Oscillations

General physics

Simple Harmonic Motion (SHM) — Physics Lesson

🎯 Learning Objectives

Learning Objectives

Understand the concepts of simple harmonic motion (SHM) and its characteristics
Learn the mathematical description of oscillations including period, frequency, and amplitude
Analyze the energy changes in oscillatory systems
Apply SHM principles to real-world examples like pendulums and springs
Calculate displacement, velocity, and acceleration in SHM

🗣️ Language Objectives

Language Objectives

Use scientific vocabulary related to oscillations and SHM accurately
Describe periodic motion using appropriate mathematical terminology
Explain energy transformations in oscillating systems using precise language
Communicate SHM concepts clearly in English using graphs and equations
Apply mathematical language to describe wave properties and motion

📚 Key Terms

Key Terms

English	Russian (Русский)	Kazakh (Қазақша)
Simple Harmonic Motion	Простое гармоническое движение	Қарапайым гармоникалық қозғалыс
Oscillation	Колебание	Тербеліс
Amplitude	Амплитуда	Амплитуда
Period	Период	Период
Frequency	Частота	Жиілік
Displacement	Смещение	Ығысу
Restoring force	Восстанавливающая сила	Қалпына келтіруші күш
Equilibrium position	Положение равновесия	Тепе-теңдік орны
Phase	Фаза	Фаза
Angular frequency	Угловая частота	Бұрыштық жиілік

🎴 Study Cards

SHM Study Cards

Displacement Equation

x = A cos(ωt + φ)

A = amplitude, ω = angular frequency

φ = phase constant

Velocity Equation

v = -Aω sin(ωt + φ)

Maximum velocity: v_max = Aω

At equilibrium position

Acceleration Equation

a = -Aω² cos(ωt + φ)

a = -ω²x

Maximum at extreme positions

Period Formula

T = 2π/ω

For spring: T = 2π√(m/k)

For pendulum: T = 2π√(l/g)

Energy in SHM

Total Energy = ½kA²

KE = ½mv²

PE = ½kx²

Frequency & Period

f = 1/T = ω/2π

Measured in Hz (s⁻¹)

Independent of amplitude

📖 Glossary

Glossary

Simple Harmonic Motion (SHM): A type of periodic motion where the restoring force is directly proportional to the displacement from equilibrium 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 — қалпына келтіруші күш тепе-теңдік орнынан ығысуға тура пропорционал және қарама-қарсы бағытта әсер ететін периодты қозғалыс түрі.

Amplitude (A): The maximum displacement from the equilibrium position during oscillation.

Translation

Russian: Амплитуда — максимальное смещение от положения равновесия во время колебания.
Kazakh: Амплитуда — тербеліс кезінде тепе-теңдік орнынан ең үлкен ығысу.

Period (T): The time required for one complete oscillation or cycle.

Translation

Russian: Период — время, необходимое для одного полного колебания или цикла.
Kazakh: Период — бір толық тербеліс немесе циклге қажетті уақыт.

Frequency (f): The number of complete oscillations per unit time, measured in Hertz (Hz).

Translation

Russian: Частота — количество полных колебаний в единицу времени, измеряется в герцах (Гц).
Kazakh: Жиілік — уақыт бірлігіндегі толық тербелістер саны, герцпен (Гц) өлшенеді.

Restoring Force: The force that acts to return an oscillating object to its equilibrium position.

Translation

Russian: Восстанавливающая сила — сила, которая действует для возвращения колеблющегося объекта в положение равновесия.
Kazakh: Қалпына келтіруші күш — тербелетін нысанды тепе-теңдік орнына қайтару үшін әсер ететін күш.

Angular Frequency (ω): The rate of change of phase of oscillation, measured in radians per second (rad/s).

Translation

Russian: Угловая частота — скорость изменения фазы колебания, измеряется в радианах в секунду (рад/с).
Kazakh: Бұрыштық жиілік — тербеліс фазасының өзгеру жылдамдығы, секундына радианмен (рад/с) өлшенеді.

🔬 Theory: Understanding Simple Harmonic Motion

Theory: Understanding Simple Harmonic Motion

What is Simple Harmonic Motion?

occurs when an object back and forth about an under the influence of a that is to the .

Translation

Kazakh: Қарапайым гармоникалық қозғалыс нысан тепе-теңдік орны туралы ығысуға пропорционал қалпына келтіруші күштің әсерінен алға-артқа тербелген кезде пайда болады.

Mathematical Description

The displacement x of an object in SHM can be described by:

x = A cos(ωt + φ)

Where:

A = amplitude (maximum displacement)
ω = angular frequency (rad/s)
t = time (s)
φ = phase constant (rad)

Translation

Kazakh: SHM-дағы нысанның ығысуы x мынадай формуламен сипатталады: x = A cos(ωt + φ), мұндағы A — амплитуда, ω — бұрыштық жиілік, t — уақыт, φ — фаза тұрақтысы.

Velocity and Acceleration

The velocity in SHM is:

v = -Aω sin(ωt + φ)

The acceleration in SHM is:

a = -Aω² cos(ωt + φ) = -ω²x

Translation

Kazakh: SHM-дағы жылдамдық v = -Aω sin(ωt + φ) формуласымен, ал үдеу a = -ω²x формуласымен анықталады. Үдеу ығысуға пропорционал және қарама-қарсы бағытталған.

Period and Frequency

The period T is the time for one complete oscillation:

T = 2π/ω

The frequency f is the number of oscillations per second:

f = 1/T = ω/2π

Translation

Kazakh: Период T — бір толық тербелісге кететін уақыт, T = 2π/ω формуласымен табылады. Жиілік f — секундтағы тербелістер саны, f = 1/T формуласымен анықталады.

Energy in SHM

In SHM, energy constantly transforms between kinetic and potential forms:

Kinetic Energy: KE = ½mv² = ½mA²ω²sin²(ωt + φ)
Potential Energy: PE = ½kx² = ½kA²cos²(ωt + φ)
Total Energy: E = KE + PE = ½kA² = ½mA²ω² (constant)

Translation

Kazakh: SHM-да энергия кинетикалық және потенциалдық түрлер арасында үздіксіз айналады. Жалпы энергия тұрақты болып қалады және амплитудаға тәуелді.

Examples of SHM

Mass-Spring System: T = 2π√(m/k)
Simple Pendulum: T = 2π√(l/g) (for small angles)
Physical Pendulum: T = 2π√(I/mgd)

Translation

Kazakh: SHM мысалдары: серіппелі жүйе, қарапайым маятник (кішкене бұрыштар үшін) және физикалық маятник. Әрқайсысының өзіндік период формуласы бар.

Practice Questions

1. Easy: What is the relationship between period and frequency?

Answer

Period and frequency are inversely related: T = 1/f or f = 1/T. If the period increases, the frequency decreases, and vice versa.

2. Medium: A mass oscillates with amplitude 0.1 m and angular frequency 2 rad/s. Calculate the maximum velocity.

Answer

Maximum velocity occurs at equilibrium position: v_max = Aω = 0.1 × 2 = 0.2 m/s

3. Medium: If the mass of a spring-mass system is quadrupled, how does the period change?

Answer

Using T = 2π√(m/k), if mass is quadrupled (4m), then T_new = 2π√(4m/k) = 2 × 2π√(m/k) = 2T_original. The period doubles.

4. Hard (Critical Thinking): Explain why the amplitude does not affect the period in SHM. What does this tell us about the nature of the restoring force?

Answer

The amplitude doesn’t affect the period because SHM requires a restoring force that is directly proportional to displacement (F = -kx). This linear relationship means the system oscillates with the same frequency regardless of how far it’s displaced initially. This property (isochronism) is unique to linear restoring forces and breaks down when the force becomes non-linear (e.g., large angle pendulum motion).

🧠 Memory Exercises

Exercises on Memorizing SHM Terms

Exercise 1: Complete the Equations

1. Displacement in SHM: x = A cos(______ + φ)

2. Period formula: T = ______/ω

3. For a spring-mass system: T = 2π√(______/k)

4. Total energy in SHM: E = ½k______²

Answer

1. ωt
2. 2π
3. m
4. A

Exercise 2: True or False

1. In SHM, acceleration is maximum at the equilibrium position. (T/F)

2. The period of a simple pendulum depends on its mass. (T/F)

3. Energy is conserved in ideal SHM. (T/F)

4. Frequency and angular frequency are the same. (T/F)

Answer

1. False — acceleration is zero at equilibrium, maximum at extreme positions
2. False — period depends only on length and gravity for small angles
3. True — total mechanical energy remains constant
4. False — ω = 2πf, they differ by a factor of 2π

Exercise 3: Unit Matching

Match each quantity with its unit:

Amplitude → ?
Angular frequency → ?
Period → ?
Frequency → ?

Units: s, m, Hz, rad/s

Answer

Amplitude → m (meters)
Angular frequency → rad/s (radians per second)
Period → s (seconds)
Frequency → Hz (hertz)

🎥 Educational Video

Understanding Simple Harmonic Motion — Video Lesson

Additional Video Resources:

🧮 Worked Examples

Problem Solving Examples

Example 1: Spring-Mass System

Problem: A 0.5 kg mass attached to a spring with spring constant k = 20 N/m oscillates with amplitude 0.1 m. Calculate: (a) the period, (b) the maximum velocity, (c) the total energy.

Answer

Step-by-step solution:
(a) Period calculation:
T = 2π√(m/k) = 2π√(0.5/20) = 2π√(0.025) = 2π × 0.158 = 0.99 s

(b) Maximum velocity:
First find ω: ω = √(k/m) = √(20/0.5) = √40 = 6.32 rad/s
v_max = Aω = 0.1 × 6.32 = 0.632 m/s

(c) Total energy:
E = ½kA² = ½ × 20 × (0.1)² = ½ × 20 × 0.01 = 0.1 J
Alternative: E = ½mv_max² = ½ × 0.5 × (0.632)² = 0.1 J

Answer

Quick solution:
(a) T = 2π√(0.5/20) = 0.99 s
(b) v_max = A√(k/m) = 0.1√(20/0.5) = 0.63 m/s
(c) E = ½kA² = ½(20)(0.1)² = 0.1 J

Example 2: Simple Pendulum

Problem: A simple pendulum of length 1.0 m oscillates with small amplitude on Earth (g = 9.8 m/s²). Find the period and frequency of oscillation.

Answer

Detailed solution:
Given: l = 1.0 m, g = 9.8 m/s²

For a simple pendulum (small angles):
T = 2π√(l/g)

Calculating period:
T = 2π√(1.0/9.8) = 2π√(0.102) = 2π × 0.319 = 2.01 s

Calculating frequency:
f = 1/T = 1/2.01 = 0.498 Hz ≈ 0.50 Hz

Note: The period is independent of the mass and amplitude (for small angles).

Answer

Quick answer:
T = 2π√(l/g) = 2π√(1.0/9.8) = 2.01 s
f = 1/T = 0.50 Hz

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🔬 Investigation Activity

Interactive Simulation: Pendulum Lab

Use this PhET simulation to explore how different factors affect pendulum motion:

Investigation Questions:

1. How does changing the length affect the period of the pendulum?

Answer

Increasing the length increases the period. The relationship is T ∝ √l. If length is quadrupled, the period doubles.

2. Does the mass of the bob affect the period? Why or why not?

Answer

No, the mass does not affect the period for a simple pendulum. The gravitational force (mg) and the required centripetal force both depend on mass, so mass cancels out in the equation.

3. What happens to the period when you change the amplitude for small and large angles?

Answer

For small angles (< 15°), the period is approximately independent of amplitude. For large angles, the period increases with amplitude as the simple harmonic approximation breaks down.

4. How does gravity affect the pendulum’s motion?

Answer

Stronger gravity decreases the period (T ∝ 1/√g). On planets with higher gravity, pendulums oscillate faster.

👥 Group Activity

Collaborative Learning: SHM Concepts Quiz

Work in pairs or groups to complete this interactive quiz about simple harmonic motion:

Group Discussion Points:

Compare your understanding of SHM equations with your partner
Discuss real-world examples of oscillatory motion you observe daily
Analyze how damping affects real oscillations compared to ideal SHM
Create a concept map connecting all SHM terms and relationships
Present one practical application of SHM to the class

Extension Activity:

Design a simple experiment using everyday materials to demonstrate SHM principles. Consider using:

A mass hanging from a rubber band (spring system)
A ruler clamped to a table edge (cantilever beam)
A small object suspended by string (pendulum)

📝 Individual Assessment

Structured Questions — Individual Work

Question 1: Analysis and Calculation

A particle undergoes SHM with displacement given by x = 0.05 cos(4πt + π/3) meters, where t is in seconds.

a) Identify the amplitude, angular frequency, and phase constant.

b) Calculate the period and frequency of oscillation.

c) Find the displacement, velocity, and acceleration at t = 0.1 s.

d) Determine when the particle first reaches maximum displacement.

Answer

a) A = 0.05 m, ω = 4π rad/s, φ = π/3 rad
b) T = 2π/ω = 2π/(4π) = 0.5 s, f = 1/T = 2 Hz
c) At t = 0.1 s: x = 0.05 cos(4π(0.1) + π/3) = 0.05 cos(0.4π + π/3) = -0.043 m
v = -0.05(4π) sin(0.4π + π/3) = -0.628 sin(133.3°) = -0.458 m/s
a = -0.05(4π)² cos(0.4π + π/3) = -7.9 cos(133.3°) = 5.33 m/s²
d) Maximum displacement occurs when cos(4πt + π/3) = 1, so 4πt + π/3 = 0, t = -1/12 s. Since this is negative, the first maximum occurs at t = 2π/(4π) — π/3/(4π) = 5/12 s

Question 2: Synthesis and Application

A car’s suspension system can be modeled as a spring-mass system. The car (mass 1200 kg) bounces with a period of 1.5 s when loaded.

a) Calculate the effective spring constant of the suspension.

b) If the car hits a bump causing a 5 cm compression, calculate the maximum speed during the subsequent oscillation.

c) Design modifications to reduce the oscillation period by 20%. What changes would you make?

d) Explain why real car suspensions include dampers and how they affect the motion.

Answer

a) Using T = 2π√(m/k): 1.5 = 2π√(1200/k), solving: k = 4π²(1200)/(1.5)² = 21,062 N/m
b) v_max = Aω = A√(k/m) = 0.05√(21,062/1200) = 0.05 × 4.19 = 0.21 m/s
c) To reduce T by 20%: T_new = 0.8T = 1.2 s. Since T ∝ √(m/k), need to increase k or decrease m. If changing k: k_new = k/(0.8)² = 1.56k = 32,847 N/m
d) Dampers provide resistance proportional to velocity, causing oscillations to decay exponentially. This prevents prolonged bouncing and improves ride comfort and vehicle stability.

Question 3: Critical Evaluation

The simple pendulum equation T = 2π√(l/g) is an approximation valid only for small angles.

a) Derive the exact equation for large amplitude pendulum motion.

b) Calculate the percentage error when using the small angle approximation for a 30° amplitude.

c) Evaluate the practical implications of this limitation in real-world applications.

d) Suggest methods to minimize this error in precision timing devices.

Answer

a) Exact period: T = 4√(l/g) K(sin(θ₀/2)) where K is the complete elliptic integral of the first kind
b) For θ₀ = 30°: T_exact/T_approx ≈ 1.017, so error ≈ 1.7%
c) In clocks, this error accumulates over time. In seismometers, it affects sensitivity. In metronomes, it changes tempo accuracy.
d) Methods: Use smaller amplitudes, temperature compensation, electronic oscillators, or mathematical corrections in digital systems.

Question 4: Energy Analysis

A 2 kg mass oscillates on a spring with total energy 0.5 J and maximum displacement 0.1 m.

a) Calculate the spring constant and angular frequency.

b) Find the kinetic and potential energies when the displacement is 0.06 m.

c) At what displacement is the kinetic energy three times the potential energy?

d) Sketch graphs of KE, PE, and total energy versus displacement.

Answer

a) E = ½kA²: 0.5 = ½k(0.1)², so k = 100 N/m; ω = √(k/m) = √(100/2) = 7.07 rad/s
b) At x = 0.06 m: PE = ½kx² = ½(100)(0.06)² = 0.18 J; KE = E — PE = 0.5 — 0.18 = 0.32 J
c) When KE = 3PE and KE + PE = E: 3PE + PE = 0.5, so PE = 0.125 J; ½kx² = 0.125, x = 0.05 m
d) PE increases as x² (parabola), KE = E — PE (inverted parabola), total E is constant horizontal line

Question 5: Research and Innovation

Modern technologies often utilize controlled oscillations. Research and analyze one of the following applications:

• Atomic force microscopy (AFM) cantilevers

• Quartz crystal oscillators in watches

• MEMS (Micro-Electro-Mechanical Systems) accelerometers

• Earthquake detection seismometers

a) Explain how SHM principles apply to your chosen technology.

b) Analyze the advantages and limitations of using oscillatory motion.

c) Propose improvements or alternative approaches.

d) Discuss future developments in this field.

Answer

Example for Quartz Crystal Oscillators:
a) Quartz crystals exhibit piezoelectric effect, oscillating at precise frequencies when voltage is applied. The mechanical oscillation follows SHM principles with extremely stable frequency.
b) Advantages: High precision, temperature stability, low power consumption. Limitations: Frequency drift with age, sensitivity to mechanical shock, limited frequency range.
c) Improvements: Temperature compensation circuits, better mounting techniques, atomic clocks for ultra-precision applications.
d) Future: Chip-scale atomic clocks, optical frequency standards, quantum oscillators for unprecedented accuracy.

🔗 Additional Resources

Useful Links for Further Study

🤔 Lesson Reflection

Reflection Questions

Self-Assessment:

Rate your understanding (1-5 scale):

I can identify the characteristics of SHM: ⭐⭐⭐⭐⭐
I understand the mathematical relationships in SHM: ⭐⭐⭐⭐⭐
I can analyze energy transformations in oscillations: ⭐⭐⭐⭐⭐
I can solve problems involving springs and pendulums: ⭐⭐⭐⭐⭐
I can apply SHM concepts to real-world situations: ⭐⭐⭐⭐⭐

Critical Thinking:

1. What was the most challenging concept in this lesson and why?

2. How does understanding SHM help explain everyday phenomena you observe?

3. What questions do you still have about oscillatory motion?

4. How might climate change affect the precision of pendulum clocks?

5. What role does SHM play in modern technology and engineering?

Language Learning Reflection:

1. Which technical terms were most difficult to understand initially?

2. How confident are you in explaining SHM concepts in English?

3. What mathematical language and symbols need more practice?

4. How well can you connect SHM vocabulary to real-world examples?

Practical Applications:

1. Identify three examples of oscillatory motion in your daily life.

2. How would you explain SHM to someone without a physics background?

3. What careers or fields heavily utilize principles of oscillatory motion?

4. How might understanding SHM influence your approach to problem-solving?

Next Steps:

Based on your reflection, what topics would you like to explore further?

☐ Damped oscillations and resonance phenomena
☐ Wave motion and wave equations
☐ Coupled oscillators and normal modes
☐ Applications in engineering and technology
☐ Quantum harmonic oscillator concepts
☐ Fourier analysis of complex oscillations

Learning Goals for Next Lesson:

Based on today’s learning, write 2-3 specific goals for your next physics lesson:

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