C.1 Simple harmonic motion flashcards: IB Physics SL and HL

All 69 C.1 Simple harmonic motion cards

1. DefinitionSL & HL

   Define simple harmonic motion and state the two conditions that must be satisfied.

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   A body performs simple harmonic motion when its acceleration is proportional to its displacement from a fixed equilibrium position and is always directed towards that equilibrium position, a = −ω²x. Equivalently the resultant restoring force obeys F = −mω²x. Exam tip: both halves must appear — "acceleration proportional to displacement" alone scores zero; the second mark needs "directed towards the equilibrium position", or "opposite in direction to the displacement", or an explicit account of the minus sign. Displacement is a vector measured from equilibrium (m); acceleration is a vector (m s⁻²). Sinusoidal x–t motion is a consequence of this definition, not the definition itself.

2. DefinitionSL & HL

   Define displacement in the context of an oscillation and state how it differs from distance travelled.

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   Displacement x is the vector distance of the oscillating body from its equilibrium position at a given instant, measured along the line of oscillation. SI unit m (or rad for angular oscillation); it is a vector and changes sign each half cycle. Exam tip: the mark is lost by writing "distance from the start" or "distance moved" — the reference point is the equilibrium position, not the release point. In one complete oscillation the displacement returns to its initial value while the distance travelled is 4x₀. Displacement is the quantity plotted on the y-axis of the standard SHM graph and is the x in a = −ω²x.

3. DefinitionSL & HL

   Define amplitude of an oscillation and state one property it has for ideal SHM.

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   The amplitude x₀ is the maximum magnitude of the displacement from the equilibrium position. SI unit m; it is a scalar (a magnitude) and is always positive. Exam tip: amplitude is measured from equilibrium to one extreme, NOT peak-to-peak — reading a full peak-to-trough height of 8.0 cm as the amplitude instead of 4.0 cm is one of the most common data-analysis errors in Paper 2. For ideal (undamped) SHM the amplitude is constant, and the period is independent of amplitude, so changing x₀ does not change T or f. Energy stored is proportional to x₀², so doubling the amplitude quadruples the total energy.

4. DefinitionSL & HL

   Define period and define one complete oscillation.

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   The period T is the time taken for one complete oscillation, that is, the time for the body to return to the same position moving in the same direction. SI unit second (s); it is a scalar. Exam tip: students lose the mark by omitting "moving in the same direction" — a body passing through equilibrium twice per cycle passes the same point after only T/2. For a mass on a spring released from rest, one oscillation is top → bottom → top. To measure T accurately, time N oscillations (say 20) from the centre of the motion, where the speed is greatest, and divide: T = t/N. Period is independent of amplitude for SHM (isochronous).

5. DefinitionSL & HL

   Define frequency of an oscillation, state its unit, and relate it to period.

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   The frequency f is the number of complete oscillations per unit time, f = 1/T. SI unit hertz (Hz), equivalent to s⁻¹; it is a scalar. Exam tip: write Hz, not "cycles" or "oscillations per second" in a numerical answer, and never confuse f (Hz) with ω (rad s⁻¹) — they differ by the factor 2π. A pendulum of period 2.0 s has f = 0.50 Hz and ω = π ≈ 3.1 rad s⁻¹. Frequency in ideal SHM is fixed by the system's inertia and stiffness (m and k, or l and g) and is independent of the amplitude and of the mass of the pendulum bob.

6. DefinitionSL & HL

   Define angular frequency and state how it is measured for an oscillator.

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   Angular frequency ω is the rate of change of phase of the oscillation, ω = 2π/T = 2πf; it is 2π times the number of oscillations per second. SI unit rad s⁻¹; it is a scalar. Exam tip: ω is not "angular velocity" of the mass — nothing need be rotating; it comes from the reference circle whose projection generates the SHM. Do not substitute f into an equation that requires ω. Practically, ω is found from the gradient of an a–x graph (gradient = −ω²) or from ω = 2π/T using a timed set of oscillations. Because a = −ω²x, ω² has units s⁻² and ω sets the timescale of the motion.

7. DefinitionSL & HL

   Define the equilibrium position of an oscillator and describe the restoring force.

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   The equilibrium position is the position at which the resultant force on the body is zero, so a body placed there at rest stays at rest. The restoring force is the resultant force acting when the body is displaced; for SHM it is proportional to the displacement and directed back towards equilibrium, F = −mω²x, unit N, a vector. Exam tip: for a vertical mass–spring the equilibrium position is where the spring tension already balances the weight — the extra extension is x, and the weight then cancels out, which is why gravity does not appear in T = 2π√(m/k). Stating "force is zero at the extremes" is a common and costly error.

8. DefinitionSL & HL

   Define an isochronous oscillator and state why the simple pendulum is only approximately isochronous.

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   An oscillator is isochronous when its period is independent of the amplitude of the oscillation. SI unit of the period is s. Exam tip: this is a defining property of SHM and is the reason pendulum clocks keep time as the swing decays. A simple pendulum is only approximately isochronous because the restoring force component is mg sin θ, and sin θ ≈ θ (in radians) only for small angles; for amplitudes above about 10° the true period exceeds 2π√(l/g) by roughly 0.2 % at 10° and about 1.7 % at 30°, so the motion is no longer strictly simple harmonic. Damping reduces the amplitude but not the period.

9. DefinitionSL & HL

   Define the natural frequency of a system and state what determines it.

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   The natural frequency f₀ is the frequency at which a system oscillates when it is displaced and then released, with no external periodic driving force and no damping. SI unit Hz; a scalar. Exam tip: it is fixed by the inertia and the stiffness of the system alone — f₀ = (1/2π)√(k/m) for a mass–spring and (1/2π)√(g/l) for a pendulum — and not by how far it was displaced. This is the quantity that matches the driving frequency at resonance (C.4), where the amplitude of the driven oscillation is a maximum. Do not call it "the frequency of the force"; that is the driving frequency.

10. DefinitionSL & HL

    Define a simple pendulum and list the assumptions made in the ideal model.

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    A simple pendulum is a point mass (bob) suspended by a light, inextensible string of fixed length l from a fixed frictionless pivot, free to swing in a vertical plane. Assumptions: string mass and air resistance negligible, no energy lost at the pivot, string does not stretch, and the angular amplitude is small (below about 10°) so sin θ ≈ θ. SI unit of l is m, measured from the pivot to the centre of mass of the bob. Exam tip: measuring l to the top of the bob instead of its centre is a systematic error that makes every l too small; it shows up as a positive intercept on a T² against l graph.

11. DefinitionSL & HL

    State the conditions under which a mass on a spring performs simple harmonic motion.

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    The spring must obey Hooke's law over the range used (F = −kx, elastic limit not exceeded), the spring's own mass must be negligible compared with the oscillating mass, and resistive forces must be negligible. Then the resultant force is F = −kx = ma, so a = −(k/m)x, which has the SHM form a = −ω²x with ω² = k/m. Exam tip: state Hooke's law explicitly — "the spring is elastic" is too vague. For a vertical spring the weight is cancelled by the static extension, so the same result holds and g does not enter T. If the amplitude is so large that the spring goes slack, the motion is no longer SHM.

12. DefinitionSL & HL

    Define the spring constant and state its unit and how it is determined experimentally.

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    The spring constant k is the force per unit extension of a spring obeying Hooke's law, k = F/x. SI unit N m⁻¹; it is a scalar and is a measure of the stiffness of the spring. Exam tip: k is found from the gradient of a force–extension graph (add known masses, plot mg against extension), not from a single pair of readings, and the graph must be linear over the range used. A stiffer spring (larger k) gives a shorter period, since T = 2π√(m/k). Springs in series give a smaller effective k (longer T); identical springs in parallel double k and reduce T by a factor √2.

13. EquationSL & HLData booklet: Yes

    State the defining equation of simple harmonic motion, define every symbol, and give its data-booklet status.

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    a = −ω²x (data booklet). a = acceleration of the oscillator (m s⁻²), ω = angular frequency (rad s⁻¹), x = displacement from equilibrium (m). Validity: any SHM, at every instant of the motion. The minus sign shows acceleration and displacement are always in opposite directions. Rearranged forms: a_max = ω²x₀ at the extremes, a = 0 at x = 0, and ω = √(−gradient of an a–x graph). Common misuse: substituting f for ω, which gives an answer 4π² ≈ 39.5 times too small. Sanity check: x₀ = 5.0 cm, T = 2.0 s gives ω = 3.14 rad s⁻¹ and a_max = 3.14² × 0.050 = 0.49 m s⁻².

14. EquationSL & HLData booklet: Yes

    State the relationships linking angular frequency, period and frequency, with symbols and units.

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    ω = 2π/T = 2πf (data booklet), together with T = 1/f (memorise). ω = angular frequency (rad s⁻¹), T = period (s), f = frequency (Hz). Validity: any periodic motion, and for SHM independent of amplitude. Useful rearrangements: T = 2π/ω, f = ω/2π. Common misuse: forgetting that a calculator must be in radian mode whenever ωt is used as an angle, and quoting ω in Hz. Sanity check: f = 50 Hz gives T = 0.020 s and ω = 2π × 50 = 314 rad s⁻¹. In a graph question, T is read as the time between successive peaks, or between every second zero crossing in the same direction.

15. EquationSL & HLData booklet: No – derive

    State the maximum speed and maximum acceleration of an SHM oscillator in terms of amplitude.

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    v_max = ωx₀ and a_max = ω²x₀ (derive from the booklet forms; not printed separately). v_max = maximum speed (m s⁻¹) at the equilibrium position, a_max = maximum magnitude of acceleration (m s⁻²) at the extremes, ω = angular frequency (rad s⁻¹), x₀ = amplitude (m). Validity: undamped SHM. Note a_max = ωv_max. Common misuse: assuming maximum speed and maximum acceleration occur at the same place — they are a quarter of a period apart. Sanity check: a 0.20 kg mass on a spring with T = 0.63 s and x₀ = 5.0 cm has ω = 10 rad s⁻¹, so v_max = 0.50 m s⁻¹ and a_max = 5.0 m s⁻².

16. EquationSL & HLData booklet: No – derive

    State the restoring-force equation for SHM and connect it to Newton's second law.

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    F = ma = −mω²x (derive from a = −ω²x; not printed). F = resultant restoring force (N), m = mass of the oscillator (kg), ω = angular frequency (rad s⁻¹), x = displacement from equilibrium (m). Validity: any SHM; the constant of proportionality mω² has unit N m⁻¹, so for a spring mω² = k and ω = √(k/m). Common misuse: including the weight separately for a vertical spring — it is already balanced by the static extension, and x is measured from the new equilibrium. Sanity check: m = 0.20 kg, ω = 10 rad s⁻¹, x = 0.050 m gives F = −0.20 × 100 × 0.050 = −1.0 N, directed towards equilibrium.

17. EquationSL & HLData booklet: Yes

    State the period equation for a simple pendulum, define the symbols, and give its conditions of validity.

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    T = 2π√(l/g) (data booklet). T = period (s), l = length from the pivot to the centre of mass of the bob (m), g = acceleration of free fall (m s⁻², 9.8 m s⁻² on Earth). Validity: small angular amplitude (below about 10°) so sin θ ≈ θ, light inextensible string, point mass, negligible air resistance. Rearranged: g = 4π²l/T², l = gT²/4π². Common misuse: thinking T depends on the mass of the bob or on the amplitude — it does not. Sanity check: l = 1.00 m gives T = 2π√(1.00/9.8) = 2.0 s, the classic "seconds pendulum" of half-period 1.0 s.

18. EquationSL & HLData booklet: Yes

    State the period equation for a mass–spring system and outline how it is obtained.

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    T = 2π√(m/k) (data booklet). T = period (s), m = oscillating mass (kg), k = spring constant (N m⁻¹). Derivation: resultant force −kx = ma so a = −(k/m)x; comparing with a = −ω²x gives ω² = k/m, and T = 2π/ω = 2π√(m/k). Validity: Hooke's law obeyed, spring mass negligible, no resistive forces. Rearranged: k = 4π²m/T², m = kT²/4π². Common misuse: including g for a vertical spring — the period is the same vertically and horizontally. Sanity check: m = 0.200 kg on k = 20.0 N m⁻¹ gives T = 2π√(0.0100) = 0.63 s, so f = 1.6 Hz.

19. EquationSL & HLData booklet: No – memorise

    State Hooke's law and explain why it is the reason a spring–mass system oscillates with SHM.

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    F = −kx (Hooke's law; the booklet form for elastic potential energy is E_p = ½kΔx²). F = restoring force exerted by the spring (N), k = spring constant (N m⁻¹), x = extension or compression from natural length, or displacement from equilibrium for a loaded spring (m). Because F ∝ −x, the acceleration a = F/m ∝ −x, which is exactly the SHM condition. Validity: below the elastic limit only. Common misuse: dropping the minus sign and then claiming the force acts away from equilibrium. Sanity check: k = 20.0 N m⁻¹ and x = 0.050 m gives F = 1.0 N pulling the mass back towards equilibrium.

20. Graph/diagramSL & HL

    Describe the displacement–time graph for simple harmonic motion and state what can be extracted from it.

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    Axes: displacement x/m (or cm) on y, time t/s on x. Shape: a sinusoid symmetric about x = 0, oscillating between +x₀ and −x₀. The amplitude is the peak height measured from the t-axis (half the peak-to-trough value); the period is the time between successive maxima, or twice the time between successive zero crossings. The gradient at any instant is the velocity, so the gradient is greatest in magnitude at x = 0 and zero at the extremes. Increasing the amplitude stretches the curve vertically without changing T; increasing m (spring) or l (pendulum) stretches it horizontally. If the body starts at maximum displacement the curve is a cosine; released from equilibrium it is a sine.

21. Graph/diagramSL & HL

    Describe the velocity–time graph for SHM and relate it to the displacement–time graph.

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    Axes: velocity v/m s⁻¹ on y, time t/s on x. Shape: a sinusoid of amplitude v_max = ωx₀, a quarter of a cycle (π/2 rad, or T/4) ahead of the x–t graph — v is maximum when x = 0 and zero at the extremes. The gradient of the v–t graph is the acceleration; the area between the curve and the t-axis over any interval is the change in displacement, so the area over one full period is zero. Doubling ω doubles v_max and halves T, so the peaks become both taller and closer together. Exam tip: velocity is positive whenever the x–t gradient is positive, which fixes the direction of motion at any labelled point.

22. Graph/diagramSL & HL

    Describe the acceleration–time graph for SHM and state its phase relation to displacement.

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    Axes: acceleration a/m s⁻² on y, time t/s on x. Shape: a sinusoid of amplitude a_max = ω²x₀ that is exactly π rad (T/2) out of phase with the displacement — a is a mirror image of x scaled by ω², so wherever x is a maximum a is a minimum. It leads the velocity by a quarter of a period. The area under the a–t graph over an interval gives the change in velocity. Increasing ω increases the peaks by ω² while shortening the period. Exam tip: students frequently draw a and x in phase; the minus sign in a = −ω²x means the two curves are always inverted with respect to each other.

23. Graph/diagramSL & HL

    Describe the acceleration–displacement graph for SHM and how it is used to test for SHM and find the period.

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    Axes: acceleration a/m s⁻² on y, displacement x/m on x, with x measured from equilibrium and taking both signs. Shape: a straight line through the origin with negative gradient, extending from (−x₀, +ω²x₀) to (+x₀, −ω²x₀). Straightness plus passage through the origin is the experimental proof that a ∝ −x, that is, that the motion is simple harmonic. Gradient = −ω², so ω = √(magnitude of the gradient) and T = 2π/ω. The line passes through the origin and its ends are at x = ±x₀, so the length of the line gives the amplitude. A stiffer spring or shorter pendulum makes the line steeper; changing only the amplitude lengthens the line but leaves the gradient unchanged.

24. Graph/diagramSL & HL

    Describe the linearised graph used to determine g from a simple pendulum experiment.

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    Squaring T = 2π√(l/g) gives T² = (4π²/g)l, so plot T²/s² on y against length l/m on x. Shape: a straight line through the origin. Gradient = 4π²/g, hence g = 4π²/gradient — a gradient of 4.03 s² m⁻¹ gives g = 39.5/4.03 = 9.8 m s⁻². A positive y-intercept indicates a systematic error in l, typically measuring to the top of the bob rather than to its centre of mass; the gradient, and hence g, is unaffected. Plot T² rather than T because only the squared form is linear. Uncertainty in g is found from the maximum and minimum gradients drawn through the error bars.

25. Graph/diagramSL & HL

    Describe the linearised graph used to determine the spring constant from a mass–spring oscillator.

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    Squaring T = 2π√(m/k) gives T² = (4π²/k)m, so plot T²/s² on y against oscillating mass m/kg on x. Shape: a straight line of gradient 4π²/k, so k = 4π²/gradient — a gradient of 2.0 s² kg⁻¹ gives k = 39.5/2.0 = 20 N m⁻¹. The line usually has a small positive T² intercept because the spring itself oscillates; the effective mass is m + m_spring/3, and the intercept equals (4π²/k)(m_spring/3), which allows the spring's effective mass to be found. Using a stiffer spring reduces the gradient. Compare k obtained here with k from the gradient of a static force–extension graph as a check.

26. Concept/explainSL & HL

    Explain what conditions a body must satisfy for its motion to be simple harmonic, and how these conditions are expressed by the defining equation.

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    • The body must have a single stable equilibrium position to which it returns
    • The resultant force (and hence acceleration) must be directly proportional to the displacement from that equilibrium position
    • The resultant force and acceleration must always be directed towards the equilibrium position, i.e. opposite in sign to the displacement
    • Symbolically a = −ω²x, where the constant ω² is fixed by the system (k/m for a spring, g/l for a pendulum), so ω and therefore T and f are independent of amplitude
    • Motion is then sinusoidal in time and isochronous. Exam tip: many students write only "acceleration is proportional to displacement" and lose the second mark; the direction (or the minus sign) must be stated explicitly.
27. Concept/explainSL & HLData booklet: No – derive

    Explain why a simple pendulum performs simple harmonic motion only for small amplitudes.

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    • The restoring force along the arc is the tangential component of weight, F = −mg sin θ
    • Newton's second law gives a = −g sin θ, which is not proportional to θ (or to arc displacement x = lθ) in general
    • For small θ in radians, sin θ ≈ θ, so a ≈ −gθ = −(g/l)x, which is the SHM condition with ω² = g/l
    • The approximation is good to about 1 % for θ ≤ 10°, so amplitudes must be small compared with the length
    • For large amplitudes the period lengthens slightly and depends on amplitude, so the motion is periodic but not simple harmonic. Exam tip: the small-angle approximation is only valid in radians; answers quoting sin θ ≈ θ with θ in degrees gain no credit.
28. Concept/explainSL & HLData booklet: Yes

    Explain, using Hooke's law, why a mass oscillating on a vertical spring undergoes simple harmonic motion, and why the equilibrium extension does not appear in the period.

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    • At the equilibrium position the spring tension balances the weight: ke₀ = mg
    • For a further displacement x from equilibrium the resultant force is F = −(k(e₀ + x) − mg) = −kx, since the constant terms cancel
    • Hence a = −(k/m)x, which is of the form a = −ω²x with ω² = k/m
    • The proportionality constant contains only k and m, so T = 2π√(m/k) is unaffected by gravity or by the static extension
    • The same result holds for a horizontal spring–mass system on a frictionless surface. Exam tip: students often try to include mg in the resultant force at displacement x; the mark is for showing that weight cancels the equilibrium tension.
29. Concept/explainSL & HL

    Describe the energy changes that occur during one complete oscillation of a mass–spring system, assuming no resistive forces.

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    • At maximum displacement the mass is momentarily at rest: kinetic energy is zero and the elastic potential energy stored in the spring is a maximum, equal to the total energy
    • As the mass moves towards equilibrium the restoring force does positive work, so potential energy is converted into kinetic energy
    • At the equilibrium position the displacement is zero, potential energy is a minimum and the speed and kinetic energy are maxima
    • Beyond equilibrium kinetic energy is converted back into potential energy until the mass is instantaneously at rest at the other extreme
    • The total energy stays constant; the interchange happens twice per cycle, so the energy varies at twice the frequency of the displacement. Exam tip: state where each energy is maximum, do not merely say "energy is conserved".
30. Concept/explainSL & HL

    Explain what is meant by saying that simple harmonic motion is isochronous, and outline the historical and technological importance of this property.

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    • Isochronous means the period is independent of the amplitude of oscillation
    • It follows from a = −ω²x: ω depends only on the physical constants of the system (k and m, or g and l), not on x₀
    • So a pendulum swinging through 4° and one swinging through 8° take the same time per swing, provided the small-angle condition still holds
    • Nature of Science: Galileo's observation of a swinging lamp led to pendulum timekeeping, and Huygens' pendulum clock made accurate navigation and astronomy possible
    • Modern equivalents are the quartz crystal oscillator and the caesium atomic clock, which define the second. Exam tip: isochronous is about amplitude independence, not about "always having the same period whatever you do to it" — changing l or g does change T.
31. Concept/explainSL & HL

    Describe and account for the phase relationships between the displacement, velocity and acceleration of a body in simple harmonic motion.

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    • Displacement varies sinusoidally with time; velocity is the gradient of the displacement–time graph and acceleration is the gradient of the velocity–time graph
    • Velocity leads displacement by a quarter of a period, i.e. by π/2 rad (90°): v is zero at the extremes where x is maximum, and maximum where x = 0
    • Acceleration is π rad (180°) out of phase with displacement — exactly antiphase — because a = −ω²x
    • Acceleration leads velocity by π/2 rad
    • Consequently speed is greatest at the equilibrium position and the magnitude of acceleration is greatest at the extremes. Exam tip: the commonest error is to claim v and a are in phase, or to say a is "90° out of phase" with x; a and x are always antiphase.
32. Concept/explainSL & HLData booklet: Yes

    Explain why the period of a simple pendulum is independent of the mass of the bob but depends on the local gravitational field strength.

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    • The restoring force for small angles is mg sin θ ≈ mgθ, which is proportional to the mass
    • Newton's second law gives a = F/m, so the mass cancels: a = −(g/l)x
    • Hence ω² = g/l and T = 2π√(l/g) contains no mass term
    • This is the same cancellation of inertial and gravitational mass that makes free-fall acceleration independent of mass, and is evidence for the equivalence principle
    • Because g appears, the same pendulum runs slow where g is smaller — at higher altitude, at the equator, or on the Moon — which is why pendulum gravimeters were historically used to survey g. Exam tip: say the mass cancels between the restoring force and the inertia; "heavier means more force" alone is not enough.
33. Concept/explainSL & HLData booklet: Yes

    Outline how the period of a simple pendulum changes when it is taken to the Moon, and when it oscillates inside a lift that accelerates upwards.

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    • T = 2π√(l/g), so T ∝ 1/√g at fixed length
    • On the Moon g ≈ 1.6 m s⁻², about one sixth of the Earth value, so T increases by a factor √6 ≈ 2.5 and the pendulum swings noticeably more slowly
    • In a lift accelerating upwards with acceleration a the bob experiences an effective gravitational field g_eff = g + a, so T decreases
    • If the lift accelerates downwards, g_eff = g − a and T increases; in free fall g_eff = 0 and the pendulum does not oscillate at all
    • A mass–spring system is unaffected in every case because T = 2π√(m/k) contains no g. Exam tip: students frequently claim the spring period also changes in the lift — it does not.
34. Concept/explainSL & HLData booklet: Yes

    Explain the meaning of angular frequency ω for a linear oscillator and how it is related to uniform circular motion.

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    • Angular frequency ω = 2π/T = 2πf, measured in rad s⁻¹, is the rate of change of the phase angle of the oscillation
    • It is not an angular velocity of the oscillating body itself, which moves along a straight line
    • Simple harmonic motion is mathematically the projection onto a diameter of a point moving in a circle of radius x₀ at constant angular speed ω: the reference circle
    • The projected displacement is x = x₀ sin ωt and the projected acceleration is the component of the centripetal acceleration ω²x₀, giving a = −ω²x
    • This model explains why one full oscillation corresponds to 2π rad of phase. Exam tip: a very common slip is to use ω = 2π/T with T in minutes or to confuse f (Hz) with ω (rad s⁻¹) by a factor of 2π.
35. Worked problemSL & HLData booklet: Yes

    A simple pendulum is to be used as the timing element of a clock with a period of 2.00 s. Calculate the required length, taking g = 9.8 m s⁻².

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    Principle: for small amplitudes T = 2π√(l/g), so squaring gives l = gT²/(4π²). Substitution: l = (9.8 m s⁻² × (2.00 s)²)/(4π²) = (9.8 × 4.00)/39.478. Intermediate: numerator = 39.2 m s⁻²·s², denominator = 39.478. Final answer: l = 0.993 m ≈ 0.99 m. Check/Trap: a "seconds pendulum" of length about 1 m has a period of 2 s, not 1 s — one complete oscillation is out and back, so each swing takes 1.0 s. Also check the unit: (m s⁻²)(s²) = m. Using T = 1.00 s by mistake gives 0.25 m, a factor of four out, because l ∝ T².

36. Worked problemSL & HLData booklet: Yes

    A 0.25 kg mass hung from a light vertical spring oscillates with a period of 0.40 s. Determine the spring constant.

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    Principle: T = 2π√(m/k) for a mass–spring system, so k = 4π²m/T². Substitution: k = (4π² × 0.25 kg)/(0.40 s)² = (39.478 × 0.25)/0.160. Intermediate: numerator = 9.87 kg, denominator = 0.160 s². Final answer: k = 61.7 N m⁻¹ ≈ 62 N m⁻¹ (2 s.f., matching the data). Check/Trap: the unit is kg s⁻² = N m⁻¹, so a numerical answer around 60 for a mass of a quarter of a kilogram oscillating twice per second is sensible. The common error is to forget to square the 4π² bracket or to square only π, giving k ≈ 15 N m⁻¹. Amplitude is not needed and does not affect k or T.

37. Worked problemSL & HLData booklet: Yes

    A body performs simple harmonic motion of amplitude 2.0 cm at a frequency of 5.0 Hz. Calculate the maximum magnitude of its acceleration and state where it occurs.

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    Principle: a = −ω²x with ω = 2πf, so the magnitude is greatest when x = x₀. Substitution: ω = 2π × 5.0 Hz = 31.4 rad s⁻¹, so ω² = 987 s⁻². Then |a|max = ω²x₀ = 987 s⁻² × 0.020 m. Intermediate: 987 × 0.020 = 19.7. Final answer: |a|max = 20 m s⁻² (2 s.f.), occurring at the two extremes of the motion and directed towards the equilibrium position. Check/Trap: convert 2.0 cm to 0.020 m — leaving it in centimetres gives 1974 m s⁻², an absurd value. Also do not use f in place of ω: that would give 25 × 0.020 = 0.50 m s⁻², about 40 times too small.

38. Worked problemSL & HLData booklet: Yes

    A displacement–time graph for a particle in SHM shows an amplitude of 4.0 cm and a period of 0.50 s. Determine the magnitude of the acceleration when the displacement is 2.0 cm, and the maximum acceleration.

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    Principle: read x₀ and T from the graph, then use ω = 2π/T and a = −ω²x. Substitution: ω = 2π/0.50 s = 12.6 rad s⁻¹, ω² = 158 s⁻². At x = 2.0 cm = 0.020 m: |a| = 158 × 0.020 = 3.16 m s⁻². At x = x₀ = 0.040 m: |a|max = 158 × 0.040 = 6.32 m s⁻². Final answers: 3.2 m s⁻² and 6.3 m s⁻², each directed towards equilibrium. Check/Trap: acceleration is proportional to displacement, so doubling x must double a — a quick internal consistency check. Read the period as one complete cycle (peak to next peak), not peak to trough, which would halve T and quadruple the answers.

39. Worked problemSL & HLData booklet: No – derive

    In an experiment the period of a simple pendulum is measured for several lengths and T² is plotted against l. The best-fit line passes through the origin with gradient 4.10 s² m⁻¹. Determine g.

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    Principle: T = 2π√(l/g) squares to T² = (4π²/g)l, a straight line through the origin of gradient 4π²/g. Rearranging, g = 4π²/gradient. Substitution: g = 39.478/(4.10 s² m⁻¹). Intermediate: 39.478/4.10 = 9.63. Final answer: g = 9.63 m s⁻² ≈ 9.6 m s⁻² (3 s.f.), unit check: 1/(s² m⁻¹) = m s⁻². Check/Trap: students routinely invert the relationship and quote g = 4π² × gradient = 162 m s⁻². The line must pass through the origin; a non-zero intercept indicates a systematic error such as measuring the length to the top rather than the centre of the bob.

40. Worked problemSL & HLData booklet: Yes

    A pendulum on Earth has a period of 2.0 s. The same pendulum is taken to the Moon, where g = 1.6 m s⁻². Calculate its new period. Also state the effect on a mass–spring oscillator.

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    Principle: T = 2π√(l/g) with l constant gives T ∝ 1/√g, so T_M/T_E = √(g_E/g_M). Substitution: T_M = 2.0 s × √(9.8/1.6) = 2.0 × √6.125. Intermediate: √6.125 = 2.47. Final answer: T_M = 4.95 s ≈ 5.0 s (2 s.f.). For the mass–spring system T = 2π√(m/k) is independent of g, so its period is unchanged at its Earth value. Check/Trap: weaker g means slower oscillation, so the period must increase — if your answer is smaller than 2.0 s you have inverted the ratio. Do not scale by g/g rather than by the square root: that would give 12 s.

41. Worked problemSL & HLData booklet: Yes

    A graph of acceleration against displacement for an oscillator is a straight line through the origin with gradient −25 s⁻². Determine the angular frequency, the period and the frequency of the oscillation.

    Show answer

    Principle: a = −ω²x is of the form y = mx with gradient −ω², so ω² = 25 s⁻². Substitution: ω = √25 = 5.0 rad s⁻¹. Then T = 2π/ω = 6.283/5.0 = 1.257 s and f = 1/T = ω/2π = 5.0/6.283 = 0.796 Hz. Final answers: ω = 5.0 rad s⁻¹, T = 1.3 s, f = 0.80 Hz (2 s.f.). Check/Trap: the negative gradient is essential evidence for SHM — quote ω² as the magnitude of the gradient, but never write ω = √(−25). A frequency slightly below 1 Hz with a period slightly above 1 s is self-consistent, since fT = 1.

42. Exam technique/trapSL & HL

    Identify the calculator and unit traps that most often cost marks in simple harmonic motion questions.

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    • Trap: evaluating sin(ωt + φ) or the small-angle approximation with the calculator in degree mode. Why students fall for it: ω is quoted in rad s⁻¹ but angles feel like degrees. Correct approach: set RAD mode for every SHM calculation and only convert to degrees when a phase difference is asked for in degrees
    • Trap: leaving amplitudes in cm or mm inside ω²x₀, giving answers 100 or 1000 times too large
    • Trap: using f instead of ω, an error of 2π (or (2π)² in acceleration and energy)
    • Trap: reading T from a graph as half a cycle
    • Command terms: "determine" requires working and a unit; "state" needs no working; "estimate" still needs a numerical value with a sensible s.f.
43. Exam technique/trapSL & HL

    Explain the trap in the statement "the acceleration of an oscillator is greatest where its speed is greatest", and how to answer questions about the extremes of SHM correctly.

    Show answer

    • The trap: everyday intuition links large acceleration with fast motion, and students transfer this to SHM
    • In SHM a = −ω²x, so the magnitude of acceleration is greatest at maximum displacement, where the body is instantaneously at rest and the restoring force is largest
    • Speed is greatest at the equilibrium position, where x = 0 and therefore a = 0 and the resultant force is zero
    • Correct approach: locate the quantity on the x–t graph first, then use the gradient relationships v = dx/dt and a = dv/dt
    • Note that the body is momentarily at rest at the extremes but is not in equilibrium there
    • Command term guidance: "explain" requires the causal chain force → acceleration, not merely a restatement.
44. Exam technique/trapSL & HLData booklet: No – derive

    Describe an experiment to determine g using a simple pendulum, including variables, measurements, graph and uncertainty treatment.

    Show answer

    • Apparatus: bob on inextensible thread from a split cork clamped to a stand, metre rule, digital stopwatch, fiducial mark at the equilibrium position
    • Independent variable l (centre of clamp to centre of bob), dependent variable T, controlled: same bob, amplitude below 10°, same location
    • Method: time 20 complete oscillations from the fiducial mark, repeat three times, divide by 20 to reduce the effect of reaction time; repeat for 6–8 lengths
    • Plot T² against l; gradient = 4π²/g, so g = 4π²/gradient
    • Uncertainty: ΔT²/T² = 2ΔT/T; draw max and min gradient lines through the error bars, Δg from half their difference
    • Limitations: air resistance, thread stretch, clamp movement; improve with a photogate.
45. Exam technique/trapSL & HLData booklet: No – derive

    Describe an experiment using a mass–spring system to determine the spring constant and the effective mass of the spring.

    Show answer

    • Apparatus: helical spring on a rigid clamped stand, slotted masses on a hanger, stopwatch or photogate, fiducial marker at equilibrium
    • Independent variable: suspended mass m; dependent variable: period T; controlled: same spring, small amplitude so Hooke's law holds and the spring never goes slack
    • Method: time 20 oscillations, repeat and average, for at least six masses
    • Theory: T = 2π√((m + m_s/3)/k), so T² = (4π²/k)m + (4π²/k)(m_s/3)
    • Plot T² against m: gradient = 4π²/k gives k; the positive vertical intercept gives the effective mass of the spring, a systematic effect
    • Limitations: sideways swinging and rotation, overstretching beyond the elastic limit; improve with a photogate and a guide rod.
46. Exam technique/trapSL & HL

    Give command-term guidance for SHM graph questions and explain the difference between sketching, plotting and drawing.

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    • "Sketch" a graph: show the correct shape, label both axes with quantity and unit, mark key features (amplitude, period, zeros) — a numerical scale is not required but must not be wrong
    • "Plot" means mark data points accurately on given axes with error bars
    • "Draw" a line of best fit means a smooth line with points balanced either side, not dot-to-dot
    • For SHM the marker looks for: sinusoidal shape, correct number of cycles, correct phase relative to any given graph, and axes crossing at the right places
    • "Deduce" requires a conclusion supported by stated reasoning; "suggest" allows a plausible physical explanation
    • Trap: sketching v–t with the same phase as x–t; v must be a quarter cycle ahead.
47. Exam technique/trapSL & HL

    Distinguish random from systematic error in pendulum timing and explain how each affects a T² against l graph.

    Show answer

    • Random error: variation in reaction time when starting and stopping the stopwatch. Students underestimate it because the display reads to 0.01 s while human reaction time is about 0.2 s
    • Effect: points scatter about the best-fit line; reduce by timing 20 oscillations, using a fiducial mark at the equilibrium position where the bob moves fastest, and repeating
    • Systematic error: measuring the length to the top of the bob rather than to its centre, so every l is too small by the bob's radius. Effect: the line is displaced and does not pass through the origin, giving a non-zero intercept, while the gradient and hence g are unaffected
    • Precision refers to the spread of repeats; accuracy refers to closeness to the accepted value of 9.81 m s⁻².
48. DefinitionHL only

    Define the phase angle of an oscillator and state what it represents physically.

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    The phase angle (phase constant) φ in x = x₀ sin(ωt + φ) is the phase of the oscillation at t = 0; it fixes the starting position and direction of motion of the oscillator. Unit: radian (rad), a scalar, conventionally taken in the range 0 to 2π. Exam tip: φ = 0 means the body is at equilibrium moving in the positive direction at t = 0, while φ = π/2 gives x = x₀ cos ωt, that is, release from maximum displacement. Marks are lost by working in degrees while the calculator is in radian mode. The total phase (ωt + φ) advances by 2π each period, which is exactly why ω is called the angular frequency.

49. DefinitionHL only

    Define phase difference between two oscillators and state how it is measured from graphs.

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    The phase difference between two oscillations of the same frequency is the difference in their phase angles, Δφ = 2πΔt/T, where Δt is the time lag between corresponding points on the two curves. Unit: radian (rad); a scalar. Exam tip: two oscillators are in phase when Δφ = 0 (or a multiple of 2π) and in antiphase when Δφ = π rad (T/2). Within a single SHM, velocity leads displacement by π/2, acceleration leads velocity by π/2, and acceleration is π out of phase with displacement. Quoting a phase difference as a time, or in degrees when radians are asked for, loses the mark; state whether one oscillation leads or lags.

50. DefinitionHL only

    Define the total energy of a simple harmonic oscillator and state how it depends on amplitude and frequency.

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    The total energy E_T of an SHM system is the sum of the kinetic energy of the mass and the potential energy stored in the system, E_T = ½mω²x₀². SI unit joule (J); a scalar. Exam tip: for undamped SHM E_T is constant throughout the motion — it is all kinetic at x = 0 and all potential at x = ±x₀. E_T ∝ x₀², so doubling the amplitude quadruples the energy, and E_T ∝ ω², so doubling the frequency also quadruples it. Answers claiming energy is "lost at the extremes" score zero: it is transferred to potential store, not lost. With damping E_T decreases exponentially with time as work is done against resistive forces.

51. DefinitionHL only

    Define the kinetic and potential energies of an SHM oscillator and state where each is a maximum.

    Show answer

    The kinetic energy is E_k = ½mv² = ½mω²(x₀² − x²), maximum ½mω²x₀² at the equilibrium position x = 0 and zero at the extremes. The potential energy is E_p = ½mω²x², measured from zero at the equilibrium position, maximum ½mω²x₀² at x = ±x₀ and zero at x = 0. SI unit joule (J); both are scalars and at every instant E_k + E_p = E_T. Exam tip: state the reference level — E_p is defined as zero at equilibrium, so for a vertical spring it is the combined elastic-plus-gravitational store. Each energy varies with time at twice the frequency of the displacement, because both depend on the square of a sinusoid.

52. EquationHL onlyData booklet: Yes

    State the displacement equations for SHM with phase angle, define all symbols, and give the booklet status.

    Show answer

    x = x₀ sin(ωt + φ) (data booklet); the alternative x = x₀ cos(ωt + φ) is the same motion with φ shifted by π/2. x = displacement (m), x₀ = amplitude (m), ω = angular frequency (rad s⁻¹), t = time (s), φ = phase angle (rad). Validity: undamped SHM; the calculator must be in radian mode. Choose the sine form when the body starts at equilibrium (φ = 0) and the cosine form when released from rest at maximum displacement. Common misuse: using ωt in degrees. Sanity check: x₀ = 4.0 cm, T = 2.0 s (ω = π), φ = 0; at t = 0.50 s, x = 0.040 sin(π/2) = 4.0 cm, the first extreme, as expected after T/4.

53. EquationHL onlyData booklet: Yes

    State the velocity–time equation for SHM and explain the phase relation it expresses.

    Show answer

    v = ωx₀ cos(ωt + φ) (data booklet), the time derivative of x = x₀ sin(ωt + φ). v = instantaneous velocity (m s⁻¹), ω = angular frequency (rad s⁻¹), x₀ = amplitude (m), t = time (s), φ = phase angle (rad). Because cos leads sin by π/2, velocity leads displacement by a quarter of a period. Maximum speed ωx₀ occurs whenever x = 0. Differentiating again gives a = −ω²x₀ sin(ωt + φ) = −ω²x. Common misuse: writing v = ωx₀ sin(ωt + φ), which wrongly puts v in phase with x. Sanity check: x₀ = 4.0 cm, ω = π rad s⁻¹, φ = 0; at t = 0, v = π × 0.040 = 0.13 m s⁻¹, the maximum.

54. EquationHL onlyData booklet: Yes

    State the equation giving speed in terms of displacement for SHM and give its conditions of use.

    Show answer

    v = ±ω√(x₀² − x²) (data booklet). v = speed at displacement x (m s⁻¹), ω = angular frequency (rad s⁻¹), x₀ = amplitude (m), x = displacement from equilibrium (m). The ± shows the body passes each point twice per cycle in opposite directions. Validity: undamped SHM; it is the time-independent form, so use it whenever t is neither given nor wanted. Special cases: x = 0 gives v_max = ωx₀; x = ±x₀ gives v = 0. Common misuse: writing ω√(x₀ − x) or forgetting to square. Sanity check: ω = π rad s⁻¹, x₀ = 4.0 cm, x = 2.0 cm gives v = π√(16 − 4) × 10⁻² = 0.11 m s⁻¹, about 87 % of v_max.

55. EquationHL onlyData booklet: Yes

    State the equation for the total energy of an SHM oscillator and its useful rearrangements.

    Show answer

    E_T = ½mω²x₀² (data booklet). E_T = total energy (J), m = oscillating mass (kg), ω = angular frequency (rad s⁻¹), x₀ = amplitude (m). Equivalent forms: E_T = ½mv_max² since v_max = ωx₀, and for a spring E_T = ½kx₀² because k = mω². Validity: undamped SHM, energy measured with E_p = 0 at equilibrium. Common misuse: using the instantaneous displacement x in place of the amplitude x₀. Sanity check: m = 0.20 kg, ω = 3.14 rad s⁻¹, x₀ = 5.0 cm gives E_T = 0.5 × 0.20 × 9.87 × 0.0025 = 2.5 × 10⁻³ J, matching ½mv_max² with v_max = 0.157 m s⁻¹.

56. EquationHL onlyData booklet: Yes

    State the equations for the potential and kinetic energies of an SHM oscillator at displacement x.

    Show answer

    E_p = ½mω²x² (data booklet); E_k = ½mω²(x₀² − x²) is not printed and is obtained as E_T − E_p. E_p = potential energy (J) taken as zero at equilibrium, E_k = kinetic energy (J), m = mass (kg), ω = angular frequency (rad s⁻¹), x = displacement (m), x₀ = amplitude (m). Their sum is E_T = ½mω²x₀², which is constant. Common misuse: adding a separate gravitational term for a vertical spring — it is already contained in E_p measured from the new equilibrium. Sanity check: at x = x₀/2 the potential energy is ¼E_T and the kinetic energy ¾E_T, so E_k = 3E_p there.

57. Graph/diagramHL only

    Describe the graphs of kinetic, potential and total energy against displacement for SHM.

    Show answer

    Axes: energy E/J on y, displacement x/m on x from −x₀ to +x₀. E_p = ½mω²x² is an upward parabola with its minimum of zero at x = 0, reaching E_T at x = ±x₀. E_k = ½mω²(x₀² − x²) is the inverted parabola, maximum E_T at x = 0 and zero at the extremes. E_T is a horizontal straight line at ½mω²x₀². The two parabolas cross at x = ±x₀/√2, where E_k = E_p = ½E_T. The curvature of either parabola is set by mω² = k, so the gradient of an E_p against x² graph is ½mω², giving ω. Doubling the amplitude raises the horizontal line by a factor of four and widens the curves.

58. Graph/diagramHL only

    Describe how kinetic and potential energy vary with time in SHM and state the frequency of that variation.

    Show answer

    Axes: energy E/J on y, time t/s on x. Both E_k and E_p are sinusoidal curves lying entirely above the t-axis, oscillating between 0 and E_T; they are in antiphase, and their sum is a horizontal line at E_T. Each completes a full cycle in T/2, so the energy varies at twice the frequency of the displacement — the mean value of each is ½E_T. Starting from maximum displacement, E_p = E_T cos²(ωt) and E_k = E_T sin²(ωt). Exam tip: the most common error is drawing the energy curves with period T rather than T/2, or letting them go negative. With damping the horizontal E_T line decays exponentially and the peaks fall.

59. Concept/explainHL onlyData booklet: Yes

    Derive the expression a = −ω²x for a mass on a horizontal spring and identify the resulting expression for the period.

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    • Take x as the displacement from the natural (equilibrium) position of a mass m on a frictionless surface attached to a spring of constant k
    • Hooke's law gives the restoring force F = −kx, the minus sign showing it opposes the displacement
    • Newton's second law: ma = −kx, so a = −(k/m)x
    • Comparing with the SHM condition a = −ω²x identifies ω² = k/m, so ω = √(k/m)
    • Since ω = 2π/T, T = 2π√(m/k), which is amplitude independent
    • The same derivation applies to a vertical spring once the weight is shown to cancel the equilibrium tension
    • Exam tip: the mark for "deriving" requires the explicit comparison step with a = −ω²x, not just quoting the final period formula.
60. Concept/explainHL onlyData booklet: Yes

    Derive the expression for the angular frequency of a simple pendulum performing small-amplitude oscillations.

    Show answer

    • A bob of mass m hangs on a light inextensible string of length l and is displaced through a small angle θ
    • The component of weight along the arc is the restoring force: F = −mg sin θ
    • For small θ measured in radians, sin θ ≈ θ, so F ≈ −mgθ
    • The displacement along the arc is x = lθ, hence θ = x/l and F ≈ −(mg/l)x
    • Newton's second law gives a = −(g/l)x, so comparison with a = −ω²x yields ω² = g/l and T = 2π√(l/g)
    • The mass cancels, so the period is independent of the bob's mass
    • Exam tip: state clearly where the small-angle approximation enters, and that it requires radians; without it the motion is periodic but not simple harmonic.
61. Concept/explainHL onlyData booklet: Yes

    Show how v = ±ω√(x₀² − x²) follows from energy conservation, and explain the physical meaning of the ± sign.

    Show answer

    • The total energy of an undamped oscillator is constant: E_T = ½mω²x₀²
    • The potential energy at displacement x is E_p = ½mω²x², so the kinetic energy is E_k = E_T − E_p = ½mω²(x₀² − x²)
    • Writing E_k = ½mv² and equating gives v² = ω²(x₀² − x²), hence v = ±ω√(x₀² − x²)
    • The ± sign shows that the body passes through each displacement twice per cycle with equal speed but opposite direction of travel — once moving outwards and once returning
    • Setting x = 0 gives the maximum speed v = ωx₀; setting x = ±x₀ gives v = 0
    • Exam tip: if a question asks for speed, quote the magnitude; if it asks for velocity, the direction or sign must be justified.
62. Worked problemHL onlyData booklet: Yes

    A particle undergoes SHM described by x = x₀ sin(ωt + φ) with x₀ = 5.0 cm, f = 2.0 Hz and φ = π/6 rad. Calculate the displacement at t = 0.10 s.

    Show answer

    Principle: substitute directly into the booklet equation with the calculator in radian mode. Substitution: ω = 2πf = 2π × 2.0 = 12.57 rad s⁻¹, so ωt = 12.57 × 0.10 = 1.257 rad. Phase = 1.257 + 0.524 = 1.781 rad. Intermediate: sin(1.781) = 0.978. Final answer: x = 0.050 m × 0.978 = 0.0489 m ≈ 4.9 cm, i.e. close to the positive extreme. Check/Trap: 1.781 rad is just past π/2, so the sine must be just below 1 — a good sanity check. In degree mode sin(1.781°) = 0.031 would give 0.16 cm, a classic wrong answer. Note x₀ must be in metres if an SI answer is required.

63. Worked problemHL onlyData booklet: Yes

    An oscillator has amplitude 8.0 cm and period 0.40 s. Calculate its speed when the displacement is 4.0 cm, and compare this with its maximum speed.

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    Principle: use v = ±ω√(x₀² − x²) with ω = 2π/T. Substitution: ω = 2π/0.40 = 15.7 rad s⁻¹. x₀² − x² = (0.080)² − (0.040)² = 6.40 × 10⁻³ − 1.60 × 10⁻³ = 4.80 × 10⁻³ m². Intermediate: √(4.80 × 10⁻³) = 0.0693 m. Final answer: v = 15.7 × 0.0693 = 1.09 ≈ 1.1 m s⁻¹. Maximum speed v = ωx₀ = 15.7 × 0.080 = 1.26 ≈ 1.3 m s⁻¹, so the ratio is √3/2 = 0.87. Check/Trap: at half the amplitude the speed is 87 % of maximum, not 50 % — the relationship is not linear. Squaring before subtracting is essential; √(x₀ − x) is a frequent error.

64. Worked problemHL onlyData booklet: Yes

    A 0.30 kg mass oscillates with SHM of amplitude 5.0 cm at 4.0 Hz. Determine the total energy and the kinetic energy when the displacement is 3.0 cm.

    Show answer

    Principle: E_T = ½mω²x₀², E_k = ½mω²(x₀² − x²). Substitution: ω = 2π × 4.0 = 25.1 rad s⁻¹, ω² = 632 s⁻². E_T = 0.5 × 0.30 × 632 × (0.050)² = 0.15 × 632 × 2.50 × 10⁻³. Intermediate: 0.15 × 632 = 94.8; × 2.50 × 10⁻³ = 0.237 J. For E_k: x₀² − x² = 2.50 × 10⁻³ − 0.90 × 10⁻³ = 1.60 × 10⁻³ m². E_k = 94.8 × 1.60 × 10⁻³ = 0.152 J. Final answers: E_T = 0.24 J, E_k = 0.15 J, so E_p = 0.085 J. Check/Trap: E_k + E_p must equal E_T. Energy depends on x₀², so doubling the amplitude quadruples the energy — a favourite multiple-choice test.

65. Worked problemHL onlyData booklet: Yes

    At t = 0 a body in SHM is at half its maximum positive displacement and moving in the positive direction. Determine the phase angle φ in x = x₀ sin(ωt + φ).

    Show answer

    Principle: apply the initial conditions to x = x₀ sin(ωt + φ) and v = ωx₀ cos(ωt + φ). Substitution: at t = 0, x = x₀/2 gives sin φ = 0.5, so φ = π/6 rad or φ = 5π/6 rad. Selecting: v > 0 at t = 0 requires cos φ > 0, which holds for π/6 but not for 5π/6. Final answer: φ = π/6 rad = 30° (0.52 rad). Check/Trap: the sine equation always has two solutions in one cycle; the velocity condition selects the correct one. If instead the body started at maximum positive displacement, φ = π/2 and the equation reduces to x = x₀ cos ωt. Keep the calculator in radian mode throughout.

66. Worked problemHL onlyData booklet: Yes

    A 0.20 kg mass on a spring of constant 80 N m⁻¹ is pulled 6.0 cm from equilibrium and released from rest. Determine the period, the maximum speed, the time to first reach x = 3.0 cm, and the kinetic energy there.

    Show answer

    Principle: ω = √(k/m) from a = −(k/m)x. Substitution: ω = √(80/0.20) = √400 = 20 rad s⁻¹, so T = 2π/20 = 0.314 s ≈ 0.31 s. Maximum speed v = ωx₀ = 20 × 0.060 = 1.2 m s⁻¹. Released from rest at the extreme, so x = x₀ cos ωt: cos(20t) = 3.0/6.0 = 0.5 gives 20t = π/3 rad, t = 0.0524 s ≈ 0.052 s. Kinetic energy: E_k = ½mω²(x₀² − x²) = 0.5 × 0.20 × 400 × (3.60 − 0.90) × 10⁻³ = 40 × 2.70 × 10⁻³ = 0.108 J ≈ 0.11 J. Check/Trap: E_T = ½kx₀² = 0.144 J, and 0.052 s is one sixth of T, both consistent. Starting from rest requires the cosine form, not sine.

67. Worked problemHL onlyData booklet: No – derive

    Two identical oscillators have period 0.80 s. On a displacement–time graph the peak of one occurs 0.10 s after the peak of the other. Determine the phase difference in radians and as a fraction of a cycle.

    Show answer

    Principle: a time lag Δt corresponds to a phase difference Δφ = 2πΔt/T, since a full cycle is 2π rad. Substitution: Δφ = 2π × (0.10 s)/(0.80 s). Intermediate: 0.10/0.80 = 0.125 of a cycle. Final answer: Δφ = 2π × 0.125 = 0.785 rad ≈ 0.79 rad, which is π/4 rad or 45°, one eighth of a cycle. Check/Trap: compare like features — peak with peak, or positive-going zero with positive-going zero. Measuring peak to trough would add π and give 3π/4. Phase difference has no unit beyond the radian and must lie between 0 and 2π (or be quoted as the smaller equivalent angle).

68. Exam technique/trapHL onlyData booklet: Yes

    Explain the traps in choosing between the sine and cosine forms of the SHM displacement equation, and in quoting phase differences.

    Show answer

    • The booklet gives x = x₀ sin(ωt + φ); with φ = 0 the body starts at x = 0 moving positively, while φ = π/2 converts it to x = x₀ cos ωt, the case of release from rest at maximum displacement
    • Trap: students assume the sine form always applies and get zero displacement at t = 0 for a mass released from the extreme. Correct approach: read the initial conditions from the stem and choose φ accordingly
    • Trap: quoting Δφ in degrees when the question asks for radians, or as a time
    • Trap: forgetting that the ± in v = ±ω√(x₀² − x²) gives two velocities at each x
    • Command terms: "determine" needs the substitution shown; "deduce" needs the reasoning that selected the phase angle.
69. Exam technique/trapHL onlyData booklet: Yes

    Explain the common errors in SHM energy questions and how to check an energy answer quickly.

    Show answer

    • Trap: using E_p = ½kx² for a pendulum. Why: it is memorised from the spring case. Correct approach: ½mω²x² is the general form and reduces to ½kx² only when ω² = k/m
    • Trap: treating energy as proportional to amplitude rather than to amplitude squared; doubling x₀ quadruples E_T
    • Trap: drawing E_k–t and E_p–t at the same frequency as x–t; they oscillate at 2f because they depend on sin²ωt and cos²ωt
    • Trap: negative values on an energy graph
    • Quick checks: E_k + E_p = E_T at every instant; the curves cross at x = ±x₀/√2; E_k(max) = ½mv²max with v_max = ωx₀
    • "Show that" questions require an answer quoted to one more significant figure than the given data.

Practise this topic with exam-style questions: C.1 Simple harmonic motion questions (SL) · C.1 Simple harmonic motion questions (HL) · all flashcards

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