Energy in simple harmonic motion

General physics

Energy in Simple Harmonic Motion — Physics Lesson

🎯 Learning Objectives

Learning Objectives

Describe the interchange between kinetic and potential energy during simple harmonic motion
Recall and use E = ½mω²x₀² for the total energy of a system undergoing simple harmonic motion
Analyze energy transformations at different positions in SHM
Calculate kinetic and potential energies at any point in the oscillation
Apply conservation of energy principles to SHM systems

🗣️ Language Objectives

Language Objectives

Use scientific terminology related to energy and oscillations accurately
Describe energy transformations using appropriate physics vocabulary
Explain mathematical relationships between energy, amplitude, and frequency
Communicate analysis of energy graphs and diagrams clearly
Write conclusions about energy conservation in oscillatory systems

📝 Key Terms

Key Terms

English Term	Russian Translation	Kazakh Translation
Simple Harmonic Motion	Простое гармоническое движение	Қарапайым гармоникалық қозғалыс
Kinetic Energy	Кинетическая энергия	Кинетикалық энергия
Potential Energy	Потенциальная энергия	Потенциалды энергия
Total Energy	Полная энергия	Толық энергия
Amplitude	Амплитуда	Амплитуда
Angular Frequency	Угловая частота	Бұрыштық жиілік
Oscillation	Колебание	Тербеліс
Equilibrium Position	Положение равновесия	Тепе-теңдік күйі

🃏 Topic Flashcards

Interactive Flashcards

Practice with these flashcards to memorize key concepts about energy in simple harmonic motion.

📚 Glossary

Glossary

Simple Harmonic Motion (SHM): A type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium position and acts in the opposite direction.
Translation
Russian: Простое гармоническое движение — тип периодического движения, при котором восстанавливающая сила прямо пропорциональна смещению от положения равновесия и действует в противоположном направлении.
Kazakh: Қарапайым гармоникалық қозғалыс — тепе-теңдік күйінен ауытқуға тура пропорционал және қарама-қарсы бағытта әсер ететін қалпына келтіруші күші бар периодты қозғалыс түрі.
Kinetic Energy in SHM: The energy possessed by an oscillating object due to its motion. Maximum at equilibrium position, zero at maximum displacement.
Translation
Russian: Кинетическая энергия в ПГД — энергия, которой обладает колеблющийся объект благодаря своему движению. Максимальна в положении равновесия, равна нулю при максимальном смещении.
Kazakh: ҚГҚ-дағы кинетикалық энергия — тербелетін нысанның қозғалысының арқасында ие болған энергиясы. Тепе-теңдік күйінде максималды, ең үлкен ауытқуда нөлге тең.
Potential Energy in SHM: The energy stored in the system due to the displacement from equilibrium. Zero at equilibrium position, maximum at maximum displacement.
Translation
Russian: Потенциальная энергия в ПГД — энергия, запасенная в системе из-за смещения от равновесия. Равна нулю в положении равновесия, максимальна при максимальном смещении.
Kazakh: ҚГҚ-дағы потенциалды энергия — тепе-теңдіктен ауытқуының арқасында жүйеде сақталған энергия. Тепе-теңдік күйінде нөлге тең, ең үлкен ауытқуда максималды.
Total Energy in SHM: The sum of kinetic and potential energies, which remains constant throughout the motion (in the absence of damping). Equal to E = ½mω²x₀².
Translation
Russian: Полная энергия в ПГД — сумма кинетической и потенциальной энергий, которая остается постоянной на протяжении всего движения (при отсутствии затухания). Равна E = ½mω²x₀².
Kazakh: ҚГҚ-дағы толық энергия — кинетикалық және потенциалды энергиялардың қосындысы, бұл барлық қозғалыс кезінде тұрақты болып қалады (сөну болмаған жағдайда). E = ½mω²x₀² тең.
Amplitude (x₀): The maximum displacement from the equilibrium position during oscillation. Determines the total energy of the SHM system.
Translation
Russian: Амплитуда — максимальное смещение от положения равновесия во время колебания. Определяет полную энергию системы ПГД.
Kazakh: Амплитуда — тербелу кезінде тепе-теңдік күйінен ең үлкен ауытқу. ҚГҚ жүйесінің толық энергиясын анықтайды.
Angular Frequency (ω): The rate of change of phase with time, measured in radians per second. Related to frequency by ω = 2πf.
Translation
Russian: Угловая частота — скорость изменения фазы со временем, измеряется в радианах в секунду. Связана с частотой соотношением ω = 2πf.
Kazakh: Бұрыштық жиілік — фазаның уақытпен өзгеру жылдамдығы, секундына радианмен өлшенеді. Жиіликпен ω = 2πf қатынасымен байланысты.

📖 Theory: Energy in Simple Harmonic Motion

Theory: Energy Transformations in SHM

Introduction to Energy in SHM

In simple harmonic motion, energy continuously transforms between kinetic and potential forms. This interchange is fundamental to understanding oscillatory systems.

Kazakh Translation

Қарапайым гармоникалық қозғалыста энергия кинетикалық және потенциалды түрлер арасында үздіксіз түрленеді. Бұл өзара алмасу тербелетін жүйелерді түсінудің негізі болып табылады.

Animation showing simple harmonic motion of a mass-spring system

Types of Energy in SHM

1. Kinetic Energy (KE)

The kinetic energy of an oscillating object depends on its velocity:

KE = ½mv²

For SHM with displacement x = x₀cos(ωt), the velocity is:

v = -x₀ω sin(ωt)

Therefore, kinetic energy becomes:

KE = ½m(x₀ω)² sin²(ωt)

Kazakh Translation

Тербелетін нысанның кинетикалық энергиясы оның жылдамдығына байланысты. x = x₀cos(ωt) орын ауыстыруы бар ҚГҚ үшін жылдамдық v = -x₀ω sin(ωt) болады. Сондықтан кинетикалық энергия KE = ½m(x₀ω)² sin²(ωt) болады.

2. Potential Energy (PE)

The potential energy stored in the system due to displacement:

PE = ½kx²

Since k = mω² for SHM, and x = x₀cos(ωt):

PE = ½mω²x₀² cos²(ωt)

Kazakh Translation

Ауытқуының арқасында жүйеде сақталған потенциалды энергия PE = ½kx² формуласымен беріледі. ҚГҚ үшін k = mω², ал x = x₀cos(ωt) болғандықтан, PE = ½mω²x₀² cos²(ωt) болады.

Graph showing kinetic, potential, and total energy variations with displacement

Total Energy Formula

The total energy in SHM is conserved:

E = KE + PE = ½mω²x₀²

Derivation:

E = KE + PE = ½m(x₀ω)² sin²(ωt) + ½mω²x₀² cos²(ωt)

E = ½mω²x₀²[sin²(ωt) + cos²(ωt)]

Since sin²(ωt) + cos²(ωt) = 1:

E = ½mω²x₀²

Kazakh Translation

ҚГҚ-дағы толық энергия сақталады және E = ½mω²x₀² формуласымен беріледі. Бұл кинетикалық және потенциалды энергиялардың қосындысынан алынады, мұнда sin²(ωt) + cos²(ωt) = 1 тригонометриялық тепе-теңдігі қолданылады.

Energy Interchange During Oscillation

Position	Displacement (x)	Velocity (v)	Kinetic Energy	Potential Energy
Equilibrium	x = 0	v = ±x₀ω (maximum)	Maximum = ½mω²x₀²	Zero
Maximum displacement	x = ±x₀	v = 0	Zero	Maximum = ½mω²x₀²
Intermediate	0 < \|x\| < x₀	0 < \|v\| < x₀ω	Intermediate	Intermediate

Graph showing how kinetic and potential energies vary with time in SHM

Practice Questions

Question 1 (Easy):

A mass of 0.2 kg oscillates with amplitude 0.05 m and angular frequency 10 rad/s. Calculate the total energy of the system.

Answer

Using E = ½mω²x₀²
E = ½ × 0.2 kg × (10 rad/s)² × (0.05 m)²
E = ½ × 0.2 × 100 × 0.0025
E = 0.025 J
The total energy is 0.025 J.

Question 2 (Medium):

In the system from Question 1, find the kinetic and potential energies when the displacement is 0.03 m.

Answer

Potential Energy:
PE = ½mω²x² = ½ × 0.2 × (10)² × (0.03)² = 0.009 J

Kinetic Energy:
Since E = KE + PE, and E = 0.025 J
KE = E — PE = 0.025 — 0.009 = 0.016 J

Check: PE + KE = 0.009 + 0.016 = 0.025 J ✓

Question 3 (Medium):

At what fraction of the amplitude does the kinetic energy equal the potential energy in SHM?

Answer

When KE = PE, each equals E/2
PE = ½mω²x² = E/2 = ½mω²x₀²/2 = ¼mω²x₀²
Therefore: ½mω²x² = ¼mω²x₀²
Simplifying: x² = x₀²/2
So: x = x₀/√2 ≈ 0.707x₀

KE equals PE when displacement is about 70.7% of amplitude.

Question 4 (Critical Thinking):

A pendulum oscillates with decreasing amplitude due to air resistance. Analyze how the total energy, maximum kinetic energy, and maximum potential energy change over time. What remains constant and what changes?

Answer

Analysis of damped oscillation:

What decreases:
— Total energy E decreases due to energy loss to friction
— Maximum KE decreases (occurs at equilibrium with reduced speed)
— Maximum PE decreases (occurs at reduced amplitude)
— Amplitude x₀ decreases exponentially

What remains constant:
— Angular frequency ω (for light damping)
— Period T (approximately, for light damping)
— Energy conservation principle still applies: energy lost = work done against friction

Key insight: At any instant, KE + PE + Energy dissipated = Initial total energy
The energy «leaks out» of the mechanical system but total energy (including thermal) is still conserved.

🧠 Memorization Exercises

Exercises on Memorizing Terms

Exercise 1: Fill in the Blanks

The total energy in SHM is given by E = ½m_____²x₀²
Kinetic energy is maximum at the _______ position.
Potential energy is maximum at _______ displacement.
At any point in SHM: Total Energy = _______ Energy + _______ Energy
When displacement is zero, kinetic energy equals the _______ energy.

Answer

1. ω (omega/angular frequency)
2. equilibrium
3. maximum
4. Kinetic, Potential
5. total

Exercise 2: Energy Position Matching

Match each position with the correct energy description:

Positions:

x = 0 (equilibrium)
x = +x₀ (positive amplitude)
x = -x₀ (negative amplitude)
x = x₀/2 (half amplitude)

Energy States:

Maximum PE, zero KE
Maximum KE, zero PE
Equal KE and PE (approximately)
Maximum PE, zero KE

Answer

1-B: x = 0 → Maximum KE, zero PE
2-A: x = +x₀ → Maximum PE, zero KE
3-D: x = -x₀ → Maximum PE, zero KE
4-C: x = x₀/2 → Mixed KE and PE (not quite equal, but closer than other positions)

Exercise 3: Formula Components

For the total energy formula E = ½mω²x₀², identify what each symbol represents:

E = _______
m = _______
ω = _______
x₀ = _______
½ = _______

Answer

1. Total energy of the SHM system
2. Mass of the oscillating object
3. Angular frequency (2πf)
4. Amplitude (maximum displacement)
5. Constant factor from integration/energy derivation

📺 Educational Video

Video Tutorial: Energy in Simple Harmonic Motion

Additional Resources:

🔬 Problem Solving Examples

Worked Examples

Example 1: Mass-Spring System Energy

Problem: A 0.5 kg mass attached to a spring oscillates with amplitude 8 cm and frequency 2 Hz. Calculate:

The angular frequency
The total energy of the system
The maximum speed of the mass
The speed when displacement is 5 cm

🎤 Audio Solution

Detailed Solution with Pronunciation

Step 1: Angular frequency (pronounced: ANG-yoo-lar FREE-kwen-see)

ω = 2πf = 2π × 2 Hz = 4π rad/s ≈ 12.57 rad/s

Step 2: Total energy

E = ½mω²x₀²

E = ½ × 0.5 kg × (4π)² × (0.08 m)²

E = 0.25 × 16π² × 0.0064 = 0.253 J

Step 3: Maximum speed (at equilibrium)

v_max = x₀ω = 0.08 m × 4π rad/s = 1.01 m/s

Step 4: Speed at x = 5 cm

Using energy conservation: KE = E — PE

½mv² = E — ½mω²x²

v² = (2E/m) — ω²x² = (2×0.253/0.5) — (4π)²×(0.05)²

v = 0.79 m/s

📝 Quick Solution

Brief Solution

Given: m = 0.5 kg, x₀ = 0.08 m, f = 2 Hz

1. Angular frequency:

ω = 2πf = 4π rad/s

2. Total energy:

E = ½mω²x₀² = ½(0.5)(4π)²(0.08)² = 0.253 J

3. Maximum speed:

v_max = x₀ω = 0.08 × 4π = 1.01 m/s

4. Speed at x = 0.05 m:

v = ω√(x₀² — x²) = 4π√(0.08² — 0.05²) = 0.79 m/s

Example 2: Simple Pendulum Energy Analysis

Problem: A simple pendulum of length 1 m oscillates with maximum angular displacement 15°. The bob has mass 0.2 kg. Calculate the total energy and the speed at the lowest point.

🎤 Audio Solution

Detailed Solution with Pronunciation

Step 1: Convert angle to radians

θ₀ = 15° × π/180 = 0.262 radians

Step 2: For small angles, x₀ = Lθ₀

x₀ = 1 m × 0.262 = 0.262 m

Step 3: Find angular frequency

For simple pendulum: ω = √(g/L) = √(9.8/1) = 3.13 rad/s

Step 4: Calculate total energy

E = ½mω²x₀² = ½ × 0.2 × (3.13)² × (0.262)²

E = 0.1 × 9.8 × 0.0686 = 0.067 J

Step 5: Speed at lowest point

At lowest point, all energy is kinetic:

½mv² = E, so v = √(2E/m) = √(2×0.067/0.2) = 0.82 m/s

📝 Quick Solution

Brief Solution

Given: L = 1 m, θ₀ = 15°, m = 0.2 kg

Convert to radians: θ₀ = 0.262 rad

Linear amplitude: x₀ = Lθ₀ = 0.262 m

Angular frequency: ω = √(g/L) = 3.13 rad/s

Total energy:

E = ½mω²x₀² = 0.067 J

Speed at bottom:

v = √(2E/m) = 0.82 m/s

🔬 Investigation Task

Interactive Simulation

Use this PhET simulation to investigate energy transformations in oscillating systems:

Investigation Questions:

How does increasing the amplitude affect the total energy?
At what position is the kinetic energy equal to the potential energy?
How does changing the mass affect the period and energy?
What happens to the energy when you add damping?

Brief Answers

1. Total energy increases with the square of amplitude (E ∝ x₀²)
2. KE = PE when displacement is about 70.7% of amplitude (x = x₀/√2)
3. Increasing mass increases period (T ∝ √m) and energy (E ∝ m) for same amplitude
4. Damping causes energy to decrease exponentially over time due to energy loss

👥 Group/Pair Activity

Collaborative Learning Activity

Work with your partner or group to complete this energy analysis challenge:

Discussion Points:

Why is energy conservation important in understanding SHM?
How do energy graphs help visualize oscillatory motion?
What practical applications use energy principles in oscillating systems?
How does damping affect the energy analysis of real oscillators?

Group Challenge Activities:

Create energy vs. time graphs for different SHM systems
Design experiments to measure energy in oscillating systems
Compare energy storage in different types of oscillators
Investigate energy harvesting from oscillatory motion

✏️ Individual Assessment

Structured Questions — Individual Work

Question 1 (Analysis):

A horizontal mass-spring system has a mass of 2 kg and spring constant 50 N/m. The mass is displaced 0.2 m from equilibrium and released.

Calculate the angular frequency of oscillation.
Determine the total energy of the system.
Find the kinetic and potential energies when x = 0.1 m.
Calculate the speed of the mass when it passes through equilibrium.
At what displacement is the kinetic energy three times the potential energy?

Answer

a) ω = √(k/m) = √(50/2) = 5 rad/s
b) E = ½mω²x₀² = ½(2)(5)²(0.2)² = 1 J
c) PE = ½kx² = ½(50)(0.1)² = 0.25 J; KE = E — PE = 1 — 0.25 = 0.75 J
d) v = √(2E/m) = √(2×1/2) = 1 m/s
e) If KE = 3PE, then E = KE + PE = 3PE + PE = 4PE, so PE = E/4
½kx² = E/4 = 0.25 J, so x² = 0.01, x = 0.1 m

Question 2 (Synthesis):

Two identical masses are attached to springs with different spring constants. Spring A has k₁ = 100 N/m, spring B has k₂ = 400 N/m. Both masses are given the same initial energy of 2 J.

Compare the amplitudes of oscillation for both systems.
Compare the frequencies of oscillation.
Compare the maximum speeds achieved.
Which system has the larger amplitude and why?
Design an experiment to verify these theoretical predictions.

Answer

a) E = ½kx₀², so x₀ = √(2E/k)
x₀A = √(4/100) = 0.2 m; x₀B = √(4/400) = 0.1 m
b) ωA = √(k₁/m) = √(100/m); ωB = √(k₂/m) = √(400/m) = 2√(100/m)
So ωB = 2ωA (B oscillates twice as fast)
c) vmax = x₀ω
vmaxA = 0.2√(100/m) = 2√(100/m); vmaxB = 0.1×2√(100/m) = 0.2×2√(100/m)
Both have same maximum speed!
d) System A has larger amplitude because softer spring allows more displacement for same energy
e) Use identical masses, measure periods and amplitudes, verify E = ½mω²x₀² for both

Question 3 (Evaluation):

A simple pendulum clock is designed to keep accurate time. The pendulum has length 1 m and amplitude 5°.

Calculate the total energy per unit mass of the pendulum.
Analyze how the energy changes if the amplitude decreases to 3° due to air resistance.
Evaluate the effect on timekeeping accuracy.
Propose methods to maintain constant amplitude.
Compare this system with a modern quartz clock in terms of energy requirements.

Answer

a) E/m = ½ω²x₀² = ½(g/L)(Lθ₀)² = ½gLθ₀² = ½(9.8)(1)(5π/180)² = 0.037 J/kg
b) New energy: E’/m = ½gL(3π/180)² = 0.013 J/kg (65% decrease)
c) For small angles, period is independent of amplitude, so minimal effect on accuracy
d) Use escapement mechanism to add energy each swing, automatic winding systems
e) Pendulum: continuous energy input needed to overcome friction; Quartz: much lower power (battery lasts years), electronic oscillator more stable

Question 4 (Critical Thinking):

An engineer designs a shock absorber using a spring-mass-damper system. The goal is to absorb energy efficiently while minimizing oscillations.

Analyze the energy flow in an undamped vs. damped oscillator.
Explain why critical damping is often preferred for shock absorbers.
Calculate the energy dissipated in one complete oscillation for a lightly damped system with decay constant γ = 0.1 s⁻¹.
Design considerations: How would you optimize the system for different applications (car suspension vs. building earthquake protection)?
Evaluate the trade-offs between energy absorption and response time.

Answer

a) Undamped: energy oscillates between KE and PE with no loss; Damped: energy continuously converts to heat through friction
b) Critical damping returns to equilibrium fastest without overshooting, optimal for shock absorption
c) For light damping: E(t) = E₀e⁻²γt; Energy lost in one period T = E₀(1 — e⁻²γT)
If T = 2π/ω ≈ 1s, energy lost ≈ E₀(1 — e⁻⁰·²) ≈ 0.18E₀
d) Car: moderate damping for comfort; Building: high damping capacity for large energy absorption
e) Higher damping = more energy absorption but slower response; Lower damping = faster response but less energy dissipation

Question 5 (Application):

A renewable energy company wants to harvest energy from ocean waves using oscillating buoys. Each buoy has mass 1000 kg and oscillates with amplitude 2 m and period 8 s.

Calculate the total energy stored in one buoy’s oscillation.
Determine the average power that could theoretically be extracted.
Analyze the efficiency challenges in real energy harvesting.
Compare this with other renewable energy sources in terms of energy density.
Propose improvements to increase energy yield from the system.

Answer

a) ω = 2π/T = 2π/8 = 0.785 rad/s
E = ½mω²x₀² = ½(1000)(0.785)²(2)² = 1,230 J
b) If all energy extracted per period: P = E/T = 1,230/8 = 154 W per buoy
c) Real efficiency ~20-40% due to mechanical losses, variable wave conditions, power conversion losses
d) Solar: ~150 W/m²; Wind: ~300-600 W/m²; Wave: ~50-100 W/m² (highly variable)
e) Optimize buoy design for local wave conditions, use resonance tuning, improve power take-off systems, deploy arrays for consistent output

🔗 Additional Resources

Useful Links and References

📚 Study Materials:

🎥 Video Resources:

🧮 Interactive Tools:

📖 Advanced Reading:

🤔 Lesson Reflection

Reflection Questions

Think about your learning today:

💡 Understanding:

Can you visualize how energy transforms between kinetic and potential forms during oscillation?
How does the total energy formula E = ½mω²x₀² connect amplitude, frequency, and mass?
What insights do energy graphs provide about oscillatory motion?
How does energy conservation help you understand SHM systems?

🎯 Application:

How would you explain energy transformations in a playground swing to a friend?
What practical applications of energy in oscillating systems can you think of?
How might this knowledge apply to engineering vibration problems?
Which problem-solving strategies were most effective for energy calculations?

🔄 Next Steps:

What aspects of oscillations and waves would you like to explore further?
How confident do you feel about analyzing energy in complex oscillating systems?
What questions do you still have about energy conservation in SHM?
How might this knowledge connect to other areas of physics you’ve studied?

📝 Self-Assessment Scale (1-5):

Rate your confidence in:

Describing energy interchange in SHM: ___/5
Using the total energy formula E = ½mω²x₀²: ___/5
Calculating kinetic and potential energies at any position: ___/5
Analyzing energy graphs and diagrams: ___/5
Applying energy conservation to solve SHM problems: ___/5

🎯 Learning Goals Achieved:

☐ I can describe the interchange between kinetic and potential energy during SHM
☐ I can recall and use E = ½mω²x₀² for total energy calculations
☐ I understand energy transformations at different positions in oscillation
☐ I can apply conservation of energy to analyze SHM systems
☐ I can solve complex problems involving energy in oscillating systems

🌟 Key Insights:

«Energy in SHM is like a pendulum of energy itself — constantly swinging between kinetic and potential forms, but the total always remains constant. The amplitude determines how much energy the system has, while the frequency determines how quickly this energy transforms back and forth. It’s a beautiful demonstration of conservation principles in nature!»