Q: What does Singapore Physics Olympiad (SPhO): Parent & Student Guide cover? A: Evergreen reference to SPhO structure, scoring, timeline and prep strategy for IP and JC candidates.
TL;DR SPhO rewards deep conceptual fluency in mechanics, E&M and thermal/modern topics, plus disciplined practical skills. Build a June→November plan that blends past papers, experiment design, and error analysis. Distinctions open doors to APhO/IPhO training and strengthen H2/H3 Physics outcomes.
Registration quick answer (Singapore): Schools nominate candidates to SPhO; individual sign‑up is not offered. Yearly circulars and dates are announced via official organisers — check NUS Physics’ SPhO page for current instructions: https://www.physics.nus.edu.sg/olympiad/spho
1 What is SPhO?
Organiser: Institute of Physics, Singapore (IPS)
Format: 4-h Theory + 4-h Practical (on separate days)
Typical cadence: November sittings; awards/ceremony thereafter
Eligibility: JC/IP candidates via school nomination (confirm yearly details)
2 Why SPhO matters for IP students
Curriculum bridge: accelerates H2 mastery; primes H3 research modules
Portfolio signalling: nationally recognised achievement; pipeline to APhO/IPhO
Metacognition: planning experiments under time pressure mirrors A-Level Paper 3/4 skills
3 Paper structure and scoring
Theory: long structured problems; vector fluency, calculus in context
Note: IPS releases yearly details on format and eligibility via schools. Treat examples above as orientation; your school circular is authoritative for your cohort.
9 Worked examples (with solutions)
9.1 Induction and energy consistency (classic “loop entering B-field”)
Problem. A conducting rectangular loop of width w and total resistance R is pulled at constant speed v into a region of uniform magnetic field B pointing into the page. The field edge is sharp and vertical. When a length x of the loop is inside the field, find (i) the induced emf, (ii) current, and (iii) the pulling force F required to keep speed constant. Show that the mechanical power input equals the electrical power dissipated.
Solution. Flux Φ = B·(w·x). The flux changes because x increases at rate v.
Faraday: emf magnitude ε = |dΦ/dt| = B w v. Direction by Lenz: current opposes entry.
Ohm's law: I = ε / R = (B w v)/R.
Magnetic force on the “leading” segment (length w) in field: F = I w B = (B w v / R)·w·B = B² w² v / R, opposing motion. To keep v constant, the applied pull must match this magnitude.
Power balance. Mechanical input P_mech = F v = (B² w² v / R)·v = B² w² v² / R. Electrical dissipation P_elec = I² R = (B² w² v²)/R. Hence P_mech = P_elec as expected (energy conservation). ∎
Exam habits. Draw a clear diagram, mark the “active” segment in field, and state Lenz's direction verbally to avoid sign confusion.
9.2 Mechanics — two-block system with friction
Problem. A block (mass m) rests on a rough horizontal table (coefficient of kinetic friction μ) and is connected by a light, inextensible string over a smooth pulley at the table's edge to a hanging mass M. The system is released from rest and moves with acceleration a. Find a and the string tension T, assuming motion occurs and friction is kinetic.
Solution. Take rightward motion of m and downward motion of M as positive. For block m: friction f_k = μ m g opposes motion (leftward). Newton's second law:
Table block: T - μ m g = m a. Hanging block: M g - T = M a. Add equations: (T - μ m g) + (M g - T) = (m + M) a ⇒ a = (M g - μ m g)/(m + M).
Back-substitute to get T, e.g. from table block: T = m a + μ m g = m (M g - μ m g)/(m + M) + μ m g = (m M g + μ m² g + μ m g (m + M) - μ m² g)/(m + M) simplifies to T = (m M g + μ m M g)/(m + M) = m M g (1 + μ)/(m + M).
Sanity checks. • If μ = 0, a = M g/(m + M) (frictionless Atwood variant). • If M ≪ μ m, acceleration tends to negative (no motion); consistent with the “assuming motion” clause—use inequality M > μ m to ensure sliding.
