Hang a mass on a spring, give it a little pull, and let go — it bounces up and down forever (near enough). That bouncing is simple harmonic motion, and there’s one tidy formula that tells you exactly how long each bounce takes.
📘 What you need to know
A mass–spring system is simply a mass fixed to the end of a spring, then pulled from rest and released
The spring supplies the restoring force, and it obeys Hooke’s law: F = −kx
The time period is T = 2π√(m/k)
The same formula works whether the spring is vertical or horizontal
The period does not depend on gravity — a mass–spring clock keeps the same time on the Earth and the Moon
A stiffer spring (bigger k) gives a shorter period; a heavier mass (bigger m) gives a longer period
The Restoring Force Comes From the Spring
When you stretch or squash a spring, it pushes back — and the harder you stretch it, the harder it pushes back. That’s exactly the “proportional to displacement” rule that simple harmonic motion needs. We write it as Hooke’s law:
Restoring force (Hooke’s law)F = −kx
Here k is the spring constant (in N m⁻¹), telling you how stiff the spring is, and x is how far the mass sits from its resting (equilibrium) position. The minus sign is doing the important job: it says the force always points back towards equilibrium, opposite to the way you pulled. Pull the mass down and the spring tugs it up; push it up and the spring pushes it down.
Left: the mass sitting at equilibrium. Right: pulled down by a displacement x (grey), the spring’s restoring force F (violet) points straight back towards equilibrium.
The Time Period Formula
Feed that restoring force through Newton’s second law and the maths spits out a lovely result — the time for one full up-and-down cycle depends only on the mass and the spring’s stiffness:
Period of a mass–spring systemT = 2π√(m/k)
where T is the time period in seconds, m is the mass on the spring in kilograms, and k is the spring constant in N m⁻¹. Read it like a story: put a heavier mass on and it’s harder to shift, so it bounces more slowly (longer T); use a stiffer spring (bigger k) and it snaps back harder, so it bounces faster (shorter T).
Two details that examiners love to test. First, the same formula works whether the spring hangs vertically or lies horizontally. Second — and this catches people out — there is no g anywhere in the formula, so gravity has no say in the period. Take your mass–spring system to the Moon and it keeps exactly the same rhythm.
Same mass, same spring, same period. Because g never appears in T = 2π√(m/k), the orientation — and gravity — make no difference.
🧭 Working out a mass–spring period
Convert units first — mass into kilograms, and check k is in N m⁻¹
Drop the values straight into T = 2π√(m/k)
Do the divisionm/k inside the root first, then take the square root, then multiply by 2π
Need the frequency? Just use f = 1/T
If a question mentions Hooke’s law, remember k is the same spring constant that appears here
Quick recap: the spring gives a restoring force F = −kx, and the bounce time is T = 2π√(m/k) — heavier is slower, stiffer is faster, and gravity never gets a vote.
WE 1
A 250 g mass hangs from a spring of spring constant 40 N m⁻¹ and is set oscillating vertically.
(a) Calculate the time period of the oscillation.
(b) Hence find the frequency.
Part (a)
m = 250 g = 0.25 kg
T = 2π√(m/k) = 2π√(0.25 / 40)T = 2π × √0.00625 = 2π × 0.0791T ≈ 0.50 sPart (b)f = 1/T = 1 / 0.4967f ≈ 2.0 HzNotice we never needed g — the same spring on the Moon would give the same 0.50 s.
WE 2
A 200 g toy is attached to a horizontal spring of spring constant 90 N m⁻¹ and pulled 5.0 cm from its equilibrium position.
(a) Calculate the restoring force on the toy at that point.
(b) Calculate its acceleration there.
Part (a)
Using the size of F = kx, with x = 5.0 cm = 0.050 m
F = 90 × 0.050 = 4.5 NF ≈ 4.5 N (towards equilibrium)Part (b)
Newton’s second law: a = F/m, with m = 0.20 kg
a = 4.5 / 0.20a ≈ 23 m s⁻²This is the acceleration at full stretch (the amplitude) — the biggest it ever gets.
💡 Top tips
Grams to kilograms, centimetres to metres — sort units out before touching the formula
Inside the root it’s m over k, not k over m — a quick check: heavier mass should make T bigger
No g in the formula means Earth and Moon give the same period — a favourite exam trap
The spring constant k here is the very same one from Hooke’s law, so those two topics often turn up together
⚠ Common mistakes
Flipping the fraction and using √(k/m) instead of √(m/k)
Leaving the mass in grams or the extension in centimetres
Thinking the period changes on the Moon — it doesn’t, because gravity isn’t in the formula
Taking the square root of only m or only k instead of the whole ratio m/k
Forgetting the factor of 2π out front
Up next: we swap the spring for a piece of string and look at the Time Period of a Simple Pendulum — where, unlike here, gravity is suddenly back in charge.
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