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SHM systems: pendulums and springs

Two real oscillators carry the theory, a mass on a spring and a pendulum on a string. Each has a period formula with two variables, a continuous exchange between kinetic and potential energy, and damping, which removes energy over time.

Year 13AQA 3.6.1.3 CIE 17.2.1, 17.2.2, 17.3.1, 17.3.2 OCR A 5.3.1(c)(ii), 5.3.2(a), 5.3.2(b), 5.3.3(b)(i)EDEXCEL 2, 181, 183, 188

Builds on Simple harmonic motion and Density and Hooke's law.

IN THIS TOPIC

Use T = 2π√(m/k) for a mass-spring system, and say why g never appears in it.
Use T = 2π√(l/g) for a simple pendulum, with the small-angle condition stated.
Describe how E_k, E_p and the total energy vary with displacement and with time.
Put a number on the total with E = ½kA², and get CIE's E = ½mω²x₀² out of it through ω² = k/m.
Tell light, heavy and critical damping apart by what each does to the motion.
Get k or g from the gradient of a T² graph, the way required practical 7 does.

COMMON MISCONCEPTION

Heavier pendulums swing slower.

The mass-spring clock

Hang a mass on a spring, displace it, and Hooke's law supplies exactly the restoring force SHM demands. From F = −kx comes a = −(k/m)x, the defining equation with ω² = k/m. The period follows.

T = 2\pi\sqrt{\frac{m}{k}}

ON THE AQA DATA SHEET

FIG. 1A mass-spring oscillator. Its period depends on m and k alone: four times the mass, twice the period; amplitude and g never appear.

Read the formula like an ingredients list. More mass makes it sluggish, and quadrupling m doubles T. A stiffer spring runs quicker, since k sits underneath. Then there are two famous absences. Amplitude is missing because SHM is isochronous. g is missing because gravity only shifts the equilibrium point the mass bounces about, leaving the bounce itself untouched. The same clock keeps the same time in orbit.

WORKED EXAMPLE

Reading a spring clock

A 0.20 kg mass hangs from a spring of stiffness 32 N m⁻¹. Find the period and frequency of its oscillations.

Write it with the square root spelled out. T = 2π × the square root of m/k = 2π × the square root of 0.20/32.

That gives T = 2π × 0.079 = 0.50 s, and f = 1/T = 2.0 Hz.

Check the levers. A stiffer spring or a lighter mass would tick faster, and both sit exactly where the formula puts them, k below and m above.

The pendulum clock

A pendulum's restoring force is the component of gravity along its arc, and for small angles that component is proportional to displacement, giving approximate SHM with

T = 2\pi\sqrt{\frac{l}{g}}

ON THE AQA DATA SHEET

FIG. 2A simple pendulum. The period depends on l and g; the bob's mass cancels, and the SHM only holds for small swings.

The mass is absent. Restoring force and inertia both scale with m, so it cancels, exactly as it did for objects in free fall. What does appear is g, and a pendulum therefore doubles as a portable gravity meter. Do not skate over the small-angle approximation. Both the formula and the simple harmonic motion behind it are approximations that hold for small swings only. Other oscillators can appear in questions, liquid in a U-tube being the classic, but with all the necessary information provided.

Required practical 7 tests both formulas. Time twenty oscillations and divide, so your reaction time is spread across twenty periods instead of one, then plot T² against m for the spring or against l for the pendulum. Each gives a straight line through the origin, of gradient 4π²/k and 4π²/g respectively, so a stiffness or a value of g comes out of a whole graph instead of a single reading.

The energy see-saw

FIG. 3Energy against displacement: potential grows as x², kinetic fills the remainder, and the total sits constant across the swing.

The oscillation is a continuous trade. Potential energy is maximal at the extremes and zero at the centre, growing as the square of displacement; kinetic energy is its mirror, all at the centre, none at the ends; the total stays constant. Against time, each energy oscillates at twice the motion's frequency, since both extremes of a cycle look identical to the energy books. Sketching either picture, energy against displacement or against time, is a standard question.

Putting a number on that flat total takes one line. At an extreme every joule is potential. For a horizontal mass-spring that potential is the spring's own elastic energy, ½kx². For the vertical spring this lesson derived, x is measured from the hanging equilibrium, and the quantity growing as ½kx² is the combined spring-plus-gravity potential measured from its minimum there, not the spring's elastic energy alone; the two effects fold together into the same expression, which is why one formula serves both arrangements. So an oscillation of amplitude A carries

E = \frac{1}{2}kA^{2}

and that is the whole energy of the motion, at every instant of it. CIE asks for the same quantity in a form that mentions no spring at all, writing x₀ for the amplitude,

E = \frac{1}{2}m\omega^{2}x_{0}^{2}

so on that specification it is a recall item, printed in no booklet. These are one equation and not two. The mass-spring result at the top of this lesson was ω² = k/m, which rearranges to k = mω², and substituting that into ½kA² hands back ½mω²A² on the spot. Check the pair against each other once and you will trust them afterwards. Take the 0.20 kg mass on the 32 N m⁻¹ spring, where ω² = 32/0.20 = 160 s⁻², and ½ × 0.20 × 160 × 0.050² = 0.040 J, which is the ½kA² answer below to the last digit.

The second form is also the more general one. Every simple harmonic oscillator obeys it, spring or not, because ω and the amplitude are all simple harmonic motion has; ½kA² needs a stiffness and so applies only to springs. Reach for it when a question gives you a mass and a period but never a value of k, which is the shape most pendulum questions arrive in.

GUIDED PRACTICE

The see-saw with numbers on it

The same oscillator, k = 32 N m⁻¹ and m = 0.20 kg, swings with amplitude 0.050 m. Find the total energy, then the speed at the centre.

Show the working

At the ends the energy is all potential, so E = ½kA² = ½ × 32 × 0.050² = 0.040 J.

At the centre it is all kinetic, so ½mv² = 0.040 and v = 0.63 m s⁻¹. The same answer follows from v = ωA, a cross-check worth the thirty seconds.

INDEPENDENT PRACTICE

A pendulum abroad

A 1.0 m pendulum ticks on Earth (g = 9.81) and then on Mars (g = 3.71 m s⁻²). Find both periods, and state the general rule your answers illustrate.

Show the working

On Earth, T = 2π × the square root of 1.0/9.81 = 2.0 s. On Mars, T = 2π × the square root of 1.0/3.71 = 3.3 s.

Weaker gravity, slower clock. The period scales as one over the square root of g, which makes a pendulum a gravimeter that happens to tell the time.

Damping

FIG. 4A damped oscillation: the amplitude decays inside an exponential envelope while the period barely changes.

Real oscillators leak. Damping is the loss of energy to resistive forces, air resistance and internal friction among them, and its signature is an amplitude that decays, exponentially in the common case, while the period barely changes. The energy leaves as internal energy in the surroundings. The rhythm survives, quieter each cycle, until it fades entirely.

Three grades of damping are examined by name. Light damping lets a system oscillate many times over, the amplitude shrinking slowly inside its envelope, which is a pendulum swinging in air. Heavy damping smothers the oscillation, so the system creeps back to equilibrium over a long time and never crosses it, which is that pendulum in treacle. Between them lies critical damping, the case that brings a system back to equilibrium in the shortest possible time with no oscillation at all. Car suspensions and the needle of a moving-coil meter are tuned for it, so that a bump or a reading settles once and stays settled.

ASSESSMENT FOCUS

Match the formula to the system, T = 2π√(m/k) for springs and T = 2π√(l/g) for pendulums, and notice that neither one contains the amplitude.
The pendulum's period is independent of mass; the spring's is independent of g. Each absence gets asked about on its own.
Both formulas hide a square root, so quadrupling m or l doubles T. Halving T needs a quarter of the length, not half of it.
Energy against time oscillates at twice the frequency of the motion, and the total draws a flat line. Sketches are marked on both features.
CIE prints neither form of the total energy, so learn E = ½mω²x₀² and remember that ω² = k/m turns it straight back into ½kA². Quote whichever form is built from the quantities the question actually gave you.
State the small-angle condition whenever you use the pendulum formula. For large swings the motion stops being simple harmonic and the formula overreaches.
Critical damping means the quickest return to equilibrium with no oscillation. Heavy damping also avoids oscillating but takes longer, and mixing the two up is the standard loss of a mark here.

CHECK YOURSELF

What length of simple pendulum has a period of 2.0 s? Take g = 9.81 m s⁻². Would the answer change on the Moon?

Show a hint

Rearrange for l, and remember what sits inside the square root.

Show the answer

Rearranging: l = g(T/2π)² = 9.81 × (2.0/2π)² = 0.99 m, the metre-long “seconds pendulum” of old clocks.

On the Moon, yes. There g is about a sixth of ours, so the same 2.0 s period needs l = 1.62 × (0.318)² ≈ 0.16 m. A pendulum reads its local gravity.

The bob's mass, note, was never asked for and never needed.

Springs count m and k. Pendulums count l and g.

The energy see-saws; the total holds still.

Damping drains the amplitude; light damping stretches the period only slightly.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

19 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the shm systems: pendulums and springs questions page.

6 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Required practical 7: simple harmonic motion: mass-spring and pendulum puts this topic in the lab, and the written papers ask about it.
Uncertainty arithmetic is the maths this lesson leans on, worked through from GCSE.

CHECK YOUR PROGRESS

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Use T = 2π√(m/k) for a mass-spring system, and say why g never appears in it.
Use T = 2π√(l/g) for a simple pendulum, with the small-angle condition stated.
Describe how E_k, E_p and the total energy vary with displacement and with time.
Put a number on the total with E = ½kA², and get CIE's E = ½mω²x₀² out of it through ω² = k/m.
Tell light, heavy and critical damping apart by what each does to the motion.
Get k or g from the gradient of a T² graph, the way required practical 7 does.

Open the full revision checklist to track your progress across the whole unit.