The big idea: An oscillator repeats the same motion over and over.
The period T is the time for one full oscillation (seconds).
The frequency f is how many oscillations happen each second (hertz, Hz). They are simply opposites: f = 1 ÷ T.
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).
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.
[Diagram: phys-formula-triangle] - Available in full study mode
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.
IB-style question — natural frequency of a mass-spring
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.
Solution
- It's a mass-spring, so use the given formula:
- Put in the numbers (m = 0.50, k = 200):
- Work out the inside, then the root:
- So the period is:
- Frequency is 1 ÷ T:
Final answer
T = 0.31 s and f = 3.2 Hz.
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How this is tested: Period/frequency questions are usually ratio problems — they change one quantity and ask for the new period or frequency.
- Paper 1A: a one-mark ratio — e.g. add a second spring (the combined stiffness becomes 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.
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).
IB-style question — second spring added in parallel
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.
Solution
- Use the given spring formula for both cases:
- The mass is unchanged; only k → 2k. Write the new period:
- Pull the factor of 2 out of the root — it divides T:
- So the period shrinks by √2:
Final answer
Tnew = T ÷ √2 ≈ 0.71 T — a stiffer spring (2k) oscillates faster, so the period is shorter.