The big idea: Watch a child on a swing go back… and forth… back… and forth, the same trip over and over. That repeating to-and-fro is an oscillation.
The period T is the time for one full swing (seconds); the frequency f is how many happen each second (hertz, Hz). They are simply opposites: f = 1 ÷ T.
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Two oscillators you must know: Two standard systems oscillate with simple harmonic motion (SHM) — a back-and-forth motion where the pull always points back to the middle.
- A mass on a spring — its period depends on the mass m and the stiffness k. - A simple pendulum — its period depends on the length l and gravity g.
| Quantity | Symbol | What it means | Unit |
|---|---|---|---|
| period | T | time for one full oscillation | s |
| frequency | f | oscillations per second | Hz |
| angular frequency | ω | how fast the cycle turns (2π per cycle) | rad s⁻¹ |
What 'spring constant' means: The spring constant k is the spring's stiffness — a stiffer spring (bigger k) pulls back harder, so it oscillates faster (a shorter period).
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Each oscillator has its own given formula for the period. All three formulas below are in the data booklet — you don't memorise them, you choose the right one. Tap any to see its booklet badge.
- period — time for one full oscillation (s)
- mass on the spring (kg)
- spring constant — its stiffness (N m⁻¹)
- period — time for one full swing (s)
- length of the pendulum (m)
- gravitational field strength (m s⁻², 9.8 on Earth)
- frequency — oscillations per second (Hz)
- period — seconds for one oscillation (s)
- angular frequency (rad s⁻¹)
Frequency is just 1 ÷ period: f = 1 ÷ T — and the other way round, T = 1 ÷ f. Use the formula triangle: cover the one you want and read off the rest.
f = 1 ÷ T (and T = 1 ÷ f). Cover f to get 1 ÷ T; cover T to get 1 ÷ f.
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What surprises students: The mass of a pendulum does NOT affect its period (m is not in T = 2π√(l/g)).
The stiffness/gravity swap matters too: g is not in the spring formula. Pick the formula for the system in front of you and ignore the quantities that aren't in it.
A 0.50 kg mass hangs from a spring of spring constant k = 200 N m⁻¹. Find the period of its oscillations, then its frequency.
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How this is tested — period/frequency questions are usually ratio problems that change one quantity:
Paper 1A
- A one-mark ratio — add a second spring (combined stiffness 2k), or double a pendulum's length, and find the new period.
Paper 2
- Plug numbers into T = 2π√(m/k) or T = 2π√(l/g) for a natural frequency.
The classic trap: Both formulas have a square root, so multiplying a quantity by 4 only changes the period by √4 = 2, not by 4. And changing the mass of a pendulum changes nothing.
The square-root shortcut for ratios: Because T ∝ √m (spring) and T ∝ √l (pendulum), a factor inside the root comes out as its square root.
Length ×4 → period ×√4 = ×2. Gravity ÷9 → period ×√9 = ×3 (g is on the bottom, so smaller g means a bigger period).
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A mass on a single spring (constant k) has period T. An identical second spring is added beside the first, so the two together pull with a combined stiffness of 2k. Find the new period in terms of T.
Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.