Period & Frequency
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
For a mass on a spring, T = 2π√(m/k), and f = 1/T. The period depends only on the mass and the spring constant: not on the amplitude, and not on whether the spring is horizontal or vertical.
1. What sets the period
2. Scaling and combinations
T ∝ √m: four times the mass, twice the period. T ∝ 1/√k: four times as stiff, half the period.
Two identical springs in parallel act like 2k, so T drops by √2. In series they act like k/2, so T grows by √2.
Linearizing: to find k in the lab, plot T² against m. The slope is 4π²/k, so k = 4π²/slope. This is a classic AP lab question.
3. Example, then try it
A 0.80 kg block on a spring with k = 320 N/m oscillates on a frictionless surface. Find its period and frequency.
T = 2π√(0.80/320) = 2π(0.050) ≈ 0.31 s; f = 1/T ≈ 3.2 Hz.
In the lab data below, the best-fit line of T² against m has slope 0.20 s²/kg. Find k. Then predict the period with a 0.50 kg mass, and with that mass hanging from two such springs side by side.
Spring constant: slope = 4π²/k, so k = 4π² ÷ 0.20 ≈ 200 N/m.
0.50 kg: T = 2π√(0.50/200) = 2π(0.050) ≈ 0.31 s. (Or from the graph: T² = 0.10 s².)
Two springs in parallel: k = 400 N/m, so T = 2π√(0.50/400) ≈ 0.22 s, smaller by √2.
Answer: k ≈ 200 N/m; 0.31 s; 0.22 s with two springs side by side.
4. Common mistakes
A block on one spring has period T. The same block is hung from two identical springs connected end to end (in series). What is the new period?
Two springs in series act like one spring with k/2. T ∝ 1/√k, so the period grows by √2.
Ready to practice?
5 questions on Period & Frequency, each with a hint and an explanation for every wrong answer.