Summary to follow. 9 syllabus statements (2 HL) · 20 questions · about twenty minutes.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank ·
Specialist review in progress
· How these pages are made
Assessed in Paper 1A (multiple choice), Paper 1B (data-based) and Paper 2 (short and extended response). IB Physics guide (first assessment 2025, updated November 2023).
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In preparation: 0 of 9 sections compiled and reviewed. The rest show key terms and common misconceptions from our question bank until they are.
Simple harmonic motion (SHM)
Simple harmonic motion (SHM)
Oscillation in which the acceleration of a body is proportional to its displacement from a fixed equilibrium position and is always directed towards that position. Two conditions are therefore both needed: a restoring force whose size is proportional to the displacement, and a direction always towards equilibrium. Not every periodic, back-and-forth motion is SHM. A consequence of the two conditions is that the time period does not depend on the amplitude.
Restoring force
The resultant force on an oscillating body that acts towards the equilibrium position. In SHM its magnitude is proportional to the displacement, F = −kx, where k is a constant for the system (for a mass on a spring, the spring constant). SI unit: N. For a simple pendulum the restoring force, mg sin θ, is only approximately proportional to the displacement, and only for small amplitudes.
Students often think Any motion that repeats back and forth with a regular period is simple harmonic motion. In fact No. SHM is the particular periodic motion in which the acceleration is proportional to the displacement from equilibrium and directed towards it. Many periodic motions do not meet this condition.
Students often think Any resultant force that acts towards the equilibrium position produces SHM, whatever its size at each displacement. In fact No. The force must also be proportional to the displacement. A restoring force of constant size produces oscillation, but not simple harmonic oscillation.
Defining equation of SHM, a = −ω²x
Defining equation of SHM, a = −ω²x
The acceleration a of a body moving with SHM equals −ω² multiplied by its displacement x from equilibrium, where ω is the angular frequency, a constant for the system. The magnitude of the acceleration is proportional to the magnitude of the displacement, so it is zero at equilibrium and greatest at the extremes, where it has magnitude ω²x₀.
Significance of the minus sign in a = −ω²x
ω² is positive, so the minus sign means that the acceleration always has the opposite sign to the displacement: the acceleration always points towards the equilibrium position. It does not mean that the acceleration is negative, that the body is slowing down, or that energy is being lost. Reversing the choice of positive direction reverses the signs of both a and x, so the minus sign remains.
Students often think The minus sign in a = −ω²x shows a deceleration: the oscillating body is slowing down. In fact No. It means the acceleration is opposite to the displacement. The body speeds up while moving towards equilibrium and slows down while moving away from it.
Students often think ω enters the SHM relations unsquared, so the displacement (or its square) is simply multiplied by ω. In fact ω is squared. The magnitude of the acceleration is ω² multiplied by the displacement.
Equilibrium position
Equilibrium position
The position at which the resultant force on the oscillating body, and so its acceleration, is zero. It is the midpoint of the oscillation and the origin from which displacement is measured. The body passes through it at its greatest speed; it is not the place where the body is at rest.
Displacement, x
The distance of the oscillating body from its equilibrium position, in a stated direction. It is a vector; for motion along a line it is a signed quantity, positive on one side of equilibrium and negative on the other, so it changes sign twice in each cycle. SI unit: m.
Amplitude, x₀
The maximum displacement of the oscillating body from its equilibrium position: the distance from equilibrium to either extreme. It is half the distance between the two extreme positions. For an undamped oscillator it is constant. SI unit: m.
Time period, T
The time taken for one complete oscillation, for example from one extreme to the other and back again, or between two successive passes through equilibrium in the same direction. SI unit: s.
Frequency, f
The number of complete oscillations per unit time. SI unit: Hz (s⁻¹).
Angular frequency, ω
The quantity 2πf, which measures the rate of oscillation in radians per second: one complete oscillation corresponds to 2π rad. It is the constant ω in a = −ω²x. SI unit: rad s⁻¹. It is not the same number as the frequency f.
Students often think The amplitude is the total distance between the two extreme positions of the oscillation (peak-to-peak). In fact No. The amplitude is the maximum displacement from equilibrium, which is half the distance between the two extremes.
Students often think Displacement is measured from a convenient fixed point, such as the point where the body was released or the zero of the measuring scale. In fact From the equilibrium position, not from the point of release, the end of the motion, or the zero of a measuring scale.
Relations between T, f and ω
Relations between T, f and ω
T = 1/f = 2π/ω, so f = 1/T and ω = 2πf = 2π/T. Converting between ω and f requires the factor 2π; ω and 1/T are not interchangeable.
Students often think Angular frequency and frequency are the same quantity, so ω can be found as 1/T or f used wherever ω appears. In fact No. ω = 2πf. An oscillator with f = 2.0 Hz has ω = 4π ≈ 12.6 rad s⁻¹.
Students often think The angular frequency is found as ω = 2πT (or the period as T = 2πω). In fact No. ω = 2π/T. A longer period means a smaller angular frequency.
Spring constant, k
Spring constant, k
The force per unit extension of a spring that obeys Hooke’s law, k = F/x. A stiffer spring has a larger k. SI unit: N m⁻¹.
Time period of a mass–spring system
For a body of mass m oscillating on a spring of spring constant k, T = 2π√(m/k). The period increases with mass (in proportion to √m), decreases with spring constant, and does not depend on the amplitude. Mass must be in kg, not grams and not replaced by weight in N.
Students often think The period of a mass–spring system is proportional to the mass, so quadrupling the mass quadruples the period. In fact No. T = 2π√(m/k), so T is proportional to √m. Four times the mass doubles the period.
Students often think A larger mass on the spring oscillates faster (shorter period), as though T = 2π√(k/m). In fact No. A larger mass has more inertia, so the same spring force gives it a smaller acceleration and the period is longer: T = 2π√(m/k).
Simple pendulum
Simple pendulum
A small, dense bob (treated as a point mass) on a light, inextensible string of length l, measured from the point of suspension to the centre of mass of the bob, swinging in a vertical plane. Its motion is approximately simple harmonic only for small amplitudes, where sin θ ≈ θ and the restoring force is almost proportional to the displacement.
Time period of a simple pendulum
For small amplitudes, T = 2π√(l/g). The period increases with length (in proportion to √l), decreases with the gravitational field strength g, and does not depend on the mass of the bob or, while the amplitude remains small, on the amplitude.
Students often think A heavier pendulum bob swings with a different period, usually thought to be shorter because it is pulled down more strongly. In fact No. For a simple pendulum T = 2π√(l/g), which contains no mass. A heavier bob of the same length has the same period.
Students often think A larger amplitude gives a longer period, because the bob has further to travel in each oscillation. In fact No. For small amplitudes the period does not depend on the amplitude: the bob travels further but, because the restoring force is larger, it moves faster.
Energy changes in one cycle (qualitative)
Energy changes in one cycle (qualitative)
During one cycle of an undamped oscillation, energy is transferred back and forth between kinetic energy and potential energy (elastic for a mass on a horizontal spring, gravitational for a pendulum). Kinetic energy is greatest at the equilibrium position and zero at each extreme; potential energy is least at equilibrium and greatest at each extreme. Each reaches its maximum twice per cycle.
Total energy of an undamped oscillator
The sum of the kinetic and potential energies. With no resistive forces it stays constant throughout the cycle: kinetic energy gained is potential energy lost, and vice versa. Energy is not used up by the motion itself; only resistive forces (damping) remove energy from the oscillator.
Students often think An oscillator has energy only when it is moving, so its total energy is greatest at equilibrium and zero at the extremes. In fact No. At the extremes the kinetic energy is zero, but the potential energy is at its maximum and equals the total energy.
Students often think Energy is used up in keeping the body moving, so the total energy of an oscillator falls every cycle even when there is no friction or air resistance. In fact No. With no resistive forces the total energy of the oscillator is constant. Energy is only transferred back and forth between kinetic and potential stores.
Phase angle HL
Phase angle
The argument (ωt + φ) of the sine function that describes the oscillation, which specifies the stage of the cycle the oscillator has reached. The phase constant φ is its value at t = 0 and is fixed by the displacement and the direction of motion at t = 0. One complete cycle corresponds to 2π rad. SI unit: rad; radians are used in all phase angle calculations.
Phase difference
The difference in phase angle between two oscillations of the same frequency. If one reaches a given stage of its cycle a time Δt after the other, the phase difference is 2π(Δt/T) rad. Oscillations with phase difference 0 (or 2π) rad are in phase; with π rad they are in antiphase. Two oscillators at the same displacement at the same instant are in phase only if they are also moving in the same direction.
Students often think Phase angles can be worked out in degrees, and the sine of ωt can be evaluated with the calculator in degree mode. In fact No. Phase angles are in radians: one complete cycle is 2π rad, and the argument of x = x₀ sin(ωt + φ) is in radians because ω is in rad s⁻¹.
Students often think The phase difference between two oscillations is the time lag between them. In fact No. The time lag Δt must be converted into an angle: Δφ = 2π(Δt/T) rad.
Displacement and velocity as functions of time HL
Displacement and velocity as functions of time
x = x₀ sin(ωt + φ) and v = ωx₀ cos(ωt + φ). The maximum speed is ωx₀, reached at equilibrium. The argument ωt + φ is in radians, so a calculator must be in radian mode.
Velocity at a given displacement
v = ±ω√(x₀² − x²). The speed is ωx₀ at x = 0 and zero at x = ±x₀; the ± shows that at a given displacement the body may be moving in either direction. √(x₀² − x²) is not equal to x₀ − x.
Energy in SHM (quantitative)
Total energy E_T = ½mω²x₀², constant for an undamped oscillator; potential energy E_p = ½mω²x², proportional to the square of the displacement; kinetic energy E_k = E_T − E_p = ½mω²(x₀² − x²). SI unit: J. At x = x₀/2, E_p is one quarter of E_T and E_k is three quarters.
Students often think Sine and cosine can be swapped in x = x₀ sin(ωt + φ) without changing φ, so the body's position at t = 0 is the same either way. In fact No. With x = x₀ sin(ωt + φ) and v = ωx₀ cos(ωt + φ), φ = 0 puts the body at equilibrium at t = 0. Replacing sin by cos while keeping the same φ describes a different starting point (an extreme for φ = 0); a cosine form is valid only with φ changed by π/2.
Students often think In v = ±ω√(x₀² − x²), the square root can be simplified to x₀ − x. In fact No. The square root of a difference of squares is not the difference of the numbers: √(0.080² − 0.050²) = 0.062, not 0.030.
Diagnostic a bearings check, not a test
10 questions, one per part of the topic where we can. Answer them, then see which statements you own and which to read.
1 A body is displaced from its equilibrium position and released. Which condition ensures that it then moves with simple harmonic motion?
Answer and reasoning
A resultant force of constant size acts on it, directed towards the equilibrium position. — A student who thinks any force towards equilibrium is enough picks this. A constant restoring force does make the body oscillate, but its size must be proportional to the displacement for the motion to be simple harmonic.
The resultant force on it is proportional to its displacement and acts in the same direction as it. — A student who remembers "proportional to displacement" but not the direction picks this. A force in the same direction as the displacement pushes the body further from equilibrium; it never returns, so there is no oscillation at all.
It moves back and forth along the same path, taking the same time for each cycle. — A student who equates any regular oscillation with SHM picks this. Many periodic motions, such as a ball bouncing on a hard floor, are not simple harmonic. SHM requires a restoring force proportional to the displacement.
The resultant force on it is proportional to its displacement and directed towards equilibrium. — This is the condition for SHM. The size of the resultant force is proportional to the displacement from equilibrium, and its direction is always towards equilibrium, so a = −ω²x.
2 The defining equation of simple harmonic motion is a = −ω²x. What is the significance of the minus sign?
Answer and reasoning
The body is slowing down at every point of its motion, as its acceleration is negative. — A student who reads a minus sign as "deceleration" picks this. The body speeds up whenever it moves towards equilibrium, because its acceleration and velocity then point the same way; the minus sign relates a to x, not to v.
The oscillation is losing energy, so its amplitude gradually decreases as time passes. — A student who links a minus sign with decay, as in radioactive decay, picks this. a = −ω²x describes undamped SHM of constant amplitude; the minus sign gives the direction of the acceleration.
The acceleration always points opposite to the displacement, towards the equilibrium position. — ω² is positive, so the minus sign makes a and x opposite in sign at every instant. Whichever side of equilibrium the body is on, its acceleration points back towards equilibrium.
The acceleration of the body points in the negative direction throughout the motion. — A student who reads the minus sign as a fixed direction picks this. x is itself signed: when x is negative, a = −ω²x is positive. The acceleration reverses each time the body passes equilibrium.
3 A particle moves with simple harmonic motion. Which statement about the quantities used to describe its motion is correct?
Answer and reasoning
The amplitude is greatest at each extreme and zero when the particle is at equilibrium. — A student who confuses amplitude with displacement picks this. It is the displacement that is greatest at the extremes and zero at equilibrium; the amplitude is the maximum displacement, one fixed value for the whole motion.
The equilibrium position is where the particle is momentarily at rest in each cycle. — A student who links equilibrium with being at rest picks this. The particle is momentarily at rest at the extremes; the equilibrium position is where the resultant force is zero, and the particle passes through it at its greatest speed.
Displacement is measured from equilibrium and is negative for half of each cycle. — Displacement is measured from equilibrium and is signed. The particle spends half of each cycle on each side of equilibrium, so its displacement is positive for half the cycle and negative for the other half.
The time period is the time taken to move from one extreme to the other. — A student who counts one swing as a complete oscillation picks this. Moving from one extreme to the other is half an oscillation; the period is the time to return to the starting extreme.
4 A body moving with simple harmonic motion passes through its equilibrium position 40 times in 12.0 s. What is its angular frequency?
Answer and reasoning
20.9 rad s⁻¹ — A student who counts each pass through equilibrium as one oscillation gets T = 12.0/40 = 0.300 s and ω = 2π/0.300 = 20.9 rad s⁻¹. Each oscillation includes two passes, one in each direction.
1.67 rad s⁻¹ — A student who treats ω and f as the same gives f = 1/0.600 = 1.67 Hz as the angular frequency. ω = 2πf: one oscillation is 2π rad, so ω = 10.5 rad s⁻¹.
3.77 rad s⁻¹ — A student who multiplies 2π by the period gets 2π × 0.600 = 3.77. ω = 2π/T; a longer period means a smaller angular frequency, and 2πT has the unit s, not rad s⁻¹.
10.5 rad s⁻¹ — The body passes equilibrium twice per oscillation, so 40 passes are 20 oscillations: T = 12.0/20 = 0.600 s. ω = 2π/T = 2π/0.600 = 10.5 rad s⁻¹.
Working Two passes through equilibrium per oscillation → 20 oscillations in 12.0 s. T = 12.0 s / 20 = 0.600 s. ω = 2π/T = 2π / 0.600 s = 10.47 rad s⁻¹ ≈ 10.5 rad s⁻¹.
5 A trolley hung from a newton meter gives a reading of 14.7 N. The trolley is then attached to a spring system on a horizontal track and oscillates with negligible friction. The spring constant of the system is 7.50 N m⁻¹. Take g = 9.8 m s⁻². What is the time period of the oscillation?
Answer and reasoning
8.80 s — A student who substitutes the weight, 14.7 N, for the mass gets 2π√(14.7/7.50) = 8.80 s. The period depends on the trolley’s inertia, its mass in kg: m = 14.7/9.8 = 1.50 kg.
2.81 s — The mass is m = W/g = 14.7/9.8 = 1.50 kg. T = 2π√(m/k) = 2π√(1.50/7.50) = 2π√0.200 = 2.81 s.
1.26 s — A student who leaves out the square root gets 2π × 1.50/7.50 = 1.26 s. T = 2π√(m/k), so T is proportional to √m, not to m.
14.0 s — A student who inverts the ratio gets 2π√(7.50/1.50) = 14.0 s. The equation is T = 2π√(m/k): a larger mass lengthens the period and a stiffer spring shortens it.
Working m = W/g = 14.7 N / 9.8 m s⁻² = 1.50 kg. T = 2π√(m/k) = 2π√(1.50 kg / 7.50 N m⁻¹) = 2π√(0.200 s²) = 2π × 0.447 s = 2.81 s.
6 The time period of a simple pendulum is given by T = 2π√(l/g). Which condition must be met for this equation to give an accurate value?
Answer and reasoning
The amplitude can be of any size, since the period is independent of the amplitude. — A student who over-generalises the independence of period from amplitude picks this. For a pendulum it holds only while the amplitude is small; at large amplitudes the motion is not SHM and the period is longer than 2π√(l/g).
The bob must be light, as a heavier bob is pulled harder and swings faster than predicted. — A student who thinks a heavier bob swings faster picks this. The gravitational restoring force and the bob’s inertia are both proportional to its mass, so the mass cancels and T = 2π√(l/g) contains no m. The only condition for the equation is a small amplitude.
The bob swings with a regular period, which makes its motion simple harmonic. — A student who equates any regular oscillation with SHM picks this. A pendulum swinging through a large angle is periodic but not simple harmonic, and the equation does not then apply.
The amplitude is small, so the restoring force is close to proportional to displacement. — The restoring force on the bob is mg sin θ. Only for small angles is sin θ ≈ θ, so that the force is nearly proportional to the displacement and the motion is nearly simple harmonic. The equation is derived on that basis.
7 A mass on a horizontal spring moves with simple harmonic motion on a frictionless surface; air resistance is negligible. Which statement about the energy of the oscillator is correct?
Answer and reasoning
At each extreme, E_k is greatest, because the acceleration of the mass is greatest there. — A student who links the greatest acceleration with the greatest speed picks this. The acceleration is indeed greatest at the extremes, but there the mass is momentarily at rest, so E_k is zero. E_k is greatest at the equilibrium position, where the speed is greatest and the acceleration is zero.
At each extreme, the total energy is zero, because the mass is not moving there. — A student who thinks a body has energy only while it is moving picks this. At the extremes E_k is zero, but the spring is most stretched or compressed, so it stores the greatest elastic potential energy. The total energy is the same at every point in the cycle; it is never zero.
At each extreme, all the energy is stored in the spring, because the mass is at rest. — At each extreme the mass is momentarily at rest, so E_k is zero and all of the oscillator's energy is elastic potential energy stored in the stretched or compressed spring. The surface is horizontal, so the gravitational potential energy does not change. As the mass moves back towards equilibrium, this energy is transferred to E_k, and then back again.
The total energy falls during each cycle, because energy is used up in keeping the mass moving. — A student who thinks energy is used up in keeping a body moving picks this. Energy is lost only when resistive forces transfer it to the surroundings. Here there is no friction and air resistance is negligible, so energy is only transferred between E_k and elastic potential energy, and the total stays constant.
8 Two oscillators, P and Q, each have a time period of 0.80 s. Q reaches its maximum positive displacement 0.20 s after P does. What is the phase difference between P and Q, in radians? HL
Answer and reasoning
1.57 rad — Q lags P by 0.20/0.80 = 0.25 of a cycle. One cycle is 2π rad, so the phase difference is 2π × 0.25 = π/2 = 1.57 rad.
0.20 rad — A student who takes the time lag as the phase difference gives 0.20. A time must be converted into an angle: Δφ = 2π(Δt/T) = 2π × 0.20/0.80 = 1.57 rad.
0.25 rad — A student who stops at the fraction of a cycle, Δt/T = 0.25, gives 0.25. The phase difference is that fraction of 2π rad: 0.25 × 2π = 1.57 rad.
90.0 rad — A student who works in degrees finds a quarter of 360°, 90°, and picks the matching number. The guide requires radians: a quarter cycle is π/2 = 1.57 rad; 90 rad would be more than fourteen cycles.
Working Fraction of a cycle = Δt/T = 0.20 s / 0.80 s = 0.25. Phase difference Δφ = 2π × 0.25 = π/2 rad = 1.57 rad.
9 A particle moves with simple harmonic motion with an amplitude of 0.080 m and a time period of 0.50 s. What is its speed when its displacement is 0.050 m? HL
Answer and reasoning
0.38 m s⁻¹ — A student who simplifies √(x₀² − x²) to x₀ − x gets 12.57 × 0.030 = 0.38 m s⁻¹. The square root of a difference of squares must be evaluated as it stands: √0.0039 = 0.0624.
0.78 m s⁻¹ — ω = 2π/T = 2π/0.50 = 12.57 rad s⁻¹. v = ω√(x₀² − x²) = 12.57 × √(0.080² − 0.050²) = 12.57 × 0.0624 = 0.78 m s⁻¹.
0.12 m s⁻¹ — A student who uses the frequency, f = 1/0.50 = 2.0 Hz, in place of ω gets 2.0 × 0.0624 = 0.12 m s⁻¹. The equation needs the angular frequency, ω = 2πf = 12.57 rad s⁻¹.
0.63 m s⁻¹ — A student who takes the speed as proportional to the displacement, v = ωx, gets 12.57 × 0.050 = 0.63 m s⁻¹. The speed is greatest at equilibrium and zero at the extremes: v = ω√(x₀² − x²).
Working ω = 2π/T = 2π/0.50 s = 12.57 rad s⁻¹. v = ω√(x₀² − x²) = 12.57 × √(0.0064 − 0.0025) m s⁻¹ = 12.57 × √0.0039 = 12.57 × 0.06245 = 0.78 m s⁻¹.
10 Each option describes a body that can move along a straight line. It gives the resultant force F on the body when it is held at displacements x = +0.020 m, +0.040 m and +0.060 m from its equilibrium position, in that order. At the corresponding negative displacements, F has the same size and the opposite sign. Which body will move with simple harmonic motion when displaced and released?
Answer and reasoning
F = −0.30 N, −0.60 N, −0.90 N — F doubles and triples as x doubles and triples, so F is proportional to x (F = −15x in N), and its sign is always opposite to x, so it acts towards equilibrium. Both conditions for SHM are met.
F = −0.50 N, −0.50 N, −0.50 N — A student who thinks any force towards equilibrium produces SHM picks this. The force does act towards equilibrium, so the body oscillates, but its size is constant rather than proportional to x, so the motion is not simple harmonic.
F = +0.30 N, +0.60 N, +0.90 N — A student who checks only for proportionality picks this. F is proportional to x but has the same sign, so it pushes the body away from equilibrium. Released, the body accelerates away and does not oscillate.
F = −0.90 N, −0.60 N, −0.30 N — A student who expects the acceleration to be greatest near equilibrium, where the speed is greatest, picks this. In SHM the force is proportional to the displacement, so it is smallest near equilibrium and greatest at the largest displacement.
Working Test each set for F ∝ x and F opposite to x. First set: F/x = −0.30/0.020 = −0.60/0.040 = −0.90/0.060 = −15 N m⁻¹, constant and negative, so F = −15x: SHM. Second: F constant (not ∝ x). Third: F/x = +15 N m⁻¹, force away from equilibrium, no oscillation. Fourth: |F| decreases as x increases, not ∝ x.
Read the ones marked not yet in Learn, then Verify.
Verify confirm before you go
10 more questions. Every wrong answer here is a real misconception, and you see why it is wrong straight away.
1 A body moves with simple harmonic motion along a horizontal line with angular frequency 4.0 rad s⁻¹. Displacements to the right of the equilibrium position are positive. The body is released from rest 0.050 m to the right of the equilibrium position. What is its acceleration when it has moved 0.020 m from the point of release?
Answer and reasoning
+0.48 m s⁻² — A student who treats the acceleration as proportional to the displacement in the same direction gets +0.48 m s⁻². The minus sign in a = −ω²x makes the acceleration opposite to the displacement: the body is to the right, so it accelerates to the left.
−0.48 m s⁻² — The displacement is measured from equilibrium: x = 0.050 − 0.020 = +0.030 m. a = −ω²x = −(4.0)² × 0.030 = −0.48 m s⁻². The negative sign shows the acceleration points left, towards equilibrium.
−0.12 m s⁻² — A student who multiplies by ω instead of ω² gets −4.0 × 0.030 = −0.12, whose unit would be m s⁻¹, not m s⁻². The equation is a = −ω²x, so ω must be squared: 16 × 0.030 = 0.48.
−0.32 m s⁻² — A student who measures displacement from the point of release uses x = 0.020 m and gets −16 × 0.020 = −0.32 m s⁻². In a = −ω²x, x is measured from the equilibrium position, so x = 0.030 m.
Working x is measured from equilibrium: x = 0.050 m − 0.020 m = +0.030 m. a = −ω²x = −(4.0 rad s⁻¹)² × (0.030 m) = −16 × 0.030 = −0.48 m s⁻² (0.48 m s⁻² towards the left, towards equilibrium).
2 A glider on a horizontal air track moves with simple harmonic motion alongside a metre rule. It moves back and forth between the 0.20 m mark and the 0.60 m mark on the rule. What are the amplitude and the equilibrium position of the motion?
Answer and reasoning
Amplitude 0.20 m; equilibrium position at the 0.40 m mark — The equilibrium position is midway between the extremes, at the 0.40 m mark. The amplitude is the maximum displacement from there: 0.60 − 0.40 = 0.20 m.
Amplitude 0.40 m; equilibrium position at the 0.40 m mark — A student who takes the amplitude as the distance between the extremes gets 0.60 − 0.20 = 0.40 m. That is twice the amplitude; the amplitude is measured from equilibrium to one extreme.
Amplitude 0.60 m; equilibrium position at the 0.40 m mark — A student who measures displacement from the zero of the rule takes the largest reading, 0.60 m, as the amplitude. Displacement in SHM is measured from the equilibrium position, the 0.40 m mark.
Amplitude 0.40 m; equilibrium position at the 0.20 m mark — A student who thinks equilibrium is where the glider is at rest places it at an extreme, the 0.20 m mark, and measures 0.40 m to the other extreme. The glider is at rest at the extremes; equilibrium, where the resultant force is zero, is midway, at 0.40 m.
Working Equilibrium position = midpoint of the extremes = (0.20 m + 0.60 m)/2 = 0.40 m mark. Amplitude = distance from equilibrium to an extreme = 0.60 m − 0.40 m = 0.20 m.
3 A body on a spring oscillates with time period T and amplitude A. It is replaced by a body of four times the mass, which is set oscillating on the same spring with amplitude 2A. What happens to the time period?
Answer and reasoning
It quadruples, as T is proportional to m and does not depend on the amplitude. — A student who drops the square root expects the period to rise in proportion to the mass. T = 2π√(m/k), so four times the mass gives √4 = 2 times the period.
It halves, as T depends on √(k/m), so a larger mass shortens the period. — A student who thinks a heavier body oscillates faster inverts the ratio. With the same spring force, a larger mass accelerates less, so the period increases: T = 2π√(m/k).
It doubles, as T depends on √m and does not depend on the amplitude. — T = 2π√(m/k). Multiplying m by 4 multiplies √m by 2, so the period doubles. The period of SHM does not depend on the amplitude, so the change from A to 2A has no effect.
It is unchanged, as the period of an oscillator does not depend on the mass. — A student who carries over the pendulum result picks this. A pendulum’s period is independent of the bob’s mass, but a mass–spring system’s is not: the spring force does not grow with the mass, so T = 2π√(m/k).
4 A long simple pendulum hangs in the stairwell of a school. Swinging with a small amplitude, its bob takes 2.10 s to move from one extreme position to the other. Take g = 9.8 m s⁻². What is the length of the pendulum?
Answer and reasoning
4.38 m — One extreme to the other is half an oscillation, so T = 2 × 2.10 = 4.20 s. l = gT²/(4π²) = 9.8 × 4.20² / (4π²) = 4.38 m.
1.09 m — A student who takes one swing from extreme to extreme as the period uses T = 2.10 s and gets 9.8 × 2.10²/(4π²) = 1.09 m. A complete oscillation returns the bob to its starting extreme, so T = 4.20 s.
6.55 m — A student who treats T as proportional to l, as if T = 2πl/g, gets l = gT/(2π) = 9.8 × 4.20/(2π) = 6.55 m. T is proportional to √l, so T must be squared: l = gT²/(4π²).
27.5 m — A student who squares T but not the 2π when removing the square root gets l = gT²/(2π) = 9.8 × 4.20²/(2π) = 27.5 m. Squaring T/(2π) gives T²/(4π²).
Working Extreme to extreme = half an oscillation, so T = 2 × 2.10 s = 4.20 s. T = 2π√(l/g) → l = gT²/(4π²) = (9.8 m s⁻²)(4.20 s)² / (4π²) = 172.9 / 39.48 m = 4.38 m.
5 A simple pendulum swings with a small amplitude and time period T. Its bob is replaced by one of twice the mass, keeping the same length from the point of suspension to the centre of the bob. The new pendulum is set swinging with twice the original amplitude, which is still small. What is the new time period?
Answer and reasoning
It is less than T, as a heavier bob is pulled more strongly and swings faster. — A student who thinks heavier bodies move faster under gravity picks this. The larger gravitational force on a heavier bob is matched by its larger inertia, so its acceleration at each position, and its period, are unchanged.
It is unchanged at T, as for small amplitudes the period depends only on l and g. — T = 2π√(l/g) contains neither the mass of the bob nor the amplitude. The length is unchanged and g is unchanged, so the period is unchanged. The larger amplitude is still small, so the equation still applies.
It is more than T, as the bob now travels twice as far in each oscillation. — A student who thinks a longer path must take longer picks this. Doubling the amplitude doubles the restoring force at each point of the path, so the bob moves faster and covers twice the distance in the same time.
It is √2 × T, as the period of any oscillator is proportional to √m. — A student who carries the mass–spring result T = 2π√(m/k) over to the pendulum picks this. For a pendulum the restoring force is proportional to the mass, which cancels, so T = 2π√(l/g) does not depend on the mass.
6 The displacement of an undamped oscillator varies sinusoidally with time, with time period T. Which statement describes how the kinetic energy of the oscillator varies with time?
Answer and reasoning
It rises and falls once in each period, in step with the displacement. — A student who expects every quantity to follow the displacement cycle picks this. Kinetic energy depends on the speed, not the direction, and is greatest at equilibrium. The oscillator passes through equilibrium twice per period, once in each direction, so the kinetic energy has two maxima in each period.
It is greatest at zero displacement, with two maxima in each period. — Kinetic energy depends on the speed, which is greatest at equilibrium. The oscillator passes through equilibrium twice per cycle, once in each direction, so the kinetic energy reaches its maximum twice in each period.
It is greatest at each extreme displacement, where the acceleration is greatest. — A student who links the greatest acceleration with the greatest speed picks this. At the extremes the oscillator is momentarily at rest, so its kinetic energy is zero there.
It varies between negative and positive values, just as the velocity does. — A student who treats kinetic energy as having a direction picks this. Kinetic energy depends on the square of the speed, so it is never negative; it varies between zero and a maximum.
7 The displacement of a particle moving with simple harmonic motion is x = x₀ sin(ωt + φ). At t = 0 the particle passes through its equilibrium position moving in the negative x-direction. What is the phase constant φ? HL
Answer and reasoning
φ = 0 rad, as sin 0 = 0 and so the particle starts at equilibrium — A student who fixes the phase by displacement alone picks this. φ = 0 does give x = 0 at t = 0, but then v = ωx₀ cos 0 is positive: the particle would be moving in the positive direction. The direction of motion is part of the phase.
φ = −π/2 rad, as moving in the negative direction makes x negative, so sin φ = −1 — A student who confuses the direction of motion with the side of equilibrium takes x to be negative, sets sin φ = −1 and picks this. φ = −π/2 gives x = −x₀ at t = 0: the particle would be at the negative extreme and momentarily at rest, not passing through equilibrium.
φ = π/2 rad, as cos(π/2) = 0 and the particle starts at equilibrium — A student who uses the cosine where the equation has the sine picks this. In x = x₀ sin(ωt + φ), φ = π/2 gives x = x₀ sin(π/2) = x₀: the particle would start at the positive extreme.
φ = π rad, since sin π = 0 and x becomes negative just after t = 0 — At t = 0, x = x₀ sin φ = 0, so φ = 0 or π rad. The velocity is v = ωx₀ cos(ωt + φ); for motion in the negative direction cos φ must be negative, so φ = π rad. Just after t = 0, sin(ωt + π) is negative, as required.
Working x(0) = x₀ sin φ = 0 → φ = 0 or π rad. v = ωx₀ cos(ωt + φ); v(0) = ωx₀ cos φ < 0 requires cos φ = −1, so φ = π rad.
8 A body of mass 0.40 kg moves with simple harmonic motion with angular frequency 5.0 rad s⁻¹ and amplitude 0.12 m. What is its kinetic energy when its displacement is 0.060 m? HL
Answer and reasoning
5.4 × 10⁻² J — E_T = ½mω²x₀² = ½ × 0.40 × 5.0² × 0.12² = 0.072 J. E_p = ½mω²x² = ½ × 0.40 × 25 × 0.060² = 0.018 J. E_k = E_T − E_p = 0.054 J = 5.4 × 10⁻² J.
3.6 × 10⁻² J — A student who takes the potential energy as proportional to the displacement shares the energy equally at half the amplitude: 0.072/2 = 0.036 J. E_p depends on x², so at x₀/2 it is a quarter of E_T and E_k is three quarters, 0.054 J.
1.8 × 10⁻² J — A student who takes the speed as v = ωx calculates ½m(ωx)² = ½ × 0.40 × 25 × 0.060² = 0.018 J. That is E_p, not E_k. The speed is v = ω√(x₀² − x²), so E_k = ½mω²(x₀² − x²).
1.1 × 10⁻² J — A student who uses ω rather than ω² gets ½ × 0.40 × 5.0 × (0.12² − 0.060²) = 0.011 J. The energy equations contain ω²; check the unit: kg s⁻¹ m² is not a joule.
Working E_T = ½mω²x₀² = ½(0.40 kg)(5.0 rad s⁻¹)²(0.12 m)² = 0.072 J. E_p = ½mω²x² = ½(0.40)(25)(0.060)² = 0.018 J. E_k = E_T − E_p = 0.072 J − 0.018 J = 0.054 J = 5.4 × 10⁻² J.
9 A particle moves with simple harmonic motion with a time period of 2.0 s and an amplitude of 0.050 m. At t = 0 it passes through its equilibrium position moving in the positive direction, so x = x₀ sin ωt. What is its displacement at t = 0.30 s? HL
Answer and reasoning
8.2 × 10⁻⁴ m — A student with the calculator in degree mode evaluates sin(0.942°) = 0.0164 and gets 0.050 × 0.0164 = 8.2 × 10⁻⁴ m. ωt is in radians because ω is in rad s⁻¹.
2.9 × 10⁻² m — A student who uses cosine instead of sine gets 0.050 cos(0.942 rad) = 0.050 × 0.588 = 2.9 × 10⁻² m. The particle starts at equilibrium, so x = x₀ sin ωt; the cosine form would start it at an extreme.
4.0 × 10⁻² m — ω = 2π/T = π rad s⁻¹. x = x₀ sin ωt = 0.050 sin(0.30π) = 0.050 × sin(0.942 rad) = 0.050 × 0.809 = 0.040 m = 4.0 × 10⁻² m.
7.5 × 10⁻³ m — A student who uses the frequency, f = 1/T = 0.50 Hz, in place of ω gets 0.050 sin(0.50 × 0.30) = 0.050 sin(0.15) = 7.5 × 10⁻³ m. The argument of the sine is ωt, with ω = 2π/T = π rad s⁻¹.
Working ω = 2π/T = 2π/2.0 s = π rad s⁻¹. ωt = π × 0.30 = 0.942 rad (radian mode). x = x₀ sin ωt = 0.050 m × sin(0.942) = 0.050 × 0.809 = 0.0405 m ≈ 4.0 × 10⁻² m.
10 A mass hanging from a spring oscillates with simple harmonic motion with an angular frequency of 40.0 rad s⁻¹. What is the time period of the oscillation?
Answer and reasoning
0.0250 s — A student who treats the angular frequency as the frequency uses T = 1/ω = 1/40.0 = 0.0250 s. Angular frequency is 2π times the frequency, so T = 2π/ω = 0.157 s: the oscillation takes 2π times longer than this answer suggests.
0.157 s — One oscillation corresponds to 2π rad of phase, so T = 2π/ω = 2π/40.0 = 0.157 s. Equivalently f = ω/2π = 6.37 Hz and T = 1/f = 0.157 s.
251 s — A student who multiplies 2π by ω gets 2π × 40.0 = 251. The relation is T = 2π/ω: a larger angular frequency means a shorter period, and 2πω has the unit rad s⁻¹, not seconds.
6.37 s — A student who calculates ω/2π = 6.37 and reports it as the period has found the frequency, 6.37 Hz, the number of oscillations each second. The time period is its reciprocal: T = 1/6.37 = 0.157 s.
Working T = 2π/ω = 2π / (40.0 rad s⁻¹) = 0.1571 s ≈ 0.157 s (3 s.f.). Check: f = ω/2π = 6.37 Hz, and 1/f = 0.157 s.
That was your twenty minutes. Real practice on C.1 is past-paper questions marked against the mark scheme.
Paper 1A tests it as multiple choice; Paper 1B through data you have not seen before; Paper 2 with short answers and, at HL, extended responses. Look for the command words — outline, explain, compare, evaluate — and give exactly what each asks for.
Compiled from the IB Physics guide (first assessment 2025, updated November 2023) and our question bank · Specialist review in progress. How these pages are made · Free, no account ·
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