Mass-Spring Systems
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 5 practice questions.
A mass on a spring has period T = 2π√(m/k). More mass, slower oscillation. Stiffer spring, faster oscillation. The amplitude and g don’t appear, so they don’t matter.
1. What sets the period
2. Why the square root matters
Because of the square root, doubling the mass doesn’t double the period: it multiplies T by √2 ≈ 1.41. To double the period, you need four times the mass.
Check your answer’s direction first: heavier is slower (longer T), stiffer is faster (shorter T).
A vertical spring stretches by mg/k before it oscillates. Gravity sets where the equilibrium is, not how fast the mass oscillates around it. On the Moon, the same spring and mass have the same period.
3. Example, then try it
A 0.50 kg mass is attached to a spring with k = 200 N/m. Find the period and frequency.
m/k = 0.50 ÷ 200 = 0.0025; √0.0025 = 0.050.
T = 2π(0.050) ≈ 0.31 s; f = 1/T ≈ 3.2 Hz.
A 2.0 kg mass hangs from a spring and stretches it 0.20 m. Find the spring constant and the period when the mass bobs up and down. (g = 10 m/s²) What would the period be on the Moon?
Spring constant: at equilibrium, kΔy = mg, so k = (2.0)(10) ÷ 0.20 = 100 N/m.
Period: T = 2π√(2.0/100) = 2π(0.141) ≈ 0.89 s.
On the Moon: the spring stretches less, but m and k are unchanged, so T is still 0.89 s.
Answer: k = 100 N/m; T ≈ 0.89 s on Earth and on the Moon.
4. Common mistakes
An astronaut takes a mass and spring to the Moon, where g is about one-sixth of Earth’s. How does the period of the bobbing mass compare with its period on Earth?
T = 2π√(m/k) depends only on the mass and the spring constant. Gravity changes how far the spring stretches, not the period.
Ready to practice?
5 questions on Mass-Spring Systems, each with a hint and an explanation for every wrong answer.