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title: "A block of mass \\(m\\) is suspended vertically from an ideal spring of force constant \\(k\\) in a uniform downward gravitational field \\(g\\). The block oscillates vertically with amplitude \\(A\\) about its static equilibrium position \\(y_{eq}\\) between an upper turning point at \\(y_{eq} – A\\) and a lower turning point at \\(y_{eq} + A\\). Although the gravitational force acts continuously downward throughout the oscillation, precise measurements confirm that the time required for the block to travel downward from the upper turning point to the lower turning point is strictly equal to the time required to travel upward from the lower turning point to the upper turning point. Which of the following explanations correctly accounts for why the unidirectional downward force of gravity does not cause the duration of the downward stroke to differ from the duration of the upward stroke?"
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url: "https://nerd-notes.com/ubq/124683/"
date_modified: "2026-09-28T14:09:04+00:00"
---

# A block of mass \(m\) is suspended vertically from an ideal spring of force constant \(k\) in a uniform downward gravitational field \(g\). The block oscillates vertically with amplitude \(A\) about its static equilibrium position \(y_{eq}\) between an upper turning point at \(y_{eq} – A\) and a lower turning point at \(y_{eq} + A\). Although the gravitational force acts continuously downward throughout the oscillation, precise measurements confirm that the time required for the block to travel downward from the upper turning point to the lower turning point is strictly equal to the time required to travel upward from the lower turning point to the upper turning point. Which of the following explanations correctly accounts for why the unidirectional downward force of gravity does not cause the duration of the downward stroke to differ from the duration of the upward stroke?

A block of mass \(m\) is suspended vertically from an ideal spring of force constant \(k\) in a uniform downward gravitational field \(g\). The block oscillates vertically with amplitude \(A\) about its static equilibrium position \(y_{eq}\) between an upper turning point at \(y_{eq} - A\) and a lower turning point at \(y_{eq} + A\). Although the gravitational force acts continuously downward throughout the oscillation, precise measurements confirm that the time required for the block to travel downward from the upper turning point to the lower turning point is strictly equal to the time required to travel upward from the lower turning point to the upper turning point. Which of the following explanations correctly accounts for why the unidirectional downward force of gravity does not cause the duration of the downward stroke to differ from the duration of the upward stroke?

![A vertical diagram showing an ideal spring and block suspended from a rigid horizontal ceiling. At the top, a horizontal ceiling is drawn with hatched lines above it. Suspended from the ceiling is a vertical helical spring attached at its bottom end to a rectangular block of mass \(m\). To the right of the spring-mass system, a vertical dashed reference axis is oriented downward. Three horizontal dashed tick marks on this axis indicate positions: the uppermost tick mark is labeled \(y_{eq} - A\), the middle tick mark is labeled \(y_{eq}\), and the lowest tick mark is labeled \(y_{eq} + A\). A downward arrow next to the axis is labeled \(g\). The block is shown centered at the middle tick mark \(y_{eq}\). No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790604544-XXmyII.jpg)

- **A.** The greater elastic potential energy stored at the lowest turning point produces a larger net restoring acceleration at the bottom than at the top, which compensates for the downward gravitational force that opposes the block's upward motion.
- **B.** The constant gravitational force shifts the static equilibrium position downward without altering the linear dependence of the net restoring force on displacement, resulting in symmetric acceleration and deceleration phases of identical magnitude on both strokes.
- **C.** The net impulse delivered to the block over the downward stroke must be zero because the block starts and ends at rest, which mathematically requires the duration of the downward stroke to equal the duration of the upward stroke.
- **D.** The spring releases more elastic potential energy during the upward stroke than it absorbs during the downward stroke, creating a higher maximum speed on the ascent that precisely offsets the time lost traveling against gravity.

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/124683/*