5.5 Simple harmonic motion
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A2 content. Energy in SHM, damping, forced oscillations and resonance are on the next Oscillations page.
What this page covers
6 points
- The condition for simple harmonic motion, \(F = -kx\) and \(a = -\omega^2x\), and how to test whether a motion is SHM
- Amplitude, period, frequency and angular frequency
- \(x = A\cos\omega t\), \(v = -A\omega\sin\omega t\), \(a = -A\omega^2\cos\omega t\), and the maximum speed \(A\omega\)
- 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
In one minute
6 points
- Simple harmonic motion is oscillation in which the acceleration is proportional to the displacement from a fixed point and always directed towards that point, \(a = -\omega^2x\), and everything else on this page follows from that one condition.
- It comes from a restoring force proportional to displacement, \(F = -kx\). A restoring force alone is not enough: it must be proportional.
- \(\omega = 2\pi f = 2\pi/T\). Released from an extreme: \(x = A\cos\omega t\), \(v = -A\omega\sin\omega t\), \(a = -A\omega^2\cos\omega t\). Use radians.
- Maximum speed \(A\omega\) at equilibrium; maximum acceleration \(A\omega^2\) at the extremes, where the speed is zero.
- Velocity is a quarter of a cycle (π/2) ahead of displacement; acceleration is in antiphase (π) with displacement.
- Mass on a spring: \(T = 2\pi\sqrt{m/k}\). Simple pendulum, small angles: \(T = 2\pi\sqrt{l/g}\). Neither depends on the amplitude.
Contents
14 sections
- What is simple harmonic motion?
- How can you tell whether a motion is simple harmonic?
- Which equations describe the motion?
- How are the displacement, velocity and acceleration graphs linked?
- What sets the period of a mass on a spring?
- What sets the period of a simple pendulum?
- How can an oscillation be used to measure g?
- When does a resting object leave a vibrating surface?
- Exam tips for 5.5
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
What is simple harmonic motion?
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 \(\omega^2\) gives the defining equation.
\[F = -kx \qquad a = -\omega^2x\]Both are in the formulae list supplied in the exam (\(a\) in m s⁻², \(\omega\) in rad s⁻¹).
| Term | Meaning | Symbol and unit |
|---|---|---|
| Displacement | distance from the equilibrium position, with a sign for direction | x, m |
| Amplitude | maximum displacement from equilibrium | A, m |
| Period | time for one complete oscillation | T, s |
| Frequency | number of oscillations per unit time | f, Hz |
| Angular frequency | 2π × frequency; the constant in \(a = -\omega^2x\) | ω, 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.
How can you tell whether a motion is simple harmonic?
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 \(-\omega^2\); in a table, \(a/x\) is the same negative constant for every row.
A 0.40 kg glider on a level air track is held between two stretched springs. A data logger records:
| x / m | −0.060 | −0.040 | −0.020 | +0.020 | +0.040 | +0.060 |
|---|---|---|---|---|---|---|
| a / m s⁻² | +1.50 | +1.00 | +0.50 | −0.50 | −1.00 | −1.50 |
(a) Show that the glider moves with SHM. (b) Determine its period. (c) Calculate the resultant force when x = +0.040 m.
(a) Every \(a\) has the opposite sign to \(x\), so the acceleration is towards equilibrium. \(a/x = -1.00 \div 0.040 = -0.50 \div 0.020 = \dots = -25\) s⁻² for every row, so \(a \propto -x\): 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) \(\omega^2 = 25\) s⁻², \(\omega = 5.0\) rad s⁻¹, \(T = 2\pi/\omega\): 1.26 s.
(c) \(F = ma = 0.40 \times (-1.00)\): −0.40 N (0.40 N towards equilibrium).
| System | SHM? | Reason |
|---|---|---|
| Mass on a spring (within the limit of proportionality) | yes | Hooke's law makes the resultant force proportional to displacement |
| Simple pendulum | only for small angles | the restoring force is mg sin θ, close to proportional to θ only below about 10° |
| Block sliding on a smooth V-shaped track | no | the restoring acceleration has a constant size, g sin α |
| Ball bouncing on a hard floor | no | between bounces the acceleration is g downwards whatever the height |
| Ball rolling in a shallow circular bowl | approximately, for small displacements | like a pendulum: the restoring force grows almost in proportion to the displacement |
Saying a motion is SHM "because there is a restoring force" or "because it repeats". Both are true of the bouncing ball, which is not SHM. The test is proportionality between acceleration and displacement.
Which equations describe the motion?
\[x = A\cos\omega t \qquad v = -A\omega\sin\omega t \qquad a = -A\omega^2\cos\omega t\] \[T = \frac{1}{f} = \frac{2\pi}{\omega} \qquad \omega = 2\pi f\]All of these are in the formulae list. \(A\) is the amplitude and \(t\) the time since the object was at \(x = +A\); \(\omega t\) is an angle in radians. If timing starts as the object passes through equilibrium, use \(x = A\sin\omega t\) instead. Substituting \(x = A\cos\omega t\) into the acceleration gives \(a = -\omega^2(A\cos\omega t) = -\omega^2x\), the defining condition, as it must be. Since sin and cos never exceed 1:
\[v_{\max} = A\omega \ \text{(at } x = 0\text{)} \qquad a_{\max} = A\omega^2 \ \text{(at } x = \pm A\text{)}\]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.
(a) \(\omega = 2\pi \div 4.0\): 1.57 rad s⁻¹ (1.5708).
(b) \(v_{\max} = A\omega = 0.35 \times 1.5708\): 0.550 m s⁻¹.
(c) \(\omega t = 1.5708 \times 0.60 = 0.942\) rad. \(x = 0.35\cos 0.942 = 0.35 \times 0.588\): +0.206 m. \(v = -0.550 \sin 0.942 = -0.550 \times 0.809\): −0.445 m s⁻¹ (moving down). \(a = -\omega^2x = -2.467 \times 0.206\): −0.508 m s⁻² (towards equilibrium). The buoy is above equilibrium but falling, and speeding up.
Leaving the calculator in degree mode. cos(0.942°) is almost exactly 1, which would leave the buoy at 0.350 m, still at the top after 0.60 s. If an answer says the object has barely moved, check the angle mode.
How are the displacement, velocity and acceleration graphs linked?
- 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.
| Position | Displacement | Speed | Acceleration |
|---|---|---|---|
| Equilibrium | 0 | maximum, Aω | 0 |
| Positive extreme | +A | 0 | maximum, −Aω² (towards equilibrium) |
| Negative extreme | −A | 0 | maximum, +Aω² (towards equilibrium) |
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 \(a = -\omega^2x\). So acceleration also leads velocity by a quarter of a cycle.
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 \(v = -A\omega\sin\omega t\). (c) Calculate the maximum speed and state one time at which it occurs.
(a) \(v = \dfrac{-0.0631 - (-0.00759)}{0.40 - 0.20}\): −0.278 m s⁻¹.
(b) \(\omega = 2\pi/0.80 = 7.854\) rad s⁻¹, \(\omega t = 2.356\) rad: \(v = -0.050 \times 7.854 \times \sin 2.356 = -0.278\) 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) \(v_{\max} = A\omega = 0.050 \times 7.854\): 0.393 m s⁻¹, for example at t = 0.20 s.
To sketch v–t from x–t, mark the zero-gradient points first (where \(v = 0\)), then the steepest points (largest |v|), and keep the same period. Repeat to get a–t from v–t.
What sets the period of a mass on a spring?
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 \(\omega^2 = k/m\), giving
\[T = 2\pi\sqrt{\frac{m}{k}}\]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 \(m\).
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.
(a) At equilibrium tension = weight: \(k = mg/\Delta l = 0.250 \times 9.81 \div 0.080\): 30.7 N m⁻¹.
(b) \(T = 2\pi\sqrt{0.250 \div 30.66}\): 0.567 s. Check: \(m/k = \Delta l/g\), so \(2\pi\sqrt{0.080/9.81}\) gives the same 0.567 s.
(c) The amplitude is the 0.030 m pull, not the 0.080 m extension: \(\omega = 2\pi \div 0.567 = 11.07\) rad s⁻¹, \(v_{\max} = 0.030 \times 11.07\): 0.33 m s⁻¹.
(d) \(T \propto \sqrt{m}\) and the mass is 4 times larger: \(T = 2 \times 0.567\): 1.13 s.
Using the static extension (0.080 m) as the amplitude. The amplitude is measured from the equilibrium position, so it is the extra distance the mass was pulled down, 0.030 m here.
What sets the period of a simple pendulum?
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 \(\theta\), the component of the weight along the arc, \(mg\sin\theta\), pulls the bob back; the tension and \(mg\cos\theta\) act along the string and give no restoring effect.
For small \(\theta\) in radians, \(\sin\theta \approx \theta = x/l\). The restoring force becomes \(-(mg/l)x\), proportional to \(x\), so \(a = -(g/l)x\): SHM with \(\omega^2 = g/l\). The mass cancels, because it appears in both the restoring force and the inertia.
\[T = 2\pi\sqrt{\frac{l}{g}}\]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.
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).
(a) \(l = gT^2/4\pi^2 = 9.81 \times 2.00^2 \div 4\pi^2\): 0.994 m.
(b) \(T \propto \sqrt{l}\), so \(T_{\text{new}} = 2.00 \times \sqrt{1.005} = 2.00 \times 1.00250\), about 0.25% longer. The clock counts swings, so it shows only 1/1.00250 of the real time: lost \(= 86\,400 \times (1 - 1/1.00250)\): 215 s (about 3.6 minutes) per day. A percentage change in \(l\) gives half that percentage change in \(T\).
Write \(T \propto \sqrt{m}\) and \(T \propto 1/\sqrt{k}\) for a spring, \(T \propto \sqrt{l}\) and \(T \propto 1/\sqrt{g}\) for a pendulum; then 4× in the quantity gives 2× in the period. The same pendulum on the Moon (g = 1.62 N kg⁻¹) would have \(T = 2.00 \times \sqrt{9.81/1.62} = 4.92\) s.
How can an oscillation be used to measure g?
Squaring \(T = 2\pi\sqrt{l/g}\) gives \(T^2 = (4\pi^2/g)\,l\): a graph of \(T^2\) against \(l\) is a straight line through the origin with gradient \(4\pi^2/g\).
- 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.
(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).
(a) Gradient \(= 3.22 \div 0.80 = 4.025\) s² m⁻¹ \(= 4\pi^2/g\), so \(g = 4\pi^2 \div 4.025\): 9.81 m s⁻².
(b) \(T = 35.7 \div 20 = 1.785\) s; \(g = 4\pi^2l/T^2 = 4\pi^2 \times 0.800 \div 1.785^2\): 9.91 m s⁻². The time is squared, so its percentage counts twice: \(l\) 0.25%, \(T\) 0.56%, total \(0.25 + 2 \times 0.56 = 1.4\%\): 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.
If \(l\) is measured to the top of the bob, every length is too short by the same amount. The \(T^2\)–\(l\) line shifts sideways and gains an intercept, but its gradient, and so \(g\), is unchanged. A single calculation has no such protection.
When does a resting object leave a vibrating surface?
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, \(A\omega^2\), so contact is kept throughout if \(A\omega^2 \le g\).
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.
(a) \(\omega = 2\pi \times 6.0 = 37.7\) rad s⁻¹; \(A_{\max} = g/\omega^2 = 9.81 \div 37.7^2 = 6.90 \times 10^{-3}\) m: 6.9 mm.
(b) \(a_{\max} = A\omega^2 = 0.0050 \times 37.7^2\): 7.1 m s⁻², less than g, so contact is kept. Highest frequency: \(f = \sqrt{9.81 \div 0.0050} \div 2\pi\): 7.0 Hz.
(c) \(\omega^2x = g\) at x = 6.9 mm above equilibrium, on the way up; the sensor leaves the platform there.
Exam tips for 5.5
Acceleration proportional to displacement from equilibrium, and always directed towards equilibrium. Writing \(a = -\omega^2x\) and saying what each symbol means gives both.
Every \(\omega t\) is in radians. Switch to radian mode before the first calculation, and back to degrees only for angles such as a pendulum's release angle.
\(x = A\cos\omega t\) starts at an extreme; \(x = A\sin\omega t\) starts at equilibrium. Read where timing starts and keep the matching velocity equation.
State the positive direction. "−0.445 m s⁻¹, taking upwards as positive" means moving downwards at 0.445 m s⁻¹, not a mistake.
Plot \(T^2\) against \(l\) (gradient \(4\pi^2/g\)) or \(T^2\) against \(m\) (gradient \(4\pi^2/k\)). When a quantity is squared, double its percentage uncertainty.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Any oscillation with a restoring force is SHM. | The restoring force must also be proportional to displacement. | Test that a/x is a negative constant (or a–x is a straight line through the origin). |
| At the extremes the acceleration is zero because the object stops. | Velocity is momentarily zero, but displacement, and so restoring force, is greatest there. | Acceleration is maximum, Aω², at ±A; zero at equilibrium. |
| A bigger swing takes longer. | A bigger displacement brings a proportionally bigger restoring force, so the object moves faster. | The period of SHM is independent of amplitude (for a pendulum, only at small angles). |
| A heavier pendulum bob swings more slowly. | Mass appears in both the restoring force and the inertia, and cancels. | \(T = 2\pi\sqrt{l/g}\) has no mass in it. |
| A mass on a spring would have a different period on the Moon. | Gravity only shifts the equilibrium position. | \(T = 2\pi\sqrt{m/k}\) has no g in it. |
| Velocity and displacement are in antiphase. | Velocity is greatest when displacement is zero. | Velocity is a quarter cycle (π/2) ahead of displacement; acceleration is in antiphase (π). |
How this topic connects
Hooke's law (Unit 1) and circular motion (Unit 4)
\(F = k\Delta x\) for a spring is why a mass–spring system is SHM; the same \(k\) appears in \(T = 2\pi\sqrt{m/k}\). \(\omega\) is in rad s⁻¹ because the shadow of an object moving in a circle at angular speed \(\omega\) moves with SHM of angular frequency \(\omega\).
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.
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(F = -kx\) | condition for SHM in terms of force | yes |
| \(a = -\omega^2x\) | condition for SHM; acceleration at any displacement | yes |
| \(x = A\cos\omega t\) | displacement at time t (timed from an extreme) | yes |
| \(v = -A\omega\sin\omega t\) | velocity at time t | yes |
| \(a = -A\omega^2\cos\omega t\) | acceleration at time t | yes |
| \(T = 1/f = 2\pi/\omega\), \(\omega = 2\pi f\) | linking period, frequency and angular frequency | yes |
| \(T = 2\pi\sqrt{m/k}\) | period of a mass on a spring | yes |
| \(T = 2\pi\sqrt{l/g}\) | period of a simple pendulum, small angles | yes |
| \(v_{\max} = A\omega\), \(a_{\max} = A\omega^2\) | maximum speed (at equilibrium) and acceleration (at the extremes) | no — derive it from the sin and cos equations |
| \(T^2 = (4\pi^2/g)\,l\) | straight-line graph to find g | no — square the pendulum equation |
Check your understanding
1. A ball bounces repeatedly on a hard floor, reaching the same height each time. Explain why its motion is not simple harmonic.
Between bounces its acceleration is g downwards whatever its height, so the acceleration is not proportional to its displacement from any fixed point.
2. An oscillator has a frequency of 2.5 Hz. Calculate its period and angular frequency.
\(T = 1/2.5 = 0.40\) s; \(\omega = 2\pi \times 2.5 = 15.7\) rad s⁻¹.
3. An object moves with SHM of amplitude 0.12 m and period 1.5 s. Calculate its maximum speed and maximum acceleration.
\(\omega = 2\pi/1.5 = 4.19\) rad s⁻¹; \(v_{\max} = 0.12 \times 4.19 = 0.50\) m s⁻¹; \(a_{\max} = 0.12 \times 4.19^2 = 2.11\) m s⁻².
4. At an instant, an oscillator is at its maximum positive displacement. State its velocity and the direction of its acceleration.
Velocity zero; acceleration maximum and in the negative direction (towards equilibrium).
5. A pendulum's length is increased from 0.40 m to 2.42 m. By what factor does its period change?
\(T \propto \sqrt{l}\); the length increases by a factor of 6.05, so the period increases by \(\sqrt{6.05} = 2.46\) (about 2.5).
Specification checklist
Wording is ours; numbers refer to the specification points.
- (143) I can state the condition for simple harmonic motion, \(F = -kx\) or \(a = -\omega^2x\), and decide from data or a description whether a motion is SHM.
- (144) I can use \(a = -\omega^2x\), \(x = A\cos\omega t\), \(v = -A\omega\sin\omega t\), \(a = -A\omega^2\cos\omega t\), \(T = 1/f = 2\pi/\omega\) and \(\omega = 2\pi f\), working in radians.
- (145) I can use \(T = 2\pi\sqrt{m/k}\) for a mass on a spring and \(T = 2\pi\sqrt{l/g}\) for a simple pendulum, and explain why the pendulum result needs small angles.
- (146) I can sketch and interpret displacement–time graphs for SHM and find velocity from their gradient.
- (147) I can sketch and interpret velocity–time graphs for SHM, find acceleration from their gradient, describe the phase relationships between x, v and a, and use the maximum speed \(A\omega\).