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title: "A block of mass \\(m\\) on a frictionless horizontal surface is attached to a uniform ideal spring of force constant \\(k\\) and oscillates with period \\(T_0\\). The spring is then cut into two identical halves, and the same block is attached to one of the halves. When displaced and released, the block oscillates with a new period of \\(T’ = \\dfrac{T_0}{\\sqrt{2}}\\). Which of the following statements best explains why the period decreases to this value?"
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url: "https://nerd-notes.com/ubq/122686/"
date_modified: "2026-09-28T11:02:40+00:00"
---

# A block of mass \(m\) on a frictionless horizontal surface is attached to a uniform ideal spring of force constant \(k\) and oscillates with period \(T_0\). The spring is then cut into two identical halves, and the same block is attached to one of the halves. When displaced and released, the block oscillates with a new period of \(T’ = \dfrac{T_0}{\sqrt{2}}\). Which of the following statements best explains why the period decreases to this value?

A block of mass \(m\) on a frictionless horizontal surface is attached to a uniform ideal spring of force constant \(k\) and oscillates with period \(T_0\). The spring is then cut into two identical halves, and the same block is attached to one of the halves. When displaced and released, the block oscillates with a new period of \(T' = \dfrac{T_0}{\sqrt{2}}\). Which of the following statements best explains why the period decreases to this value?

![Two horizontal mass-spring oscillator setups shown in a vertical stack, labeled Setup 1 and Setup 2. In Setup 1 at the top, a vertical wall on the left is attached to a horizontal coil spring of full length labeled with spring constant k, which connects to a rectangular block of mass m resting on a flat horizontal surface. A label below Setup 1 reads Period = T_0. In Setup 2 directly below, a vertical wall on the left is attached to a shorter coil spring with half as many coils, which connects to the same rectangular block of mass m on the same horizontal surface. A label below Setup 2 reads Period = T prime. No other labels, lines, text, or axes appear.](https://nerd-notes.com/wp-content/uploads/ubq-frq-generated/stem-fig-1-1790593360-Mxv2WW.jpg)

- **A.** The spring constant is an intrinsic property of the metal that remains \(k\), but halving the spring's physical length halves the distance the block must travel in each cycle, reducing the round-trip travel time to \(\dfrac{T_0}{\sqrt{2}}\).
- **B.** Cutting the spring in half reduces the total number of elastic coils resisting the motion, which halves the effective spring constant to \(\dfrac{k}{2}\) and decreases the period because weaker springs oscillate faster according to the inverse relationship between force and acceleration.
- **C.** The spring constant is an intrinsic property of the material that remains \(k\), but removing half of the spring's mass decreases the total oscillating inertia of the system by a factor of 2, which reduces the period to \(\dfrac{T_0}{\sqrt{2}}\).
- **D.** Compressing or stretching the half-spring by a given displacement requires each remaining coil to deform twice as much as in the original spring, doubling the effective spring constant to \(2k\) and reducing the period by a factor of \(\sqrt{2}\).

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/122686/*