yahaan. $U = \frac{1}{2} k x^{2} = \frac{1}{2} (200) (0.08)^{2} = 0.64$ J. (black bowl ki height $x = 0.08$ par.) Yeh step kyun? $U$ sirf is baat par depend karta hai ki tum kitne door ho.
KE gap hai. $K = E - U = 1.44 - 0.64 = 0.80$ J. (yeh exactly woh dashed gap hai jo figure mein draw hai.) Yeh step kyun? Energy conserved hai, toh jo stored nahi hai woh motion hai.
Speed. $ω = 200/0.5 = 400 = 20$ rad/s. $∣ v ∣ = ω A^{2} - x^{2} = 20 0.1 2^{2} - 0.0 8^{2} = 20 0.0080 = 20 (0.08944) = 1.789$ m/s. Yeh step kyun? $A^{2} - x^{2}$ $A$ ke paas jaane par shrink hota hai — mass edge ke paas slow hota hai, bilkul jaise forecast tha.

Verify: $\frac{1}{2} m ∣ v ∣^{2} = \frac{1}{2} (0.5) (1.789)^{2} = 0.80$ J $= K$ . ✓ Units: N/m·m² = J. ✓

Worked example Example 2 — C2: negative displacement

Same spring as Ex 1 ( $k = 200$ , $m = 0.5$ , $A = 0.12$ ). Ab $x = - 0.08$ m par speed aur PE find karo.

Forecast: kya answers Ex 1 se alag honge? Guess: nahi — kyunki energy sirf centre se doori ki parwah karti hai, direction ki nahi.

PE. $U = \frac{1}{2} k x^{2} = \frac{1}{2} (200) (- 0.08)^{2} = 0.64$ J. Yeh step kyun? $x$ squared hai, toh $(- 0.08)^{2} = (+ 0.08)^{2}$ . Minus sign khatam ho jaata hai. (Physically: $x < 0$ matlab spring compressed hai, lekin compression utni hi energy store karti hai jitni same amount ki stretch — isliye hum $U$ ko elastic potential energy kehte hain, "stretch" energy nahi.)
Speed magnitude. $∣ v ∣ = ω A^{2} - x^{2} = 20 0.1 2^{2} - (- 0.08)^{2} = 1.789$ m/s. Yeh step kyun? Phir se $x$ squared appear hota hai — $x$ ka sign speed ke liye invisible hai.
Sign kya BATATA hai. $v = \pm ω A^{2} - x^{2}$ mein $\pm$ velocity direction hai (left vs right), na ki tum centre ke kis taraf ho. $x = - 0.08$ par mass kisi bhi direction mein move kar sakta hai; energy $x = + 0.08$ ke barabar hai. Yeh step kyun? "Kis taraf" (sign of $x$ ) aur "kis direction mein move kar raha hai" (sign of $v$ ) ko alag karna is case ka pura point hai.

Verify: Ex 1 se symmetry ke through identical: $U = 0.64$ J, $∣ v ∣ = 1.789$ m/s. ✓ Figure mein energy bowl ek perfect parabola hai — $x = 0$ ke baare mein symmetric. ✓

Worked example Example 3 — C3 & C4: do degenerate points (

$x = 0$ aur $x = \pm A$ ) Same spring. (a) centre $x = 0$ aur (b) turning point $x = + A = 0.12$ m par energy split aur speed find karo.

Forecast: mass sabse fast kahan hai? Kahan momentarily frozen hai? Swing ko bottom vs top par imagine karo. Figure mein: gap sabse wide kahan hai, aur kahan zero ho jaata hai?

(a) Centre, $x = 0$ : PE. $U = \frac{1}{2} k (0)^{2} = 0$ J. (bowl zero ko touch karta hai — gap apni sabse wide state mein hai.) Yeh step kyun? Koi deformation nahi, koi stored energy nahi.
(a) Centre: KE = sab kuch. $K = E - U = 1.44 - 0 = 1.44$ J, toh $v_{ma x} = ω A = 20 (0.12) = 2.40$ m/s. Yeh step kyun? $x = 0$ par sara budget motion hai — yeh mass ki sabse fast speed hai.
(b) Turning point, $x = A$ : PE = sab kuch. $U = \frac{1}{2} k A^{2} = 1.44$ J $= E$ . (bowl red ceiling tak uth jaata hai — gap zero ho jaata hai.) Yeh step kyun? Fully stretched → har joule stored hai.
(b) Turning point: KE aur speed. $K = E - U = 0$ , toh $∣ v ∣ = ω A^{2} - A^{2} = 0$ . Mass momentarily ruk jaata hai. Yeh step kyun? Yeh exactly woh instant hai jab palti ho rahi hai — jaise swing ka top.

Verify: $v_{ma x} = ω A = 20 \times 0.12 = 2.40$ m/s aur $\frac{1}{2} m v_{ma x}^{2} = \frac{1}{2} (0.5) (2.4)^{2} = 1.44$ J $= E$ . ✓ $x = A$ par, $A^{2} - x^{2} = 0 \Rightarrow v = 0$ . ✓

Worked example Example 4 — C5: energy split se position solve karo

Same spring ( $A = 0.12$ m). Kis displacement par kinetic energy potential energy ki teen guni hai ( $K = 3 U$ )?

Forecast: figure mein, hum woh point chahte hain jahan gap (KE) bowl height (PE) se teen guna ho. Kya yeh $x$ centre ke paas hoga ya edge ke? Agar KE dominate kar raha hai, toh hum fast centre ke paas hain — toh chhota $x$ expect karo.

$E$ use karke condition likho. $K = 3 U$ aur $K + U = E$ dete hain $3 U + U = E \Rightarrow U = \frac{1}{4} E$ . Yeh step kyun? Ratio ko fixed total ke fraction mein badlo — solve karna bahut aasaan ho jaata hai.
Formulas substitute karo. $\frac{1}{2} k x^{2} = \frac{\frac{1}{2} k A ^{2}}{4} = \frac{1}{8} k A^{2}$ . Dono sides ko $\frac{2}{k}$ se multiply karo (common $\frac{1}{2} k$ cancel hoga): $x^{2} = \frac{1}{4} A^{2}$ . Yeh step kyun? Common $\frac{1}{2} k$ cancel karne se pata chalta hai ki answer sirf $A$ par depend karta hai, $k$ ya $m$ par nahi.
Solve karo. $x = \pm \frac{A}{2} = \pm 0.06$ m. Yeh step kyun? Dono signs valid hain — condition symmetric hai, Ex 2 ka lesson match karta hai.

Verify: $x = 0.06$ par: $U = \frac{1}{2} (200) (0.06)^{2} = 0.36$ J, $K = E - U = 1.44 - 0.36 = 1.08$ J, aur $1.08 = 3 (0.36)$ . ✓

Worked example Example 5 — C6: limiting & scaling behaviour

Ek spring oscillator ki energy $E_{1} = 1.44$ J hai. Ab tum (a) amplitude double karte ho, aur alag se (b) spring constant $k$ chaar guna karte ho. $E$ aur $v_{ma x}$ ka kya hoga?

Forecast: $E \propto A^{2}$ , toh $A$ double karne par $E$ double se zyada badh jaana chahiye — factor guess karo.

(a) $A$ double karo. $E \propto A^{2} \Rightarrow E$ $2^{2} = 4$ se multiply hota hai: new $E = 4 \times 1.44 = 5.76$ J. Yeh step kyun? $A$ $E = \frac{1}{2} k A^{2}$ mein squared hai, toh scaling quadratic hai.
(a) $v_{ma x}$ par effect. $v_{ma x} = ω A$ ; $ω = k / m$ unchanged hai, toh $v_{ma x}$ sirf double hoti hai. Yeh step kyun? $v_{ma x}$ $A$ mein linear hai jabki energy quadratic hai — yeh ek common exam trap hai.
(b) $k$ chaar guna karo ( $A$ , $m$ fix). $E = \frac{1}{2} k A^{2} \Rightarrow E$ 4 se multiply hota hai: new $E = 5.76$ J. Lekin $ω = k / m$ $4 = 2$ se multiply hota hai, toh $v_{ma x} = ω A$ double bhi hoti hai. Yeh step kyun? Stiffness energy linearly badhati hai lekin speed sirf $k$ ke through.

Verify: (a) $4 \times 1.44 = 5.76$ J aur $v_{ma x}$ new $= 2 \times 2.40 = 4.80$ m/s. (b) $4 \times 1.44 = 5.76$ J; $ω_{n e w} = 800/0.5 = 40$ , $v_{ma x} = 40 \times 0.12 = 4.80$ m/s. ✓

Worked example Example 6 — C7: real-world word problem (car suspension)

Ek car ka ek corner $k = 1.5 \times 1 0^{4}$ N/m ke spring par $m = 300$ kg carry karte hue baitha hai. Ek bump ke baad woh corner amplitude $A = 0.04$ m ke saath oscillate karta hai (shock absorbers matter karne se pehle). Total oscillation energy aur us corner ki maximum speed find karo.

Forecast: chhota amplitude lekin bada stiffness — guess karo $E$ zyada 1 J jaisi hai ya 10 J jaisi.

Words ko symbols mein translate karo. "carries $m$ " → mass, "spring $k$ ", "amplitude $A$ ". Koi nayi physics nahi — same equations. Yeh step kyun? Word problems mein skill variables ko sahi naam dene ki hai, naye formulas ki nahi.
Total energy. $E = \frac{1}{2} k A^{2} = \frac{1}{2} (1.5 \times 1 0^{4}) (0.04)^{2} = \frac{1}{2} (15000) (0.0016) = 12$ J. Yeh step kyun? $A$ aur $k$ se directly fixed ceiling.
Max speed. $ω = k / m = 15000/300 = 50 = 7.071$ rad/s. $v_{ma x} = ω A = 7.071 \times 0.04 = 0.2828$ m/s. Yeh step kyun? Fastest point equilibrium par hai, $x = 0$ .

Verify: $\frac{1}{2} m v_{ma x}^{2} = \frac{1}{2} (300) (0.2828)^{2} = 12$ J $= E$ . ✓ Units: $(N/m) / kg = 1/ s^{2} =$ rad/s. ✓

Worked example Example 7 — C8: exam twist, energy & speed diya hai,

$A$ backwards nikalo $0.25$ kg ka ek mass $k = 100$ N/m ke spring par oscillate kar raha hai. Jis instant par uski speed $∣ v ∣ = 1.2$ m/s hai, potential energy $U = 0.14$ J measure ki gayi. Amplitude $A$ aur us instant par displacement $x$ find karo.

Forecast: humein pieces diye gaye hain aur pura rebuild karna hai — Ex 1 ka ulta.

Speed se KE. $K = \frac{1}{2} m ∣ v ∣^{2} = \frac{1}{2} (0.25) (1.2)^{2} = 0.18$ J. Yeh step kyun? Speed directly motion energy deti hai.
Total energy = sum. $E = K + U = 0.18 + 0.14 = 0.32$ J. Yeh step kyun? Energy conserved hai, toh kisi bhi instant par $K + U = E$ .
$E$ se Amplitude. $E = \frac{1}{2} k A^{2} \Rightarrow A = 2 E / k = 2 (0.32) /100 = 0.0064 = 0.08$ m. Yeh step kyun? Ceiling equation ko invert karo $A$ nikalne ke liye.
$U$ se Displacement. $U = \frac{1}{2} k x^{2} \Rightarrow x = \pm 2 U / k = \pm 2 (0.14) /100 = \pm 0.0028 = \pm 0.0529$ m. Yeh step kyun? $U$ $x$ ka magnitude pin karta hai; dono signs valid hain (Ex 2 ka lesson phir se).

Verify: check karo $K = \frac{1}{2} k (A^{2} - x^{2}) = \frac{1}{2} (100) (0.0 8^{2} - 0.052 9^{2}) = \frac{1}{2} (100) (0.0064 - 0.0028) = \frac{1}{2} (100) (0.0036) = 0.18$ J $= K$ . ✓

Worked example Example 8 — C9: impossible input (boundary check)

Ek student se Ex 1 ke spring par ( $A = 0.12$ m) $x = 0.15$ m par speed poochi gayi. Kya galat ho jaata hai, aur yeh hamein kya sikhata hai?

Forecast: plug in karne ki koshish karo aur dekho maths kaise complain karta hai. Figure mein, $x = 0.15$ bowl ki opening ke bahar hai.

Plug in karo. $∣ v ∣ = ω A^{2} - x^{2} = 20 0.1 2^{2} - 0.1 5^{2} = 20 0.0144 - 0.0225 = 20 - 0.0081$ . Yeh step kyun? Humein negative number ka square root milta hai — yeh real speed nahi hai.
Interpret karo. $x = 0.15 > A = 0.12$ allowed range ke bahar hai. Mass wahan kabhi nahi pahunchta: amplitude $A$ definition se sabse door point hai. Yeh step kyun? Root ke neeche negative number nature ka tarika hai yeh kehne ka ki "yeh position kabhi nahi hoti."
Rule. Physically valid displacements $- A \leq x \leq A$ obey karte hain. Boundary $x = \pm A$ par root exactly zero hai ( $v = 0$ ), Ex 3 ka turning point. Yeh step kyun? Forbidden inputs ko pehchanna tumhe exam mein nonsense report karne se bachata hai.

Verify: $A^{2} - x^{2} = 0.0144 - 0.0225 = - 0.0081 < 0$ , confirm karta hai ki $x > A$ ke liye koi real speed exist nahi karti. ✓

Recall Quick self-test (answers cover karo)

$k = 200, m = 0.5, A = 0.12$ ke liye $x = 0.08$ par speed ::: $1.789$ m/s $K = 3 U$ kahan hoga? ::: $x = \pm A /2$ Amplitude double karne se $E$ kitne factor se change hoti hai? ::: $\times 4$ Amplitude double karne se $v_{ma x}$ kitne factor se change hoti hai? ::: $\times 2$ $x = 1.25 A$ error kyun deta hai? ::: yeh $- A \leq x \leq A$ ke bahar hai; root negative ho jaata hai, koi real speed nahi