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

WHY does energy stay constant?

WHY: In ideal SHM the only force is the restoring force $F=-kx$ . This force is conservative (it stores energy as PE, gives it back exactly, no friction). With no friction or drag to bleed energy away, total mechanical energy is conserved.

WHAT is each piece?

KE = energy because the mass is moving: $\frac12 m v^2$ .
PE = energy stored in the deformation (spring stretch / pendulum height): $\frac12 k x^2$ .

HOW do they trade off? When $x=\pm A$ (turning points) the mass is momentarily at rest → all energy is PE. When $x=0$ (equilibrium) the mass moves fastest → all energy is KE.

Derivation from first principles

Step 1 — Build PE from the force (Why? Energy is the work done against the spring). The work to stretch the spring from $0$ to $x$ against $F_{\text{spring}}=-kx$ : $U(x)=-\int_0^x F_{\text{spring}}\,dx' = \int_0^x kx'\,dx' = \tfrac12 k x^2.$ Why this step? PE is stored work; integrating $kx'$ accumulates that work.

Step 2 — Write KE. $K=\tfrac12 m v^2 = \tfrac12 m A^2\omega^2\sin^2(\omega t+\phi).$ Since $\omega^2=k/m \Rightarrow m\omega^2=k$ : $K=\tfrac12 k A^2\sin^2(\omega t+\phi).$ Why this step? Replacing $m\omega^2$ by $k$ lets KE and PE be compared in the same units of $k$ and $A$ .

Step 3 — Write PE in time. $U=\tfrac12 k x^2 = \tfrac12 k A^2\cos^2(\omega t+\phi).$

Step 4 — Add them. $E=K+U=\tfrac12 kA^2\big[\sin^2(\cdot)+\cos^2(\cdot)\big]=\tfrac12 kA^2.$ Why this step? The Pythagorean identity $\sin^2+\cos^2=1$ kills the time dependence → constant.

Deriving $K$ as a function of $x$ (very handy): $K=E-U=\tfrac12 kA^2-\tfrac12 kx^2=\tfrac12 k(A^2-x^2).$ And since $K=\tfrac12 mv^2$ : $\tfrac12 mv^2=\tfrac12 k(A^2-x^2)\;\Rightarrow\; v=\pm\sqrt{\tfrac{k}{m}}\sqrt{A^2-x^2}=\pm\omega\sqrt{A^2-x^2}.$

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

Key features (80/20 — learn these first)

Total $E\propto A^2$ : double the amplitude → 4× the energy.
$E\propto \omega^2$ (since $E=\frac12 m\omega^2A^2$ ): stiffer/faster oscillator stores more.
KE and PE each oscillate at twice the frequency of $x(t)$ (because of the $\sin^2,\cos^2$ ).
Average over a cycle: $\langle K\rangle=\langle U\rangle=\tfrac14 kA^2=\tfrac12 E$ . Energy is shared 50–50 on average.

Common mistakes

Total mechanical energy in SHM

E=\tfrac12 kA^2

(constant, set by amplitude)

Why is total energy constant in ideal SHM?

Only the conservative restoring force acts; no friction, so mechanical energy is conserved

PE as a function of displacement

U=\tfrac12 kx^2

KE as a function of displacement

K=\tfrac12 k(A^2-x^2)

Speed at displacement

x

v=\omega\sqrt{A^2-x^2}

Where is speed maximum?

x=0

(equilibrium);

v_{max}=\omega A

Where is PE maximum?

At the turning points

x=\pm A

;

U=\tfrac12 kA^2

At what

x

are KE and PE equal?

x=\pm A/\sqrt2

How does total energy scale with amplitude?

E\propto A^2

(double

A

→ 4× energy)

Frequency of KE/PE oscillation vs displacement

Twice the frequency (

2\omega

)

Average KE and PE over a cycle

Each equals

\tfrac14 kA^2 = \tfrac12 E

Express E in terms of

m,\omega,A

E=\tfrac12 m\omega^2A^2

Recall Feynman: explain to a 12-year-old

Imagine a swing. At the very top of a swing you stop for a tiny moment — you're not moving, but you're high up, "loaded" with stored energy. As you whoosh down to the bottom, all that stored energy becomes speed, and you're moving fastest at the lowest point. Then it loads back up on the other side. The total "fun energy" stays the same — it just keeps changing costume between stored (height/stretch) and moving (speed). Pull the swing back further (bigger $A$ ) and there's more total energy — in fact 4 times as much if you pull back twice as far!

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, SHM mein energy ka khel bilkul ek jhoole jaisa hai. Do jagah energy store hoti hai: ek kinetic energy (KE) jo movement ki wajah se hoti hai, aur ek potential energy (PE) jo spring ke stretch ya squeeze mein store hoti hai. Jab tak koi friction nahi hai, total energy fix rehti hai — bas KE aur PE aapas mein swap karte rehte hain. Ye total hamesha $\frac12 kA^2$ hota hai, jo sirf amplitude $A$ par depend karta hai.

Intuition simple hai: edges par (jab $x=\pm A$ ) mass ek pal ke liye ruk jaata hai, $v=0$ , toh poori energy PE ban jaati hai. Centre mein ( $x=0$ ) spring relaxed hota hai, force zero, par speed maximum — toh poori energy KE ban jaati hai. Isliye yaad rakho: "PE Edges par, KE Centre mein."

Formula derive karna bhi aasaan hai. PE nikalte ho spring force $kx$ ko integrate karke: $U=\frac12 kx^2$ . KE hai $\frac12 mv^2$ , aur $v=-A\omega\sin(\omega t)$ daalne par $K=\frac12 kA^2\sin^2$ . Inko jodo toh $\sin^2+\cos^2=1$ ki wajah se time gayab ho jaata hai aur bachta hai constant $\frac12 kA^2$ . Ek important baat: $E\propto A^2$ , yaani amplitude double karo toh energy 4 guna ho jaati hai — exam mein ye trap bahut aata hai!

Aur ek aur cheez — KE aur PE displacement se double frequency ( $2\omega$ ) par oscillate karte hain, kyunki wo $\sin^2$ aur $\cos^2$ par depend karte hain. Cycle ke average mein dono barabar: $\langle K\rangle=\langle U\rangle=\frac14 kA^2$ . Bas yahi 20% concepts 80% questions cover kar dete hain.

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

WHY does energy stay constant?

Derivation from first principles

Key features (80/20 — learn these first)

Common mistakes

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Oscillations & Waves

Connections