Learning objectives
- I can explain and apply: Simple harmonic motion.
- I can explain and apply: Period and phase.
- I can explain and apply: Energy and turning points.
Panhellenic Exams / Physics / Two-unit revision pilot
Oscillations: structured theory, worked examples, answered practice, and a mastery checklist for Panhellenic Exams.
CHAPTER PLAN
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Estimated active study time
156 minutes
Objectives, key ideas, and structured theory.
Open →Worked examples that explain every step.
Open →Graded tasks with hints and answer guidance.
Open →Assignments from core work to challenge.
Open →A timed, marked test with solution guidance.
Open →Progress is stored only in this browser and does not require AI credits.
Unit
The essential chapter ideas in a clear sequence before practice.
In ideal simple harmonic motion, restoring force is proportional and opposite to displacement: F=−Dx. For an ideal horizontal frictionless spring, D=k and ω=√(k/m). Measure displacement from equilibrium.
For x=A·sin(ωt+φ), velocity is Aω·cos(ωt+φ) and acceleration is −ω²x. Period T=2π/ω does not depend on amplitude in the ideal linear model. Numerical phase exercises here use φ=0.
Total mechanical energy is ½DA². At turning points velocity is zero but force and acceleration magnitudes are greatest. At equilibrium speed is greatest. This pilot does not include damped or driven oscillations.
Written specifically for YourFavTeacher with AI assistance, without copying past-paper prompts. Technical and numerical checks do not replace review by a human educator. It is not an official paper or endorsed by an examination body.
Physics
Follow the method step by step and check why every step is valid.
An ideal horizontal spring system has m=1 kg and k=36 N/m. Find ω and T.
ω=6 rad/s; T=2π/6 s
For SHM with ω=6 rad/s and x=0.1 m, find acceleration.
−3.6 m/s²
At an oscillation endpoint, velocity is momentarily zero. Is the restoring force also zero?
No; the force magnitude is maximal.
Oscillations
Eight graded tasks from core fluency to exam-style application. Work independently before opening a hint or answer.
An ideal horizontal spring system has m=1 kg and k=9 N/m. Find ω and T.
For an ideal spring, ω=√(k/m) and T=2π/ω.
For an ideal spring, ω=√(k/m) and T=2π/ω. ω=√9=3 rad/s; T=2π/3 s Period is independent of amplitude in the ideal model. ω=3 rad/s; T=2π/3 s
Simple harmonic motion has A=0.2 m and ω=3 rad/s. Find maximum speed.
Maximum speed occurs at equilibrium: vmax=ωA.
Maximum speed occurs at equilibrium: vmax=ωA. υmax=3·0.2=0.6 m/s Speed is zero at the endpoints, not maximum. 0.6 m/s
A spring with k=100 N/m oscillates with amplitude A=0.02 m. Find total energy.
In the ideal system E=kA²/2.
In the ideal system E=kA²/2. E=0.5·100·(0.02)²=0.02 J Amplitude is in metres and is squared. 0.02 J
At which positions in simple harmonic motion is acceleration magnitude greatest?
a=−ω²x
a=−ω²x Its magnitude increases with |x|. The greatest |x| is amplitude A. At the endpoints x=±A, directed towards equilibrium.
For SHM with ω=3 rad/s and x=0.1 m, find acceleration.
Acceleration has the opposite sign to displacement.
Acceleration has the opposite sign to displacement. a=−ω²x=−3²·0.1=−0.9 m/s² The negative sign points towards equilibrium. −0.9 m/s²
Double the amplitude of an ideal spring without changing mass or stiffness. How do period and energy change?
T=2π√(m/k)
T=2π√(m/k) E′=k(2A)²/2=4E The model is ideal, lossless, and obeys Hooke's law. Same period, four times the energy.
In SHM, x=0.2sin(3t) in SI. Find position and direction of motion at t=0.
Velocity is the derivative of position.
Velocity is the derivative of position. x(0)=0; υ(t)=0.6cos(3t); υ(0)=0.6>0 x=0 does not mean the body is stationary. x(0)=0 m; υ(0)=0.6 m/s > 0
At an oscillation endpoint, velocity is momentarily zero. Is the restoring force also zero?
F=−kx
F=−kx At an endpoint |x|=A, so |F|=kA. The force towards equilibrium reverses the motion. No; the force magnitude is maximal.
Oscillations
Six distinct assignments from core fluency to challenge, each with an estimated time, hint, and answer guide.
An ideal horizontal spring system has m=1 kg and k=16 N/m. Find ω and T.
For an ideal spring, ω=√(k/m) and T=2π/ω.
For an ideal spring, ω=√(k/m) and T=2π/ω. ω=√16=4 rad/s; T=2π/4 s Period is independent of amplitude in the ideal model. ω=4 rad/s; T=2π/4 s
Simple harmonic motion has A=0.2 m and ω=4 rad/s. Find maximum speed.
Maximum speed occurs at equilibrium: vmax=ωA.
Maximum speed occurs at equilibrium: vmax=ωA. υmax=4·0.2=0.8 m/s Speed is zero at the endpoints, not maximum. 0.8 m/s
A spring with k=100 N/m oscillates with amplitude A=0.03 m. Find total energy.
In the ideal system E=kA²/2.
In the ideal system E=kA²/2. E=0.5·100·(0.03)²=0.045 J Amplitude is in metres and is squared. 0.045 J
At which positions in simple harmonic motion is acceleration magnitude greatest?
a=−ω²x
a=−ω²x Its magnitude increases with |x|. The greatest |x| is amplitude A. At the endpoints x=±A, directed towards equilibrium.
For SHM with ω=4 rad/s and x=0.1 m, find acceleration.
Acceleration has the opposite sign to displacement.
Acceleration has the opposite sign to displacement. a=−ω²x=−4²·0.1=−1.6 m/s² The negative sign points towards equilibrium. −1.6 m/s²
Double the amplitude of an ideal spring without changing mass or stiffness. How do period and energy change?
T=2π√(m/k)
T=2π√(m/k) E′=k(2A)²/2=4E The model is ideal, lossless, and obeys Hooke's law. Same period, four times the energy.
45 minutes / 40 marks
A timed, full-mark self-assessment with model-answer guidance.
Show working, units and a final check. Equivalent correct methods are accepted. Timing is a revision guide, not official examination conditions.
1. An ideal horizontal spring system has m=1 kg and k=25 N/m. Find ω and T.
5 marksFor an ideal spring, ω=√(k/m) and T=2π/ω. ω=√25=5 rad/s; T=2π/5 s Period is independent of amplitude in the ideal model. ω=5 rad/s; T=2π/5 s
2. Simple harmonic motion has A=0.2 m and ω=5 rad/s. Find maximum speed.
5 marksMaximum speed occurs at equilibrium: vmax=ωA. υmax=5·0.2=1 m/s Speed is zero at the endpoints, not maximum. 1 m/s
3. A spring with k=100 N/m oscillates with amplitude A=0.04 m. Find total energy.
5 marksIn the ideal system E=kA²/2. E=0.5·100·(0.04)²=0.08 J Amplitude is in metres and is squared. 0.08 J
4. At which positions in simple harmonic motion is acceleration magnitude greatest?
5 marksa=−ω²x Its magnitude increases with |x|. The greatest |x| is amplitude A. At the endpoints x=±A, directed towards equilibrium.
5. For SHM with ω=5 rad/s and x=0.1 m, find acceleration.
5 marksAcceleration has the opposite sign to displacement. a=−ω²x=−5²·0.1=−2.5 m/s² The negative sign points towards equilibrium. −2.5 m/s²
6. Double the amplitude of an ideal spring without changing mass or stiffness. How do period and energy change?
5 marksT=2π√(m/k) E′=k(2A)²/2=4E The model is ideal, lossless, and obeys Hooke's law. Same period, four times the energy.
7. In SHM, x=0.2sin(5t) in SI. Find position and direction of motion at t=0.
5 marksVelocity is the derivative of position. x(0)=0; υ(t)=1cos(5t); υ(0)=1>0 x=0 does not mean the body is stationary. x(0)=0 m; υ(0)=1 m/s > 0
8. At an oscillation endpoint, velocity is momentarily zero. Is the restoring force also zero?
5 marksF=−kx At an endpoint |x|=A, so |F|=kA. The force towards equilibrium reverses the motion. No; the force magnitude is maximal.
Unit
Curriculum reference sources. Always confirm the teaching sequence with the school and tutor.
Physics
Strict methodology by question: diagram, principles/laws, equations, numerical accuracy, and qualitative explanation.
The structure follows the official textbook layout and is used to organise study.
The areas that usually create mistakes or need extra revision.
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Targeted practice before full tests so coverage is clear.
How to measure progress in this chapter and when it enters a cumulative mock.
What to do after finishing the chapter and how it connects to the next unit.
Note: for the official examinable syllabus of each school year, always confirm with the school, tutor, and current Ministry/IEP announcements.