Free Pearson Edexcel International A Level Physics lesson

Simple harmonic motion

Connect a restoring force to time graphs and periods.

What you’ll learn

  • Test linear restoring acceleration about the actual equilibrium position.
  • Read gradients and phase consistently rather than treating the three graphs as copies.
  • Choose the oscillator model and its assumptions before interpreting a period.

The restoring force must be proportional

Choose displacement x from the equilibrium position and a positive direction. The condition F = −kx says that the resultant force is proportional to displacement and points back toward equilibrium. With constant mass, Newton’s second law gives a = −(k/m)x = −ω²x. The negative sign expresses direction; it does not mean that the acceleration magnitude is negative.

At either extreme, displacement and acceleration magnitudes are largest while velocity is instantaneously zero. At equilibrium, the restoring resultant and acceleration are zero while speed is largest. Zero acceleration at that one position does not mean the object stays there: its existing velocity carries it through.

For a vertical spring, measure x from its loaded equilibrium, not from the spring’s natural length. Weight shifts the equilibrium extension; the change in resultant about that point is still −kx for a Hookean spring. Friction, nonlinear stiffness or a large-angle pendulum can make the exact SHM model inappropriate even when the motion continues to repeat.

F = −kx a = −ω²x ω² = k/m for a linear mass–spring oscillator

Symbols and units

  • x is signed displacement from the actual equilibrium position in m; F is the signed restoring resultant in N and a is acceleration in m s⁻².
  • k is linear spring stiffness in N m⁻¹ and m is oscillating mass in kg; k here is not Boltzmann’s constant.
  • ω is angular frequency in rad s⁻¹; ω² and the magnitude of the acceleration–displacement gradient have units s⁻².
  • T = 2π/ω is the oscillation period in s, not temperature. The negative sign in a = −ω²x means acceleration points opposite to displacement.

How the equation explains the idea

  1. For a linear restoring resultant, Newton’s law gives ma = −kx, hence a = −(k/m)x. Comparison with a = −ω²x identifies ω² = k/m.
  2. For the stated straight line through equilibrium, gradient = Δa/Δx = −0.80/0.020 = −40 s⁻². Therefore ω² = 40 s⁻², with the positive root ω = √40 = 6.32 rad s⁻¹.
  3. One complete cycle is 2π radians, so T = 2π/ω = 0.993 s. A straight negative gradient through equilibrium supports SHM; one restoring force reading alone would not establish linearity.
  4. For a vertical spring, the equilibrium extension already balances weight. Using additional displacement from that point leaves resultant −kx; using extension from the natural length would miss this force balance.
The acceleration–displacement gradient is −ω²

Conditions. Ideal linear mass–spring oscillator with m = 0.20 kg and k = 5.0 N m⁻¹, so ω² = k/m = 25 s⁻². Displacement is measured from loaded equilibrium; damping is neglected.

Displacement from equilibrium / m; Signed acceleration / m s⁻².

The straight line through the origin has gradient −25 s⁻². Positive displacement gives negative acceleration, and negative displacement gives positive acceleration. At x = 0, zero instantaneous acceleration does not imply zero velocity.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Read an acceleration gradient

  1. An a–x straight line passes through (0.020 m, −0.80 m s⁻²).
  2. Its gradient is −40 s⁻², equal to −ω².
  3. ω = √40 = 6.32 rad s⁻¹ and T = 2π/ω = 0.993 s.

A negative straight-line gradient through equilibrium tests SHM.

Keep this idea. Test linear restoring acceleration about the actual equilibrium position.

Displacement, velocity and acceleration have different phases

For release from the positive extreme at t = 0, x = A cosωt. The corresponding velocity is v = −Aω sinωt and acceleration is a = −Aω² cosωt. Amplitude A is nonnegative; x, v and a are signed. A different starting state needs a phase shift or an equivalent sine expression, not a changed physical model.

The tangent gradient of displacement–time gives velocity, and the tangent gradient of velocity–time gives acceleration. Displacement and acceleration are in antiphase, while velocity is a quarter cycle from displacement. At x = +A the velocity is zero and the acceleration is most negative; a quarter period later the object crosses equilibrium with its most negative velocity.

Use ω = 2πf and T = 1/f. Trigonometric calculator arguments must be radians when ω is in rad s⁻¹. Distinguish frequency in cycles per second from angular frequency in radians per second. A graph’s amplitude gives displacement amplitude, not speed amplitude; the corresponding maximum speed is Aω and maximum acceleration is Aω².

x = A cosωt v = −Aω sinωt a = −Aω² cosωt ω = 2πf = 2π/T

Symbols and units

  • A is nonnegative displacement amplitude in m; x is signed displacement in m, v is signed velocity in m s⁻¹ and a is signed acceleration in m s⁻².
  • t is elapsed time in s; ω is angular frequency in rad s⁻¹; the phase ωt is measured in radians.
  • f is frequency in Hz = cycles s⁻¹ and T = 1/f is period in s. T is not temperature and A is not activity in these equations.
  • The given cosine form chooses t = 0 at rest at x = +A. A different initial phase requires a corresponding shift in all three equations.

How the equation explains the idea

  1. Start with x = A cosωt. Its time gradient gives v = −Aω sinωt; differentiating again gives a = −Aω² cosωt = −ω²x, so the signs and phase follow from the same motion.
  2. Since ωT = 2π, at t = T/4 the phase is π/2. Thus cosωt = 0 and sinωt = 1: x = 0, v = −0.040 × 5.0 = −0.20 m s⁻¹ and a = 0.
  3. At t = T/2 the phase is π: x = −0.040 m, v = 0 and a = −(5.0)²(−0.040) = +1.0 m s⁻². Acceleration points back toward equilibrium.
  4. The maximum magnitudes are Aω for speed and Aω² for acceleration. Speed peaks at equilibrium, while acceleration peaks at the displacement extremes; they do not peak together.
Displacement, velocity and acceleration have different phases

Conditions. Undamped SHM released from +A at t = 0, with fixed A > 0 and ω > 0; T = 2π/ω. Each curve uses its own explicitly labelled amplitude scale, so unlike physical units are not equated.

Time / period, t/T; Signed normalized value.

At t/T = 0.25, displacement and acceleration are zero while velocity is maximally negative. Displacement and acceleration are opposite in phase. A quarter-cycle separates the displacement and velocity extrema; the curves’ equal plotted heights result only from normalization.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

A quarter-cycle state

  1. A = 0.040 m, ω = 5.0 rad s⁻¹, with release from +A.
  2. At t = T/4, phase is π/2: x = 0, v = −Aω = −0.20 m s⁻¹ and a = 0.
  3. At t = T/2, x = −0.040 m and a = +1.0 m s⁻².

Graph slope, sign and phase must agree.

Keep this idea. Read gradients and phase consistently rather than treating the three graphs as copies.

Mass–spring and small-angle pendulum models

For a mass m on a linear spring of stiffness k, a = −(k/m)x gives T = 2π√(m/k). The ideal period does not depend on amplitude. Doubling the mass increases the period by √2; doubling stiffness decreases it by √2. The moving spring and attachments can contribute effective inertia, so a real calibration can have an intercept rather than pass exactly through zero added mass.

For a pendulum of length l, the tangential restoring force is −mg sinθ. At small angles in radians, sinθ ≈ θ and arc displacement x ≈ lθ, giving a ≈ −(g/l)x and T = 2π√(l/g). Measure l from the pivot to the bob’s centre of mass. The bob mass cancels, but large amplitude invalidates the small-angle approximation.

To estimate g, time many complete oscillations through a fixed reference in the same direction and divide by the number. Repeat at several lengths and plot T² against l; gradient is 4π²/g. A long measured interval reduces relative reaction-time error, while a wrong pivot-to-centre length remains a systematic error. Use small amplitudes and a secure, supervised support.

Spring: T = 2π√(m/k) small-angle pendulum: T = 2π√(l/g)

Symbols and units

  • T is oscillation period in s; m is effective oscillating mass in kg and k is spring stiffness in N m⁻¹.
  • l is pendulum length from pivot to bob centre of mass in m; g is gravitational acceleration in m s⁻².
  • θ is angular displacement in radians, x is signed arc displacement in m and ω is angular frequency in rad s⁻¹. The pendulum relation assumes small enough angles that sinθ ≈ θ.
  • The gradient of T² vertically against l horizontally has units s² m⁻¹; π is dimensionless.

How the equation explains the idea

  1. For the spring, a = −(k/m)x gives ω² = k/m. Substitute into T = 2π/ω to obtain T = 2π√(m/k), assuming a linear restoring force and negligible damping.
  2. For the pendulum, tangential acceleration a = −g sinθ ≈ −gθ and arc displacement x = lθ, so a ≈ −(g/l)x. Thus ω² ≈ g/l and T ≈ 2π√(l/g); bob mass has cancelled.
  3. Square the small-angle period relation: T² = (4π²/g)l. The T²–l slope is therefore 4π²/g, so g = 4π²/(gradient).
  4. Insert 4.05 s² m⁻¹: g = 4π²/4.05 = 9.75 m s⁻². The result inherits gradient and length uncertainties; timing many cycles does not remove a systematic length error.

Worked example

Determine g from a graph

  1. A fitted T²–l gradient is 4.05 s² m⁻¹.
  2. g = 4π²/4.05.
  3. g = 9.75 m s⁻²; compare with a justified gradient uncertainty before claiming agreement.

The fitted gradient uses multiple lengths and squared periods.

Keep this idea. Choose the oscillator model and its assumptions before interpreting a period.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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