Question
A block attached to an ideal spring with spring constant oscillates on a frictionless horizontal surface. What is the period of the oscillation?
- a.
- b.
- c.
- d.
Correct answer
C.
Full reasoning
- 1Compute the ratio under the root: .
- 2Take the square root: , which is the reciprocal of the angular frequency .
- 3Multiply by for a full cycle: .
- 4Amplitude appears nowhere in the formula, so the period is amplitude-independent.
Why each choice is right or wrong
Choice A
This is alone; the factor of that converts the angular timescale to a full cycle was dropped.
Choice B
This is the half period — the time from one turning point to the opposite turning point — not a full cycle.
Choice C
Correct. for one complete oscillation.
Choice D
This is the frequency, about , mislabeled as a period instead of taking its reciprocal.
Related formula
Period of a mass-spring oscillator
Assumptions: The spring is ideal (massless and linear, obeying Hooke's law). No damping acts on the oscillator.
Related topic and practice
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Sources
Authored for this public explainer registry. It has no database question ID and is not copied from a protected or official exam bank.
- University Physics Volume 1, Section 15.1: Simple Harmonic Motion — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.
Sources and corrections
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