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
This public explainer is separate from the protected practice bank. Current practice coverage varies.
Source and public-release record
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.
Review and maintenance
Source-checked by publisher- Publisher record
- Keiko Study editorial owner
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- Reviewer record
- Technical reviewer pending
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- Last source check
- 2026-08-07
- Next scheduled review
- 2026-11-07
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