Periodic Motion Practice Set: NEB 12 Physics

Worked problem 1

Read amplitude, angular frequency and phase

Given x=0.060 cos(5t+π/3) m, identify A, ω, T, f and x at t=0. A=0.060 m, ω=5 rad s⁻¹, T=2π/5=1.257 s and f=1/T=0.796 Hz. Initial displacement x₀=0.060 cosπ/3=0.030 m.

Initial velocity v₀=−Aω sinπ/3=−0.260 m s⁻¹, so the object is moving negative. Position alone would not fully identify the state. Check with the OpenStax SHM equations.

Worked problem 2

Spring period and parameter change

A 0.80 kg mass on k=125 N m⁻¹ oscillates ideally. T=2π√(m/k)=2π√(0.80/125)=0.503 s. If mass becomes 3.20 kg, it increases by factor four, so period doubles to 1.01 s without recalculating from scratch.

If k doubles instead while m remains 0.80 kg, period changes by 1/√2. State the linear-spring and small-oscillation model; a stretched spring beyond Hooke’s-law range is not described reliably.

Worked problem 3

Energy at an intermediate displacement

A spring with k=200 N m⁻¹ has amplitude 0.10 m and mass 0.50 kg. Total energy E=½kA²=1.00 J. At x=0.060 m, U=½kx²=0.36 J and K=0.64 J. Speed follows K=½mv², so v=√(2K/m)=1.60 m s⁻¹.

Independent check: ω=√(k/m)=20 rad s⁻¹ and v=ω√(A²−x²)=20√(0.0100−0.0036)=1.60 m s⁻¹. Two routes agree.

Worked problem 4

Simple pendulum and gravity

A pendulum has length 0.90 m and small-angle period 1.90 s. Estimate g. Rearrange T=2π√(L/g) to g=4π²L/T². Substitution gives g=4π²(0.90)/(1.90²)=9.84 m s⁻².

Measure length to the bob centre and time multiple cycles. The answer is an estimate within model and measurement uncertainty, not proof that all ideal assumptions were exact. Explore controlled changes in PhET Pendulum Lab.

Worked problem 5

Translate a verbal start into graphs

An oscillator starts at x=0 moving positive with angular frequency ω. Choose x=A sinωt. Then v=Aω cosωt and a=−Aω² sinωt. At t=0, v is maximum positive and a=0. At T/4, x=A, v=0 and a is maximum negative.

Sketch x, v and a on aligned axes. The v curve leads x by a quarter cycle under this chosen sine start, while a is opposite x. Label amplitudes and time marks 0, T/4, T/2, 3T/4 and T.

Worked problem 6

Damping and resonance explanation

A lightly damped oscillator loses amplitude over time because resistive forces remove mechanical energy. If driven periodically, its steady response depends on driving frequency, natural frequency and damping. Response can become large near resonance, but real damping and nonlinear limits keep it finite.

A complete answer states the energy mechanism: the driver does positive net work efficiently when force timing/phase matches the motion. “Resonance happens when frequencies are equal” is incomplete unless the response and damping are discussed.

Independent set

Solve eight transfer questions

1. For x=0.04 sin8t, find A, T, vmax and amax.
2. At x=A/2, express speed as a fraction of vmax.
3. A spring period is 0.60 s for 0.25 kg. Find k.
4. Mass doubles and k halves. Find the period factor.
5. A 1.2 m pendulum operates at small angle. Estimate T using g=9.8 m s⁻².
6. Explain why bob mass cancels from ideal pendulum period.
7. Draw x, v and a when motion begins at +A.
8. Design a T²–L pendulum investigation with uncertainty reduction.

Answer prompts: (1) use ω=8; (2) v/vmax=√(1−x²/A²); (3) rearrange spring period; (4) factor √(m/k); (5) pendulum relation; (6) tangential equation; (7) quarter-cycle shifts; (8) multiple cycles, repeats, controlled angle and best-fit graph.

Use the study guide to repair the first wrong decision, not to copy the final number.

Review protocol

Mark model, phase and graph errors separately

Classify errors as definition, model condition, initial phase, graph slope/area, equation choice, unit, algebra, practical design or explanation. Redo closed-book, then use a fresh problem after two days and one week. Keep this chapter in the Class 12 revision roadmap.

For online or physical NEB tuition, call 9846662070. The MKS Education panel provides SAT/IELTS/PTE/DET and study-abroad pre-counselling contacts after Grade 12.

Checked sources

References and related learning

• CDC Nepal: Physics Grade 12
• CDC Nepal: Secondary Curriculum
• OpenStax: Simple Harmonic Motion
• OpenStax: Pendulums
• PhET: Pendulum Lab
• PhET: Masses and Springs

Rebuild weak concepts with the Periodic Motion guide. Curriculum and institutional pages were checked on 2 August 2026; follow current CDC, NEB and school notices if requirements change.

Worked problem 7

Recover phase from initial conditions

An oscillator has A=0.050 m and ω=6.0 rad s⁻¹. At t=0 it is at x=0.025 m and moving negative. In x=Acos(ωt+φ), cosφ=x/A=0.5, giving candidate phases ±π/3. Velocity is −Aωsinφ. Negative initial velocity requires sinφ positive, so choose φ=+π/3, not −π/3. The complete equation is x=0.050cos(6t+π/3) m.

Check v₀=−(0.050)(6)sinπ/3=−0.260 m s⁻¹. Both candidate phases reproduce position; direction selects the correct state. This is why initial displacement alone cannot determine phase.

Mixed challenge

Connect data, graph and model selection

A student records ten-cycle times for a pendulum at five lengths. Plan the processing before seeing values: divide each total time by ten, calculate T², plot T² vertically against L horizontally, draw a best-fit line and interpret gradient as approximately 4π²/g. Include units s² m⁻¹. A nonzero intercept may indicate systematic timing or effective-length issues; it should be investigated rather than forced through zero.

Now suppose short lengths give consistent values but long lengths bend away from the line. List competing explanations: angle not kept small, changing release method, support motion, length measured to the wrong point, timing/counting error or a real limitation in the chosen range. Propose a test that distinguishes them. Scientific review compares hypotheses rather than naming “human error.”

Use one final timed problem that combines amplitude, phase, energy and period. After solving, write a three-sentence audit: why the model is SHM, which initial condition fixed phase, and which independent relation checked the number. A solution without this audit may be arithmetically correct but conceptually fragile.