Prerequisites
Before studying this topic, make sure you understand:
- Simple Harmonic Motion - SHM fundamentals, equations
- Work-Energy-Power - Kinetic and potential energy concepts
- Conservation of Energy - Energy conservation principle
The Hook: The Pendulum Paradox
Ever wondered why a swing keeps going back to the same height (ignoring friction)? When you’re at the top, you’re not moving — all your energy is “stored.” At the bottom, you’re moving fastest — all energy is in motion. Where does the energy go and come from?
The Marvel Connection: In Doctor Strange in the Multiverse of Madness, when magical energy shields form and flex, they’re like giant springs — storing energy when compressed, releasing it when expanding. That’s energy transformation in SHM!
The Question: If a pendulum has zero velocity at the top and maximum velocity at the bottom, where exactly is half the energy kinetic and half potential?
Energy in SHM: The Core Idea
In SHM, energy continuously transforms between two forms:
- Kinetic Energy (KE) - energy of motion
- Potential Energy (PE) - energy of position
But the total mechanical energy remains constant (in ideal SHM without damping).
In simple terms: SHM is like a perfect energy exchange. When you’re moving fast (high KE), you’re near equilibrium (low PE). When you stop at extremes (zero KE), all energy is stored (maximum PE).
The Golden Rule:
$$\boxed{E_{total} = KE + PE = \text{constant}}$$Interactive Demo
Watch energy transformation during SHM oscillation:
Kinetic Energy in SHM
At position $x$ with velocity $v$:
$$KE = \frac{1}{2}mv^2$$Since $v = \omega\sqrt{A^2 - x^2}$ in SHM:
$$\boxed{KE = \frac{1}{2}m\omega^2(A^2 - x^2)}$$Alternative form using spring constant $k = m\omega^2$:
$$\boxed{KE = \frac{1}{2}k(A^2 - x^2)}$$Special Cases
At equilibrium ($x = 0$):
$$KE_{max} = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}kA^2$$At extremes ($x = \pm A$):
$$KE_{min} = 0$$KE is ZERO at EXTREME positions
When displacement is maximum, velocity (and KE) is zero.
Potential Energy in SHM
For Spring-Mass System
The restoring force is $F = -kx$, so work done against this force stores potential energy:
$$PE = \frac{1}{2}kx^2$$Using $k = m\omega^2$:
$$\boxed{PE = \frac{1}{2}m\omega^2 x^2}$$Special Cases
At equilibrium ($x = 0$):
$$PE_{min} = 0$$At extremes ($x = \pm A$):
$$PE_{max} = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2$$PE is MAXIMUM at EXTREME positions
When displacement is maximum, all energy is potential.
Total Mechanical Energy
Adding KE and PE:
$$E = KE + PE$$ $$E = \frac{1}{2}m\omega^2(A^2 - x^2) + \frac{1}{2}m\omega^2 x^2$$ $$E = \frac{1}{2}m\omega^2 A^2$$Total Energy (constant):
$$\boxed{E = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}kA^2}$$Key Insights:
- Total energy depends only on amplitude $A$
- Energy is independent of position $x$
- Energy is independent of time $t$
- Larger amplitude → more energy
At any position:
$$\frac{1}{2}m\omega^2(A^2 - x^2) + \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}m\omega^2 A^2$$✓
At equilibrium ($x = 0$):
$$KE_{max} = \frac{1}{2}m\omega^2 A^2 = E$$✓
At extreme ($x = A$):
$$PE_{max} = \frac{1}{2}m\omega^2 A^2 = E$$✓
Energy Distribution at Different Positions
| Position | Displacement | KE | PE |
|---|---|---|---|
| Equilibrium | $x = 0$ | $E$ (100%) | $0$ (0%) |
| Midpoint | $x = \frac{A}{2}$ | $\frac{3E}{4}$ (75%) | $\frac{E}{4}$ (25%) |
| Equal energy point | $x = \frac{A}{\sqrt{2}}$ | $\frac{E}{2}$ (50%) | $\frac{E}{2}$ (50%) |
| Extreme | $x = A$ | $0$ (0%) | $E$ (100%) |
Question: At what position is KE = PE?
Answer: When $x = \frac{A}{\sqrt{2}} = 0.707A$
Proof:
$$KE = PE$$ $$\frac{1}{2}k(A^2 - x^2) = \frac{1}{2}kx^2$$ $$A^2 - x^2 = x^2$$ $$A^2 = 2x^2$$ $$x = \frac{A}{\sqrt{2}}$$This is a very common JEE question!
Average Kinetic and Potential Energy
Over one complete cycle:
Average KE:
$$\langle KE \rangle = \frac{1}{T}\int_0^T KE \, dt = \frac{E}{2}$$Average PE:
$$\langle PE \rangle = \frac{1}{T}\int_0^T PE \, dt = \frac{E}{2}$$Over a complete oscillation:
- Half the time energy is predominantly kinetic
- Half the time energy is predominantly potential
- Average KE = Average PE = $\frac{E}{2}$
This is true for all SHM, regardless of amplitude or frequency!
Energy in Different SHM Systems
1. Spring-Mass System (Horizontal)
Total energy:
$$\boxed{E = \frac{1}{2}kA^2}$$Where $k$ is spring constant and $A$ is amplitude.
2. Spring-Mass System (Vertical)
When a spring hangs vertically with mass $m$:
- Equilibrium position shifts by $x_0 = \frac{mg}{k}$
- Energy is measured from new equilibrium
- Same formula: $E = \frac{1}{2}kA^2$
Wrong: Including gravitational PE in total energy
Right: Choose equilibrium (stretched position) as reference. Then:
$$E = KE + PE_{spring}$$The gravitational PE is already accounted for in the equilibrium position shift!
3. Simple Pendulum
For small angles, a pendulum approximates SHM.
At angle $\theta$ from vertical:
Height above lowest point: $h = l(1 - \cos\theta) \approx \frac{l\theta^2}{2}$ (for small $\theta$)
Since $x = l\theta$ (arc length):
$$PE = mgh = mg\frac{x^2}{2l}$$Comparing with $PE = \frac{1}{2}kx^2$:
$$k_{effective} = \frac{mg}{l}$$Total energy:
$$\boxed{E = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}\frac{mg}{l}A^2}$$Where $A$ is the maximum angular displacement (in radians) times $l$.
Or using maximum angle $\theta_{max}$:
$$\boxed{E = mgl(1 - \cos\theta_{max}) \approx \frac{1}{2}mgl\theta_{max}^2}$$Graphical Representation
KE vs Position
Graph of $KE = \frac{1}{2}k(A^2 - x^2)$:
- Inverted parabola
- Maximum at $x = 0$
- Zero at $x = \pm A$
PE vs Position
Graph of $PE = \frac{1}{2}kx^2$:
- Upward parabola
- Minimum (zero) at $x = 0$
- Maximum at $x = \pm A$
Total Energy vs Position
Graph of $E = \frac{1}{2}kA^2$:
- Horizontal line (constant)
- Independent of position
Energy vs Time Graphs
If $x = A\sin(\omega t)$:
KE vs time:
$$KE = \frac{1}{2}m\omega^2 A^2\cos^2(\omega t) = \frac{E}{2}[1 + \cos(2\omega t)]$$PE vs time:
$$PE = \frac{1}{2}m\omega^2 A^2\sin^2(\omega t) = \frac{E}{2}[1 - \cos(2\omega t)]$$Observations:
- Both oscillate at frequency $2f$ (double the displacement frequency)
- KE and PE are out of phase by $\frac{\pi}{2}$
- When one is max, the other is min
- Average value of both is $\frac{E}{2}$
Power in SHM
Power is the rate of energy transfer:
$$P = F \cdot v = (-kx) \cdot v = -kxv$$Using $x = A\sin(\omega t)$ and $v = A\omega\cos(\omega t)$:
$$P = -kA^2\omega\sin(\omega t)\cos(\omega t)$$ $$\boxed{P = -\frac{1}{2}kA^2\omega\sin(2\omega t)}$$Average power over complete cycle:
$$\langle P \rangle = 0$$In ideal SHM (no damping), average power is zero because:
- Half the time, spring does positive work (giving energy to mass)
- Half the time, mass does positive work (giving energy to spring)
- Net energy transfer over complete cycle = 0
When to Use Energy Methods
Use Energy Method when:
- Asked about velocity at a given position (not time)
- Asked about position where KE = PE
- Finding amplitude from given conditions
- Comparing energies at different positions
Use Equation of Motion when:
- Asked about position/velocity at a specific time
- Need to find phase constant
- Calculating acceleration
Energy Method Advantage: Often faster! No need to find $\omega$, phase, or time.
Common Mistakes to Avoid
Wrong: $E = kA^2$ or $E = m\omega^2 A^2$
Right: $E = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2$
Always remember the $\frac{1}{2}$ factor!
Wrong: Higher frequency → more energy
Right: Energy depends only on amplitude and spring constant (or mass and $\omega$):
$$E = \frac{1}{2}kA^2$$Two oscillators with same $k$ and $A$ but different $m$ (hence different $\omega$) have the same energy!
But $E = \frac{1}{2}m\omega^2 A^2$ shows energy increases with $\omega^2$ if $m$ and $A$ are constant.
Wrong: KE = PE at $x = \frac{A}{2}$
Right: KE = PE at $x = \frac{A}{\sqrt{2}} \approx 0.707A$
Many students forget the $\sqrt{2}$!
Wrong: Including $mgh$ in total energy formula
Right: Choose equilibrium as reference (where spring is stretched by $\frac{mg}{k}$). Then only spring PE matters:
$$E = \frac{1}{2}kA^2$$The gravitational PE is constant and can be set to zero at equilibrium.
Memory Tricks & Patterns
The “KEPE Dance”
Memory Trick: “KE and PE dance opposite — when one peaks, the other sleeps”
- At $x = 0$: KE peaks, PE sleeps (zero)
- At $x = A$: PE peaks, KE sleeps (zero)
The “Half-Energy Formula”
“Square root of two halves the amplitude”
$$x = \frac{A}{\sqrt{2}} \rightarrow KE = PE = \frac{E}{2}$$Energy Amplitude Relationship
“Energy is Amplitude Squared”
$$E \propto A^2$$Double the amplitude → quadruple the energy!
Practice Problems
Level 1: Foundation (NCERT Style)
A particle in SHM has total energy 100 J and amplitude 10 cm. Find the force constant.
Solution:
$$E = \frac{1}{2}kA^2$$ $$100 = \frac{1}{2}k(0.1)^2$$ $$100 = \frac{1}{2}k \times 0.01$$ $$k = \frac{200}{0.01} = 20000$$N/m = 20 kN/m
A spring-mass system has total energy 50 J. At what displacement is the KE equal to 30 J? Amplitude is 20 cm.
Solution:
$$KE + PE = E$$ $$30 + PE = 50$$ $$PE = 20$$J
$$PE = \frac{1}{2}kx^2$$Also, $E = \frac{1}{2}kA^2 = 50$
$$\frac{1}{2}k(0.2)^2 = 50$$ $$k = 2500$$N/m
Now: $20 = \frac{1}{2}(2500)x^2$
$$x^2 = \frac{40}{2500} = 0.016$$ $$x = 0.126$$m = 12.6 cm
Or use ratio: $\frac{PE}{E} = \frac{x^2}{A^2}$
$$\frac{20}{50} = \frac{x^2}{(0.2)^2}$$ $$x^2 = 0.4 \times 0.04 = 0.016$$ $$x = 0.126$$m = 12.6 cm
A particle executes SHM with amplitude $A$. At what position is KE = PE?
Solution: At the equal energy position:
$$KE = PE = \frac{E}{2}$$ $$\frac{1}{2}kx^2 = \frac{1}{2} \times \frac{1}{2}kA^2$$ $$x^2 = \frac{A^2}{2}$$ $$x = \frac{A}{\sqrt{2}} = 0.707A$$Answer: $x = \pm\frac{A}{\sqrt{2}}$
Level 2: JEE Main
A particle of mass 0.5 kg executes SHM with amplitude 0.1 m and time period 2 s. Find: (a) Total energy (b) KE and PE at $x = 0.05$ m
Solution: $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi$ rad/s
(a) $E = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}(0.5)(\pi)^2(0.1)^2$
$$E = \frac{1}{2}(0.5)(9.87)(0.01) = 0.0246$$J ≈ 24.6 mJ or 0.025 J
(b) At $x = 0.05$ m:
$$PE = \frac{1}{2}m\omega^2 x^2 = \frac{1}{2}(0.5)(\pi)^2(0.05)^2$$ $$PE = \frac{1}{2}(0.5)(9.87)(0.0025) = 6.17 \times 10^{-3}$$J ≈ 6.2 mJ
$$KE = E - PE = 24.6 - 6.2 = 18.4$$mJ ≈ 18.4 mJ
Or: $KE = \frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}(0.5)(\pi)^2(0.01 - 0.0025)$
$$= \frac{1}{2}(0.5)(9.87)(0.0075) = 18.5$$mJ ✓
Two particles execute SHM with the same amplitude but different frequencies $f_1 = 2$ Hz and $f_2 = 4$ Hz. If both have mass $m$, find the ratio of their total energies.
Solution:
$$E = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}m(2\pi f)^2 A^2$$ $$E \propto f^2$$ $$\frac{E_1}{E_2} = \frac{f_1^2}{f_2^2} = \frac{4}{16} = \frac{1}{4}$$Answer: $E_1 : E_2 = 1 : 4$
A spring with spring constant 100 N/m has a 1 kg mass. The mass is displaced 10 cm and released. Find the maximum velocity.
Solution: At maximum velocity (equilibrium):
$$E = KE_{max} = \frac{1}{2}mv_{max}^2$$Also: $E = \frac{1}{2}kA^2 = \frac{1}{2}(100)(0.1)^2 = 0.5$ J
$$\frac{1}{2}(1)v_{max}^2 = 0.5$$ $$v_{max}^2 = 1$$ $$v_{max} = 1$$m/s
Or use: $v_{max} = A\omega = A\sqrt{\frac{k}{m}} = 0.1\sqrt{\frac{100}{1}} = 0.1 \times 10 = $ 1 m/s
Level 3: JEE Advanced
A block of mass $m$ attached to a spring executes SHM on a frictionless horizontal surface. When the block passes through equilibrium, a lump of putty of mass $m$ is dropped on it and sticks. Find the new amplitude in terms of original amplitude $A_0$.
Solution: Before collision (at equilibrium):
- Velocity: $v_0 = A_0\omega_0 = A_0\sqrt{\frac{k}{m}}$
- Energy: $E_0 = \frac{1}{2}kA_0^2$
During collision (momentum conservation):
$$mv_0 = (m + m)v$$ $$v = \frac{v_0}{2}$$After collision: New mass = $2m$ New energy: $E = \frac{1}{2}(2m)v^2 = \frac{1}{2}(2m)\frac{v_0^2}{4} = \frac{1}{4}(mv_0^2) = \frac{1}{4}kA_0^2$
New amplitude:
$$E = \frac{1}{2}kA^2$$ $$\frac{1}{4}kA_0^2 = \frac{1}{2}kA^2$$ $$A^2 = \frac{A_0^2}{2}$$ $$\boxed{A = \frac{A_0}{\sqrt{2}}}$$Energy is lost to heat during the inelastic collision!
A particle of mass $m$ is in SHM with amplitude $A$ and frequency $f$. What is the average kinetic energy over one time period?
Solution: Total energy: $E = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}m(2\pi f)^2 A^2 = 2\pi^2 mf^2 A^2$
Over one complete cycle:
$$\langle KE \rangle = \frac{E}{2} = \pi^2 mf^2 A^2$$Or: $\boxed{\langle KE \rangle = \frac{1}{2}m\omega^2 A^2 \times \frac{1}{2} = \frac{1}{4}m\omega^2 A^2}$
Answer: $\langle KE \rangle = \pi^2 mf^2 A^2 = \frac{1}{4}m\omega^2 A^2$
A simple pendulum of length $l$ is released from angle $\theta_0$. Find the velocity at the lowest point. (For any angle, not just small angles)
Solution: Using energy conservation:
At highest point: $E = PE = mgl(1 - \cos\theta_0)$ (taking lowest point as reference) At lowest point: $E = KE = \frac{1}{2}mv^2$
$$mgl(1 - \cos\theta_0) = \frac{1}{2}mv^2$$ $$v^2 = 2gl(1 - \cos\theta_0)$$ $$\boxed{v = \sqrt{2gl(1 - \cos\theta_0)}}$$For small angles: $1 - \cos\theta_0 \approx \frac{\theta_0^2}{2}$
$$v \approx \sqrt{gl\theta_0^2} = \theta_0\sqrt{gl}$$Since $A = l\theta_0$ and $\omega = \sqrt{\frac{g}{l}}$:
$$v = A\omega$$✓ (matches SHM formula)
Quick Revision Box
| Position | Displacement | KE | PE | Total E |
|---|---|---|---|---|
| Equilibrium | $0$ | $\frac{1}{2}kA^2$ | $0$ | $\frac{1}{2}kA^2$ |
| $x = \frac{A}{\sqrt{2}}$ | $\frac{A}{\sqrt{2}}$ | $\frac{1}{4}kA^2$ | $\frac{1}{4}kA^2$ | $\frac{1}{2}kA^2$ |
| Extreme | $\pm A$ | $0$ | $\frac{1}{2}kA^2$ | $\frac{1}{2}kA^2$ |
Key Formulas:
- Total energy: $E = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2$
- At position $x$: $KE = \frac{1}{2}k(A^2 - x^2)$, $PE = \frac{1}{2}kx^2$
- Equal energy point: $x = \frac{A}{\sqrt{2}}$
- Average KE = Average PE = $\frac{E}{2}$
Related Topics
Within Oscillations & Waves
- Simple Harmonic Motion — Foundation for energy analysis
- Wave Motion — Energy propagation in waves
- Damped Oscillations — Energy dissipation
Connected Chapters
- Work-Energy-Power — Energy concepts
- Conservation of Energy — Energy conservation principle
- Collisions — Momentum and energy in collisions
Math Connections
- Integration — For average energy calculations
- Trigonometric Identities — For energy vs time
Teacher’s Summary
Total energy is constant: $E = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2 A^2$
Energy transformation:
- At equilibrium: All kinetic ($KE = E$, $PE = 0$)
- At extremes: All potential ($KE = 0$, $PE = E$)
- At $x = \frac{A}{\sqrt{2}}$: Half-and-half ($KE = PE = \frac{E}{2}$)
Energy depends on:
- Amplitude (squared): $E \propto A^2$
- NOT on mass (for given $k$ and $A$)
- NOT on time period
Average energies: Over complete cycle, $\langle KE \rangle = \langle PE \rangle = \frac{E}{2}$
JEE Strategy: Energy method is often faster than solving equations of motion. Use $E = KE + PE$ to find velocities at different positions without needing time!
“In SHM, energy never dies — it just dances between motion and position, forever.”