Spring Oscillation Period
Period of a mass on a spring.
Formula
Variables
Example
Heavier mass, slower oscillation
How it works
This formula gives the time for one complete back-and-forth cycle of a mass attached to an ideal spring. It comes from the fact that the spring’s restoring force follows Hooke’s law, so the motion is simple harmonic: the mass accelerates toward equilibrium more strongly when the spring is stretched or compressed more. The period depends on the mass and spring stiffness only; a larger mass takes longer to reverse direction, while a stiffer spring pulls the mass back faster.
When to use it
Use this for ideal mass-spring systems in physics problems, lab experiments, and models of vibrating objects that behave like a spring-mass oscillator. It is especially useful when you need the oscillation period, frequency, or how changes in mass or spring constant affect the motion.
Worked example, step by step
Given: m = 0.50 kg, k = 200 N/m Step 1: Write the formula: T = 2\pi\sqrt{m/k} Step 2: Substitute the values: T = 2\pi\sqrt{0.50/200} Step 3: Simplify inside the square root: T = 2\pi\sqrt{0.0025} Step 4: Take the square root: T = 2\pi(0.05) Step 5: Multiply: T = 0.10\pi s Step 6: Final answer: T \approx 0.314 s
Common mistakes to avoid
- Using kilograms for mass but forgetting that k must be in newtons per meter
- Mixing up period and frequency; period is time per cycle, not cycles per second
- Forgetting the square root, which makes the result far too large
Did You Know?
A pendulum's period does not depend on how heavy the bob is, a fact Galileo reportedly noticed watching a swinging lamp.
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