Hang a mass on a spring, pull it down, and let go — it bobs up and down forever (near enough). How long does each bob take? It turns out the answer depends on just two things: how heavy the mass is and how stiff the spring is. One tidy formula ties them together, and — surprisingly — gravity doesn’t get a look-in. Here’s how it works.
📘 What you need to know
A mass–spring system is a mass on the end of a spring, oscillating in SHM
The restoring force comes from the spring: F = −kx (Hooke’s law), where k is the spring constant
The time period is T = 2π√(m/k)
A heavier mass gives a longer period; a stiffer spring (bigger k) gives a shorter period
The period does not depend on gravity — it’s the same on the Earth and the Moon
The formula works for both vertical and horizontal spring systems
What drives the oscillation?
A mass on a spring oscillates because the spring always tries to return to its natural length. Stretch it and it pulls back; compress it and it pushes back. That’s the restoring force, and for a spring it follows Hooke’s law:
Restoring force (Hooke’s law)F = −kx
Here k is the spring constant (in N m−1), a measure of stiffness, and x is the displacement from equilibrium. The minus sign tells you the force always points back towards equilibrium — which is exactly the condition for SHM. So the force grows in proportion to displacement, and the mass moves with simple harmonic motion.
Notice that Hooke’s law F = −kx has the same shape as the SHM restoring-force rule. That’s not a coincidence — it’s the whole reason a mass on a spring does SHM in the first place. The spring is nature’s perfect SHM machine.
The time period formula
Working through the SHM maths for a spring gives one of the neatest results in the topic — the equation for the time period of a mass–spring system:
Time period of a mass–spring systemT = 2π√(m / k)
Where:
T = time period (s)
m = mass on the end of the spring (kg)
k = spring constant (N m−1)
Whether the spring hangs vertically or lies flat, the same equation T = 2π√(m/k) gives the period.
WE 1
A 0.25 kg mass is attached to a spring of spring constant 40 N m⁻¹ and set oscillating. Calculate the time period of the oscillation.
Step 1 — write down the formulaT = 2π√(m/k)
Step 2 — substitute m = 0.25 kg, k = 40 N m⁻¹T = 2π√(0.25 / 40) = 2π√(0.00625)Step 3 — evaluateT = 2π × 0.0791 = 0.497 sT = 0.50 sDivide inside the root first, then square-root, then multiply by 2π. Order matters.
What changes the period?
The formula tells a clear story. Because m is on top and k is on the bottom (both inside a square root):
🎯 Reading the formula
Bigger mass → longer period. A heavier mass has more inertia, so it’s slower to speed up and slow down. It sloshes back and forth more lazily.
Stiffer spring (bigger k) → shorter period. A stiff spring pulls back harder, whipping the mass through its cycle faster.
It’s a square-root relationship. To double the period you must quadruple the mass, not just double it.
That square-root link means T is proportional to √m. Plot period against the square root of mass and you get a straight line through the origin:
A straight line through the origin confirms T ∝ √m. Four times the mass gives twice the period.
WE 2
A mass–spring system has a time period of 1.2 s. The mass is replaced with one four times heavier, using the same spring. What is the new time period?
Step 1 — T is proportional to √m
so multiplying m by 4 multiplies T by √4 = 2
Step 2 — scale the periodT_new = 1.2 × 2 = 2.4 sT = 2.4 sYou don’t need k or the actual mass — the ratio does all the work.
WE 3
A 0.30 kg mass on a spring oscillates with a time period of 0.80 s. Calculate the spring constant.
Step 1 — start from T = 2π√(m/k) and square both sides
T² = 4π² (m/k)
Step 2 — rearrange for k
k = 4π²m / T²
Step 3 — substitute m = 0.30 kg, T = 0.80 sk = 4π² × 0.30 / (0.80)² = 11.84 / 0.64k = 18.5 N m⁻¹Square the whole period (0.80² = 0.64), not just part of it.
Why gravity doesn’t matter
Here’s the part that catches everyone out. Look again at the formula: T = 2π√(m/k). There’s no g anywhere in it. That means the time period does not depend on gravitational field strength.
Take the whole system to the Moon and it oscillates with exactly the same period as on Earth. On a vertical spring, gravity does shift the equilibrium position lower (the spring stretches more to support the weight), but it doesn’t change the stiffness of the spring or the mass — and those are the only things the period depends on.
Key contrast: a mass–spring period is independent of gravity, but a simple pendulum’s period depends on g. Don’t mix the two up — it’s a favourite exam trap.
Heavier mass m up
T = 2π√(m/k)
Longer period T up
stiffer spring k up → T down
Shorter period
Mass–spring vs simple pendulum at a glance
Feature
Mass–spring
Simple pendulum
Period formula
T = 2π√(m/k)
T = 2π√(L/g)
Depends on mass?
Yes
No
Depends on gravity?
No
Yes
Restoring force from
The spring (Hooke’s law)
Gravity along the arc
💡 Top tips
T = 2π√(m/k) — mass on top, spring constant on bottom, all inside the root.
To find k, square the whole equation first: k = 4π²m/T².
No g in the formula — the period is the same on the Moon as on Earth.
It’s a square-root relationship: to double T, multiply m by 4.
The same formula covers vertical and horizontal springs — don’t look for a different one.
⚠ Common mistakes
Putting g into the mass–spring formula — there is no gravity term
Forgetting the square root, or rooting only m instead of the whole fraction m/k
When finding k, forgetting to square the period — use T², not T
Thinking doubling the mass doubles the period — it multiplies it by √2, not 2
Mixing up the spring formula with the pendulum formula T = 2π√(L/g)
Quick recap: A mass on a spring oscillates in SHM because the spring gives a Hooke’s-law restoring force (F = −kx). Its period is T = 2π√(m/k): a bigger mass lengthens it, a stiffer spring shortens it, and gravity has no effect at all. Rearranged, k = 4π²m/T².
You’ve now met the first of the two SHM systems the IB expects you to handle. The second is the simple pendulum, where the restoring force comes from gravity instead of a spring — and this time gdoes appear in the formula. That’s the next page: the time period of a simple pendulum.
Springs and periods still bouncing around?
Book a free meeting and we’ll work through the mass–spring formula, Hooke’s law and past-paper questions together.
