⚡ KE and PE Exchange in SHM

Part 1 of 7 — Energy in Simple Harmonic Motion

In SHM, energy continuously transforms between kinetic and potential forms. The total mechanical energy remains constant (no friction), but the split between KE and PE changes throughout the motion.

Energy Forms in SHM

Mass-Spring System

$KE = \frac{1}{2}mv^2 \qquad PE = \frac{1}{2}kx^2$

Total Mechanical Energy

$E = KE + PE = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{constant}$

Energy at Special Points

At the endpoints, $v = 0$ so all energy is potential. At equilibrium, $x = 0$ so all energy is kinetic.

Energy vs. Position Graphs

What the Graphs Look Like

• PE curve: Parabola $U = \frac{1}{2}kx^2$ (upward-opening, minimum at $x = 0$)
• KE curve: Inverted parabola (maximum at $x = 0$, zero at $x = \pm A$)
• Total E: Horizontal line at $E = \frac{1}{2}kA^2$

Energy vs. Time

• PE oscillates as $\cos^2(\omega t)$
• KE oscillates as $\sin^2(\omega t)$
• Both oscillate at twice the frequency of the position oscillation
• When one is at maximum, the other is at minimum

Pendulum Energy

For a pendulum, $PE = mgh$ where $h$ is the height above the lowest point:

• At the bottom: $KE$ is max, $PE$ is min
• At the sides: $KE = 0$, $PE$ is max

Energy Exchange Quiz 🎯

Energy Calculations 🧮

A 0.40 kg block on a spring ($k = 160$ N/m) oscillates with amplitude $A = 0.10$ m.

1. What is the total energy? (in J)

2. What is the maximum speed? (in m/s)

3. What is the PE when $x = 0.06$ m? (in J, round to 3 significant figures)

Energy Review 🔍

Exit Quiz — KE and PE Exchange ✅

🔋 Total Energy $E = \frac{1}{2}kA^2$

Part 2 of 7 — Energy in Simple Harmonic Motion

The total mechanical energy of a mass-spring system depends only on the spring constant and the amplitude. This simple but powerful result connects initial conditions to all energy calculations.

Deriving Total Energy

At the endpoints ($x = \pm A$), velocity is zero, so:

$E = KE + PE = 0 + \frac{1}{2}kA^2 = \frac{1}{2}kA^2$

At equilibrium ($x = 0$), displacement is zero, so:

$E = KE + PE = \frac{1}{2}mv_{\text{max}}^2 + 0 = \frac{1}{2}mv_{\text{max}}^2$

Setting them equal:

$\frac{1}{2}kA^2 = \frac{1}{2}mv_{\text{max}}^2$

$v_{\text{max}} = A\sqrt{\frac{k}{m}} = A\omega$

What Total Energy Depends On

Equivalent Expressions for Total Energy

The total energy can be written several ways:

$E = \frac{1}{2}kA^2 = \frac{1}{2}mv_{\text{max}}^2 = \frac{1}{2}m\omega^2 A^2$

Since $\omega^2 = k/m$:

$E = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}m \cdot \frac{k}{m} \cdot A^2 = \frac{1}{2}kA^2 \checkmark$

For a Pendulum

$E = mgh_{\text{max}} = \frac{1}{2}mv_{\text{max}}^2$

where $h_{\text{max}}$ is the maximum height above the lowest point. For small angles:

$h_{\text{max}} = L(1 - \cos\theta_{\text{max}}) \approx \frac{L\theta_{\text{max}}^2}{2}$

Total Energy Quiz 🎯

Energy Drill 🧮

1. A 0.50 kg mass on a spring ($k = 200$ N/m) has a total energy of 1.0 J. What is the amplitude? (in m)

2. With the same system, what is $v_{\text{max}}$? (in m/s)

3. A different spring has $k = 80$ N/m and $A = 0.25$ m. What is the total energy? (in J)

Round all answers to 3 significant figures.

Total Energy Concepts 🔍

Exit Quiz — Total Energy ✅

📐 Energy at Any Position

Part 3 of 7 — Energy in Simple Harmonic Motion

Using energy conservation, we can find the speed of an oscillating object at any position — not just at the endpoints or equilibrium.

The Master Energy Equation

At any displacement $x$:

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

Solving for velocity at position $x$:

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

Special Cases

Energy Fractions

At $x = A/\sqrt{2}$:

• $PE = \frac{1}{2}k(A/\sqrt{2})^2 = \frac{1}{4}kA^2 = E/2$
• $KE = E/2$
• This is where KE = PE (energy is split equally!)

Problem-Solving Strategy

Finding Speed at a Given Position

1. Find total energy: $E = \frac{1}{2}kA^2$
2. Find PE at position: $PE = \frac{1}{2}kx^2$
3. Find KE: $KE = E - PE$
4. Find speed: $v = \sqrt{2 \cdot KE/m}$

Finding Position at a Given Speed

1. Find total energy: $E = \frac{1}{2}kA^2$
2. Find KE: $KE = \frac{1}{2}mv^2$
3. Find PE: $PE = E - KE$
4. Find position: $x = \sqrt{2 \cdot PE/k}$

Energy Bar Charts

At each position, you can draw a bar chart:

• Total bar height is constant ($= E$)
• KE bar shrinks as PE bar grows (moving away from equilibrium)
• PE bar shrinks as KE bar grows (moving toward equilibrium)

Speed at a Position Quiz 🎯

Position-Energy Calculations 🧮

A 2.0 kg block on a spring ($k = 200$ N/m) oscillates with amplitude $A = 0.20$ m.

1. What is the total energy? (in J)

2. What is the speed at $x = 0.12$ m? (in m/s)

3. At what displacement is the speed 1.0 m/s? (in m)

Round all answers to 3 significant figures.

Energy at Any Position Review 🔍

Exit Quiz — Energy at Any Position ✅

🚀 Maximum Velocity $v_{\text{max}} = A\omega$

Part 4 of 7 — Energy in Simple Harmonic Motion

The maximum velocity is one of the most frequently tested quantities in AP Physics 1 SHM problems. It connects amplitude, angular frequency, and energy in a powerful way.

Deriving $v_{\text{max}}$

From Energy Conservation

At equilibrium ($x = 0$), all energy is kinetic:

$\frac{1}{2}mv_{\text{max}}^2 = \frac{1}{2}kA^2$

$v_{\text{max}} = A\sqrt{\frac{k}{m}} = A\omega$

From the Velocity Equation

$v(t) = -A\omega\sin(\omega t)$

The maximum value of $|\sin(\omega t)|$ is 1, so:

$v_{\text{max}} = A\omega$

Multiple Forms

$v_{\text{max}} = A\omega = 2\pi f A = \frac{2\pi A}{T} = A\sqrt{\frac{k}{m}}$

How $v_{\text{max}}$ Depends on Parameters

Connecting $v_{\text{max}}$ to Energy

$E = \frac{1}{2}mv_{\text{max}}^2 = \frac{1}{2}kA^2$

This means:

$v_{\text{max}} = \sqrt{\frac{2E}{m}}$

Speed at Any Position (revisited)

$v(x) = v_{\text{max}}\sqrt{1 - \frac{x^2}{A^2}}$

This elegant form shows that:

• At $x = 0$: $v = v_{\text{max}}$
• At $x = A$: $v = 0$
• At $x = A/2$: $v = v_{\text{max}}\sqrt{3/4} = v_{\text{max}}(\sqrt{3}/2) \approx 0.866 v_{\text{max}}$

Maximum Acceleration Connection

$a_{\text{max}} = \omega \cdot v_{\text{max}} = \omega^2 A$

So $\omega = a_{\text{max}}/v_{\text{max}}$ — a useful relationship for AP problems!

Maximum Velocity Quiz 🎯

Maximum Velocity Calculations 🧮

1. A 0.25 kg block on a spring ($k = 100$ N/m) oscillates with $A = 0.08$ m. Find $v_{\text{max}}$. (in m/s)

2. An oscillator has $v_{\text{max}} = 5.0$ m/s and $\omega = 10$ rad/s. Find the amplitude. (in m)

3. A mass has $v_{\text{max}} = 2.0$ m/s and $m = 0.50$ kg. Find the total energy. (in J)

Round all answers to 3 significant figures.

$v_{\text{max}}$ Concepts 🔍

Exit Quiz — Maximum Velocity ✅

🌊 Damped Oscillations (Conceptual)

Part 5 of 7 — Energy in Simple Harmonic Motion

Real oscillating systems always lose energy to friction, air resistance, or other dissipative forces. This causes the amplitude to decrease over time — a process called damping.

What Is Damping?

Damping is the gradual loss of mechanical energy from an oscillating system, typically due to:

• Friction (surface contact)
• Air resistance (drag)
• Internal friction (deformation of materials)

Effect on Motion

With damping:

• Amplitude decreases with each cycle
• Period stays approximately the same (for light damping)
• Total energy decreases over time
• The object eventually comes to rest at the equilibrium position

Energy Perspective

Without damping: $E = \frac{1}{2}kA^2 = \text{constant}$

With damping: $E$ decreases over time. Since $E \propto A^2$, the amplitude also decreases. The "lost" mechanical energy is converted to thermal energy (heat).

Types of Damping

1. Underdamped

• The system oscillates with decreasing amplitude
• Most common in AP problems
• Example: a pendulum in air

2. Critically Damped

• The system returns to equilibrium as quickly as possible without oscillating
• Example: car shock absorbers (ideally)

3. Overdamped

• The system returns to equilibrium slowly without oscillating
• Takes longer than critical damping
• Example: a door closer set too tight

What the Graphs Look Like

On the AP exam, you mostly need to recognize underdamped motion and understand that energy is gradually converted to thermal energy.

Damping Concepts Quiz 🎯

Energy Loss in Damped Systems

After Each Cycle

If a damped oscillator loses a fraction $f$ of its energy each cycle:

• After 1 cycle: $E_1 = (1-f)E_0$
• After 2 cycles: $E_2 = (1-f)^2 E_0$
• After $n$ cycles: $E_n = (1-f)^n E_0$

Amplitude Decay

Since $E \propto A^2$:

$A_n = A_0(1-f)^{n/2}$

Quality Factor (Conceptual)

The Q-factor measures how "good" an oscillator is at maintaining its energy:

• High Q → low damping → many oscillations before stopping
• Low Q → high damping → few oscillations before stopping

Examples: tuning fork (high Q), pendulum in honey (low Q)

Damped Energy Calculations 🧮

A damped oscillator starts with amplitude $A_0 = 0.20$ m and loses 10% of its energy each cycle.

1. What fraction of the original energy remains after 3 cycles? (as a decimal, round to 3 significant figures)

2. What is the amplitude after 3 cycles? (in m, round to 3 significant figures)

3. After how many complete cycles is the energy reduced to less than half? (give the smallest integer)

Damping Review 🔍

Exit Quiz — Damped Oscillations ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Energy in Simple Harmonic Motion

Time to tackle comprehensive energy problems that combine everything: $E = \frac{1}{2}kA^2$, speed at any position, $v_{\text{max}} = A\omega$, and damping concepts.

Energy Problem-Solving Strategy

1. Identify what you know: $m$, $k$, $A$, $v_{\text{max}}$, $E$, position, speed, etc.
2. Write the energy conservation equation: $\frac{1}{2}kA^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2$
3. Use the appropriate form:
   • Need $v_{\text{max}}$? → $v_{\text{max}} = A\omega = A\sqrt{k/m}$
   • Need $v$ at position $x$? → $v = \omega\sqrt{A^2 - x^2}$
   • Need energy? → $E = \frac{1}{2}kA^2 = \frac{1}{2}mv_{\text{max}}^2$
4. Check: Does the answer make physical sense?

Quick Reference

$\omega = \sqrt{\frac{k}{m}} = \frac{2\pi}{T} = 2\pi f$

$v_{\text{max}} = A\omega \qquad a_{\text{max}} = A\omega^2 \qquad \omega = \frac{a_{\text{max}}}{v_{\text{max}}}$

Problem 1 🎯

A 0.30 kg block attached to a spring ($k = 120$ N/m) is pulled 0.15 m from equilibrium and released from rest on a frictionless surface.

Problem 2 🧮

A pendulum bob ($m = 0.50$ kg) swings through its lowest point with a speed of 1.2 m/s.

1. What is the kinetic energy at the lowest point? (in J, round to 3 significant figures)

2. What is the maximum height above the lowest point? (in m, round to 3 significant figures, use $g = 9.8$ m/s²)

3. What is the total mechanical energy of the system? (in J, round to 3 significant figures)

Problem 3 — Energy Ratios 🎯

A block on a spring oscillates with amplitude $A$ and total energy $E$.

Problem 4 — Conceptual Problem Solving 🔍

A mass-spring system oscillates on a frictionless surface with energy $E$ and amplitude $A$. Various changes are described below.

Challenge Problem 🧮

A 1.0 kg mass on a spring ($k = 400$ N/m) has total energy $E = 8.0$ J.

1. What is the amplitude? (in m)

2. At what position is the speed 3.0 m/s? (in m, round to 3 significant figures)

3. At what position is $KE = 2PE$? (in m, round to 3 significant figures)

Exit Quiz — Problem Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Energy in Simple Harmonic Motion

This final part brings together all energy concepts in SHM — KE/PE exchange, total energy, speed at any position, maximum velocity, and damping — for a comprehensive AP exam review.

Complete Energy Summary

Core Equations

$E = \frac{1}{2}kA^2 = \frac{1}{2}mv_{\text{max}}^2 = \frac{1}{2}m\omega^2 A^2$

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

$v = \omega\sqrt{A^2 - x^2} \qquad v_{\text{max}} = A\omega$

Energy at Special Positions

Key Proportionalities

• $E \propto A^2$ (double amplitude → 4× energy)
• $E \propto k$ (double spring constant → 2× energy)
• $v_{\text{max}} \propto A$ and $v_{\text{max}} \propto \omega$
• Energy oscillates at frequency $2f$

Common AP Mistakes to Avoid

❌ "Energy is proportional to amplitude" — No, $E \propto A^2$!

❌ "Speed is maximum at the endpoints" — Speed is ZERO at endpoints, maximum at equilibrium.

❌ "KE = PE at x = A/2" — Actually, $KE = PE$ at $x = A/\sqrt{2} \approx 0.707A$.

❌ "Total energy depends on mass" — $E = \frac{1}{2}kA^2$ has no mass! (Mass affects $v_{\text{max}}$, not $E$.)

❌ "Damping changes the period" — Light damping barely affects the period; it mainly reduces amplitude.

❌ "Energy is lost in damped motion" — Total energy is conserved; mechanical energy converts to thermal energy.

AP-Style Questions — Set 1 🎯

AP Calculation Practice 🧮

1. A spring ($k = 800$ N/m) stores 4.0 J of energy when compressed. What is the compression distance? (in m)

2. A 0.20 kg block oscillates with $k = 80$ N/m and $A = 0.05$ m. What is the speed at $x = 0.03$ m? (in m)

3. A damped oscillator starts with $A_0 = 0.30$ m and amplitude decreases to $0.15$ m. What fraction of the original energy remains? (as a decimal)

Round all answers to 3 significant figures.

Comprehensive Energy Review 🔍

Final Exit Quiz — Energy in SHM ✅