The condition for simple harmonic motion, F = -kx and a = -ω²x, and how to test whether a motion is SHM

Amplitude, period, frequency and angular frequency

x = Acos ωt, v = -Aωsin ωt, a = -Aω²cos ωt, and the maximum speed Aω

Displacement–time, velocity–time and acceleration–time graphs, their gradients and phase differences

The period of a mass–spring system and of a simple pendulum (small angles), and using them in experiments

When an object resting on a vibrating surface loses contact

1. Simple harmonic motion is oscillation in which the acceleration is proportional to the displacement from a fixed point and always directed towards that point, a = -ω²x, and everything else on this page follows from that one condition.

2. It comes from a restoring force proportional to displacement, F = -kx. A restoring force alone is not enough: it must be proportional.

3. ω= 2πf = 2π/T. Released from an extreme: x = Acos ωt, v = -Aωsin ωt, a = -Aω²cos ωt. Use radians.

4. Maximum speed Aω at equilibrium; maximum acceleration Aω² at the extremes, where the speed is zero.

5. Velocity is a quarter of a cycle (π/2) ahead of displacement; acceleration is in antiphase (π) with displacement.

6. Mass on a spring: T = 2π√(m/k). Simple pendulum, small angles: T = 2π√(l/g). Neither depends on the amplitude.

The equilibrium position is where the resultant force on the object is zero; displacement x is measured from it. Whenever the object is displaced, a restoring force acts. For SHM it obeys F = -kx, where k is a constant for the system (N m⁻¹) and the minus sign means "opposite to the displacement". Newton's second law turns this into an acceleration, a = F/m = -(k/m)x; writing the positive constant k/m as ω² gives the defining equation.

Both are in the formulae list supplied in the exam (a in m s⁻², ω in rad s⁻¹).

The period of SHM does not depend on the amplitude. An oscillator with this property is called isochronous, which is why pendulums and springs make good timekeepers.

Check both parts of the condition: the acceleration (or resultant force) always points towards equilibrium, and its size is proportional to the displacement. On a graph of a against x, SHM gives a straight line through the origin with gradient -ω²; in a table, a/x is the same negative constant for every row.

WORKED EXAMPLE · QUESTION

A 0.40 kg glider on a level air track is held between two stretched springs. A data logger records:

(a) Show that the glider moves with SHM. (b) Determine its period. (c) Calculate the resultant force when x = +0.040 m.

SOLUTION

(a) Every a has the opposite sign to x, so the acceleration is towards equilibrium. a/x = -1.00 ÷0.040 = -0.50 ÷0.020 = …= -25 s⁻² for every row, so a ∝-x: SHM. (Two rows are not enough for "show that": test several, or show the a–x graph is a straight line through the origin.)

(b) ω² = 25 s⁻², ω= 5.0 rad s⁻¹, T = 2π/ω: 1.26 s.

(c) F = ma = 0.40 ×(-1.00): −0.40 N (0.40 N towards equilibrium).

All of these are in the formulae list. A is the amplitude and t the time since the object was at x = +A; ωt is an angle in radians. If timing starts as the object passes through equilibrium, use x = Asin ωt instead. Substituting x = Acos ωt into the acceleration gives a = -ω²(Acos ωt) = -ω²x, the defining condition, as it must be. Since sin and cos never exceed 1:

WORKED EXAMPLE · QUESTION

A buoy moves up and down with SHM of amplitude 0.35 m and period 4.0 s. Timing starts at its highest point; upwards is positive. Calculate (a) the angular frequency, (b) the maximum speed, and (c) the displacement, velocity and acceleration at t = 0.60 s.

SOLUTION

(a) ω= 2π÷4.0: 1.57 rad s⁻¹ (1.5708).

(b) v_(max) = Aω= 0.35 ×1.5708: 0.550 m s⁻¹.

(c) ωt = 1.5708 ×0.60 = 0.942 rad. x = 0.35cos 0.942 = 0.35 ×0.588: +0.206 m. v = -0.550 sin 0.942 = -0.550 ×0.809: −0.445 m s⁻¹ (moving down). a = -ω²x = -2.467 ×0.206: −0.508 m s⁻² (towards equilibrium). The buoy is above equilibrium but falling, and speeding up.

Gradient of the x–t graph = velocity: zero at the peaks and troughs (momentarily at rest at ±A), steepest where the graph crosses the time axis (maximum speed at equilibrium).

Gradient of the v–t graph = acceleration: steepest where v = 0, at the extremes, where the acceleration is largest.

For a curve, find the gradient at an instant by drawing a tangent and using a large triangle on it.

Phase relationships

Velocity leads displacement by a quarter of a cycle (π/2 rad, 90°): each peak of the v–t graph comes T/4 before the next peak of the x–t graph. Acceleration is in antiphase with displacement (π rad, 180°): when x is most positive, a is most negative, the graphical form of a = -ω²x. So acceleration also leads velocity by a quarter of a cycle.

WORKED EXAMPLE · QUESTION

The graph above shows the displacement of a mass on a spring. (a) Use the tangent to determine the velocity at t = 0.30 s. (b) Check with v = -Aωsin ωt. (c) Calculate the maximum speed and state one time at which it occurs.

SOLUTION

(a) v = (-0.0631 - (-0.00759))/(0.40 - 0.20): −0.278 m s⁻¹.

(b) ω= 2π/0.80 = 7.854 rad s⁻¹, ωt = 2.356 rad: v = -0.050 ×7.854 ×sin 2.356 = -0.278 m s⁻¹ ✓. The minus sign matters: the mass is below equilibrium and still moving in the negative direction, towards the trough at 0.40 s.

(c) v_(max) = Aω= 0.050 ×7.854: 0.393 m s⁻¹, for example at t = 0.20 s.

For a mass m on a spring of stiffness k, the resultant force for a displacement x from equilibrium is -kx, for a vertical spring too, because at equilibrium the weight is already balanced by the tension. So a = -(k/m)x: SHM with ω² = k/m, giving

In the formulae list (m in kg, k in N m⁻¹). It is independent of amplitude and of g: the same spring and mass have the same period on the Moon. Valid while the spring obeys Hooke's law and its own mass is small compared with m.

WORKED EXAMPLE · QUESTION

A 0.250 kg mass hung on a light spring extends it by 0.080 m. It is pulled down a further 0.030 m and released; g = 9.81 N kg⁻¹. Calculate (a) the spring constant, (b) the period, (c) the maximum speed. (d) The mass is replaced by a 1.00 kg mass. State the new period.

SOLUTION

(a) At equilibrium tension = weight: k = mg/Δl = 0.250 ×9.81 ÷0.080: 30.7 N m⁻¹.

(b) T = 2π√(0.250 ÷30.66): 0.567 s. Check: m/k = Δl/g, so 2π√(0.080/9.81) gives the same 0.567 s.

(c) The amplitude is the 0.030 m pull, not the 0.080 m extension: ω= 2π÷0.567 = 11.07 rad s⁻¹, v_(max) = 0.030 ×11.07: 0.33 m s⁻¹.

(d) T ∝√m and the mass is 4 times larger: T = 2 ×0.567: 1.13 s.

A simple pendulum is a small dense bob on a light, inextensible string of length l, measured from the point of suspension to the centre of the bob. Displaced by angle θ, the component of the weight along the arc, mgsin θ, pulls the bob back; the tension and mgcos θ act along the string and give no restoring effect.

For small θ in radians, sin θ≈θ= x/l. The restoring force becomes -(mg/l)x, proportional to x, so a = -(g/l)x: SHM with ω² = g/l. The mass cancels, because it appears in both the restoring force and the inertia.

In the formulae list (l in m, g in N kg⁻¹).

Valid only for small amplitudes (below about 10°); independent of the bob's mass and of the amplitude within that range.

WORKED EXAMPLE · QUESTION

A pendulum clock is designed to have a period of 2.00 s; g = 9.81 N kg⁻¹. (a) Calculate the length of the pendulum. (b) On a hot day the length increases by 0.50%. Determine how much time the clock loses in one day (86 400 s).

SOLUTION

(a) l = gT²/4π² = 9.81 ×2.00² ÷4π²: 0.994 m.

(b) T ∝√l, so T_new = 2.00 ×√1.005 = 2.00 ×1.00250, about 0.25% longer. The clock counts swings, so it shows only 1/1.00250 of the real time: lost = 86 400 ×(1 - 1/1.00250): 215 s (about 3.6 minutes) per day. A percentage change in l gives half that percentage change in T.

Squaring T = 2π√(l/g) gives T² = (4π²/g) l: a graph of T² against l is a straight line through the origin with gradient 4π²/g.

Clamp the string between two small blocks so the pivot is fixed; measure l to the centre of the bob (metre rule plus half the bob diameter from vernier calipers).

Displace by under 10° and time 20 oscillations; repeat, average and divide by 20. Timing many oscillations makes reaction-time uncertainty a small fraction of the measured time.

Start and stop timing as the bob passes equilibrium, marked by a fiducial mark behind the string: the bob is fastest there, so the instant is sharpest.

Take six or more lengths over as wide a range as practical.

WORKED EXAMPLE · QUESTION

(a) Use the gradient triangle on the graph to determine g. (b) For the 0.800 m length, 20 oscillations took 35.7 s; length ± 0.002 m, time for 20 oscillations ± 0.2 s. Calculate g from this single reading and its percentage uncertainty, and comment on the result from (a).

SOLUTION

(a) Gradient = 3.22 ÷0.80 = 4.025 s² m⁻¹ = 4π²/g, so g = 4π² ÷4.025: 9.81 m s⁻².

(b) T = 35.7 ÷20 = 1.785 s; g = 4π²l/T² = 4π² ×0.800 ÷1.785²: 9.91 m s⁻². The time is squared, so its percentage counts twice: l 0.25%, T 0.56%, total 0.25 + 2 ×0.56 = 1.4%: g = 9.91 ± 0.14 m s⁻². The range 9.77 to 10.05 m s⁻² includes 9.81 m s⁻², so the two agree; the graph uses all six readings and reduces the effect of random error in any one, so its value is the better one.

An object resting on (not fixed to) a surface moving with vertical SHM is pushed up by the contact force but can only be pulled down by its weight, so its largest possible downward acceleration is g. It stays in contact only while the surface's downward acceleration is no more than g. That acceleration is largest at the top of the motion, Aω², so contact is kept throughout if Aω² ≤g.

WORKED EXAMPLE · QUESTION

A sensor rests, unfixed, on a platform vibrating vertically with SHM at 6.0 Hz; g = 9.81 m s⁻². (a) Calculate the largest amplitude for which it stays in contact. (b) At 5.0 mm amplitude, calculate the maximum acceleration of the platform and the highest frequency at which the sensor still stays in contact. (c) At 10 mm amplitude, state where the sensor first loses contact.

SOLUTION

(a) ω= 2π×6.0 = 37.7 rad s⁻¹; A_(max) = g/ω² = 9.81 ÷37.7² = 6.90 ×10⁻³ m: 6.9 mm.

(b) a_(max) = Aω² = 0.0050 ×37.7²: 7.1 m s⁻², less than g, so contact is kept. Highest frequency: f = √(9.81 ÷0.0050) ÷2π: 7.0 Hz.

(c) ω²x = g at x = 6.9 mm above equilibrium, on the way up; the sensor leaves the platform there.

Hooke's law (Unit 1) and circular motion (Unit 4)

F = kΔx for a spring is why a mass–spring system is SHM; the same k appears in T = 2π√(m/k). ω is in rad s⁻¹ because the shadow of an object moving in a circle at angular speed ω moves with SHM of angular frequency ω.

Waves (Unit 2)

Each particle of a medium carrying a sinusoidal wave oscillates with SHM; the phase difference between particles is the same idea as the phase difference between x, v and a here.

Energy, damping and resonance (next Oscillations page)

Kinetic and potential energy are exchanged during each cycle; real oscillators lose energy, and a driven oscillator responds most strongly near its natural frequency 1/T.

1. A ball bounces repeatedly on a hard floor, reaching the same height each time. Explain why its motion is not simple harmonic.

2. An oscillator has a frequency of 2.5 Hz. Calculate its period and angular frequency.

3. An object moves with SHM of amplitude 0.12 m and period 1.5 s. Calculate its maximum speed and maximum acceleration.

4. At an instant, an oscillator is at its maximum positive displacement. State its velocity and the direction of its acceleration.

5. A pendulum's length is increased from 0.40 m to 2.42 m. By what factor does its period change?

1. Between bounces its acceleration is g downwards whatever its height, so the acceleration is not proportional to its displacement from any fixed point.

2. T = 1/2.5 = 0.40 s; ω= 2π×2.5 = 15.7 rad s⁻¹.

3. ω= 2π/1.5 = 4.19 rad s⁻¹; v_(max) = 0.12 ×4.19 = 0.50 m s⁻¹; a_(max) = 0.12 ×4.19² = 2.11 m s⁻².

4. Velocity zero; acceleration maximum and in the negative direction (towards equilibrium).

5. T ∝√l; the length increases by a factor of 6.05, so the period increases by √6.05 = 2.46 (about 2.5).