⚡ Energy in Simple Harmonic Motion

Total Energy in SHM

In an ideal oscillating system with no friction, total mechanical energy is conserved:

$E_{total} = KE + PE = \text{constant}$

Energy continuously transforms between kinetic and potential, but the total remains constant.

💡 Key Idea: Energy sloshes back and forth between kinetic (motion) and potential (position) forms. At any instant, total energy equals the maximum potential energy (at amplitude).

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Potential Energy in Mass-Spring System

Elastic potential energy stored in a spring:

$PE = \frac{1}{2}kx^2$

where:

• $k$ = spring constant (N/m)
• $x$ = displacement from equilibrium (m)

Key Points:

• Maximum at amplitude: $PE_{max} = \frac{1}{2}kA^2$ (when $x = \pm A$)
• Zero at equilibrium: $PE = 0$ (when $x = 0$)
• Always positive (depends on $x^2$, not $x$)
• Same value at $+x$ and $-x$

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Kinetic Energy in SHM

$KE = \frac{1}{2}mv^2$

where $v$ is the instantaneous speed.

Key Points:

• Maximum at equilibrium: $KE_{max} = \frac{1}{2}mv_{max}^2 = \frac{1}{2}m(A\omega)^2$ (when $x = 0$)
• Zero at amplitude: $KE = 0$ (when $x = \pm A$, momentarily at rest)
• Always positive

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Total Energy

Since energy is conserved:

$E_{total} = KE + PE = \text{constant}$

At any position: $E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$

Finding Total Energy:

Method 1: At amplitude ($x = A$, $v = 0$): $E_{total} = 0 + \frac{1}{2}kA^2 = \frac{1}{2}kA^2$

Method 2: At equilibrium ($x = 0$, $v = v_{max}$): $E_{total} = \frac{1}{2}mv_{max}^2 + 0 = \frac{1}{2}m(A\omega)^2$

Since $\omega^2 = \frac{k}{m}$: $E_{total} = \frac{1}{2}m \cdot A^2 \cdot \frac{k}{m} = \frac{1}{2}kA^2$

Both methods give the same result! ✓

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Energy at Any Position

At position $x$ with speed $v$:

$\frac{1}{2}kA^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$

Solving for $v$: $v = \sqrt{\frac{k}{m}(A^2 - x^2)} = \omega\sqrt{A^2 - x^2}$

This gives speed at any position!

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Energy Transformations During Oscillation

At maximum displacement ($x = A$):

• All potential energy: $E = \frac{1}{2}kA^2$
• No kinetic energy: $v = 0$
• Object momentarily at rest

At equilibrium ($x = 0$):

• All kinetic energy: $E = \frac{1}{2}mv_{max}^2$
• No potential energy
• Maximum speed

Midway ($x = \frac{A}{2}$):

• Mix of KE and PE
• $PE = \frac{1}{2}k\left(\frac{A}{2}\right)^2 = \frac{1}{8}kA^2$ (1/4 of maximum)
• $KE = \frac{1}{2}kA^2 - \frac{1}{8}kA^2 = \frac{3}{8}kA^2$ (3/4 of maximum)

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Energy Graphs

Potential Energy: Parabola

• Minimum at $x = 0$
• Maximum at $x = \pm A$
• $PE = \frac{1}{2}kx^2$

Kinetic Energy: Inverted parabola

• Maximum at $x = 0$
• Zero at $x = \pm A$
• $KE = \frac{1}{2}k(A^2 - x^2)$

Total Energy: Horizontal line

• Constant at $E = \frac{1}{2}kA^2$
• Independent of position

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Pendulum Energy

For a simple pendulum:

Potential Energy: Gravitational $PE = mgh$

where $h$ is height above lowest point.

For small angles: $h \approx \frac{L\theta^2}{2}$

Total Energy: $E = \frac{1}{2}mv^2 + mgh$

At maximum angle $\theta_0$: $E = mgL(1 - \cos\theta_0) \approx \frac{1}{2}mgL\theta_0^2$

(for small angles)

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Using Energy to Find Speed

Problem type: "Find speed at position $x$"

Method:

1. Find total energy: $E = \frac{1}{2}kA^2$
2. At position $x$: $E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$
3. Set equal and solve for $v$: $v = \omega\sqrt{A^2 - x^2}$

Check:

• At $x = 0$: $v = \omega A = v_{max}$ ✓
• At $x = A$: $v = 0$ ✓

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Effect of Damping

In real systems, friction causes energy loss:

Damped Oscillation:

• Amplitude decreases over time
• Period slightly longer
• Eventually stops

Energy is converted to:

• Heat (friction)
• Sound
• Air resistance

Not in AP Physics 1 scope: Detailed damping equations (that's AP Physics 2/C)

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⚠️ Common Mistakes

Mistake 1: Forgetting Total Energy is Constant

In ideal SHM (no friction), $E_{total} = \frac{1}{2}kA^2 =$ constant at ALL positions.

Mistake 2: Negative Potential Energy

Spring PE is always positive: $PE = \frac{1}{2}kx^2$ (even when $x < 0$, we square it!)

Mistake 3: KE and PE at Wrong Positions

• Max KE at equilibrium ($x = 0$), NOT at amplitude
• Max PE at amplitude ($x = \pm A$), NOT at equilibrium

Mistake 4: Confusing x and A

• $x$ = current position (variable)
• $A$ = amplitude (maximum displacement, constant)

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Problem-Solving Strategy

Energy Method for SHM:

1. Find amplitude $A$ (from initial conditions)
2. Calculate total energy: $E = \frac{1}{2}kA^2$
3. At any position $x$:
   • Potential: $PE = \frac{1}{2}kx^2$
   • Kinetic: $KE = E - PE = \frac{1}{2}k(A^2 - x^2)$
4. Find speed: $v = \sqrt{\frac{k}{m}(A^2 - x^2)}$
5. Check answer: Does it make sense?
   • $v$ should be max at $x = 0$
   • $v$ should be zero at $x = A$

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Comparing Two Oscillators

Same amplitude, different masses:

• Same total energy: $E = \frac{1}{2}kA^2$ (independent of mass!)
• Lighter mass → higher maximum speed
• Heavier mass → lower maximum speed

Same mass and spring, different amplitudes:

• Larger amplitude → more total energy ($E \propto A^2$)
• Larger amplitude → higher maximum speed ($v_{max} = A\omega$)
• But same period! ($T$ independent of $A$)

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Applications

Shock Absorbers

Convert kinetic energy to heat through damping:

• Car hits bump (KE increases)
• Spring compresses (converts to PE)
• Damper dissipates energy
• Smooth ride

Energy Storage

Springs can store energy:

• Mechanical watches (mainspring)
• Pogo sticks
• Trampolines

Molecular Vibrations

Chemical bonds act like springs:

• Bond energy = spring PE
• Vibrational energy quantized
• Infrared spectroscopy

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Key Formulas Summary

| Quantity | Formula | Notes | |----------|---------|-------| | Potential Energy (spring) | $PE = \frac{1}{2}kx^2$ | Max at $x = \pm A$ | | Kinetic Energy | $KE = \frac{1}{2}mv^2$ | Max at $x = 0$ | | Total Energy | $E = \frac{1}{2}kA^2$ | Constant, independent of mass | | Speed at position x | $v = \omega\sqrt{A^2 - x^2}$ | From energy conservation | | Maximum speed | $v_{max} = A\omega$ | At equilibrium | | Maximum KE | $KE_{max} = \frac{1}{2}kA^2$ | Equals total energy |

Key Relationships:

• $E_{total} = KE + PE = \frac{1}{2}kA^2$
• Energy oscillates between KE and PE
• Total energy $\propto A^2$ (doubles amplitude → 4× energy)