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

1.6.5 · Physics › Oscillations & Waves

Energy constant kyun rehti hai?

KYUN: Ideal SHM mein sirf ek hi force hoti hai — restoring force $F=-kx$. Yeh force conservative hai (energy ko PE ke roop mein store karti hai, phir exactly wapas deti hai, koi friction nahi). Jab friction ya drag energy nahi chura rahi, toh total mechanical energy conserved rehti hai.

KYA hai har piece?

• KE = energy kyunki mass move kar raha hai: $\frac{1}{2}m{v}^2$.
• PE = deformation mein stored energy (spring stretch / pendulum height): $\frac{1}{2}k{x}^2$.

KAISE trade off hoti hai? Jab $x=\pm A$ (turning points) hota hai toh mass ek pal ke liye ruk jaata hai → saari energy PE hai. Jab $x=0$ (equilibrium) hota hai toh mass sabse tez chalti hai → saari energy KE hai.

First principles se derivation

Step 1 — PE ko force se nikalo (Kyun? Energy woh work hai jo spring ke against ki gayi hai). Spring ko $0$ se $x$ tak $F_{\text{spring}}=-kx$ ke against stretch karne ka work: $U(x)=-{\int }_0^x{F}_{\text{spring}}d{x}^{\prime }={\int }_0^{x}k{x}^{\prime }d{x}^{\prime }=\frac{1}{2}k{x}^2.$ Yeh step kyun? PE stored work hai; $k{x}^{\prime }$ ko integrate karne se woh work accumulate hoti hai.

Step 2 — KE likho. $K=\frac{1}{2}m{v}^2=\frac{1}{2}m{A}^2{\omega }^2{\sin }^2(\omega t+\varphi ).$ Kyunki ${\omega }^2=k/m\Rightarrow m{\omega }^2=k$: $K=\frac{1}{2}k{A}^2{\sin }^2(\omega t+\varphi ).$ Yeh step kyun? $m{\omega }^2$ ko $k$ se replace karne se KE aur PE ko $k$ aur $A$ ki same units mein compare kiya ja sakta hai.

Step 3 — PE ko time mein likho. $U=\frac{1}{2}k{x}^2=\frac{1}{2}k{A}^2{\cos }^2(\omega t+\varphi ).$

Step 4 — Dono ko add karo. $E=K+U=\frac{1}{2}k{A}^2[{\sin }^2(\cdot )+{\cos }^2(\cdot )]=\frac{1}{2}k{A}^2.$ Yeh step kyun? Pythagorean identity ${\sin }^2+{\cos }^2=1$ time dependence ko khatam kar deti hai → constant.

$K$ ko $x$ ke function ke roop mein derive karna (bahut kaam ka): $K=E-U=\frac{1}{2}k{A}^2-\frac{1}{2}k{x}^2=\frac{1}{2}k(A^2-x^2).$ Aur kyunki $K=\frac{1}{2}m{v}^2$: $\frac{1}{2}m{v}^2=\frac{1}{2}k(A^2-x^2)\Rightarrow v=\pm \sqrt{\frac{k}{m}}\sqrt{A^2-x^2}=\pm \omega \sqrt{A^2-x^2}.$

Key features (80/20 — pehle yeh seekho)

• Total $E\propto A^2$: amplitude double karo → energy 4× ho jaati hai.
• $E\propto {\omega }^2$ (kyunki $E=\frac{1}{2}m{\omega }^2{A}^2$): stiffer/faster oscillator zyada energy store karta hai.
• KE aur PE dono $x(t)$ ki frequency se double frequency pe oscillate karte hain (${\sin }^2,{\cos }^2$ ki wajah se).
• Ek cycle pe average: $⟨K⟩=⟨U⟩=\frac{1}{4}k{A}^2=\frac{1}{2}E$. Average mein energy 50–50 share hoti hai.

Common mistakes

Concept Map

Deep Dive

• D1 · Foundations — har symbol zero se
• D2 · Visual walkthrough — derivation pictures mein
• D3 · Worked examples — har scenario
• D4 · Exercises — graded, full solutions ke saath
• D5 · Question bank — concept traps