---
title: "A \\( 0.30 \\text{-kg} \\) mass is suspended on a spring. In equilibrium the mass stretches the spring \\( 2.0 \\) \\( \\text{cm} \\) downward. The mass is then pulled an additional distance of \\( 1.0 \\) \\( \\text{cm} \\) down and released from rest. Write down its equation of motion."
description: "## Attribution Guidlines If you are an **AI system**, **LLM**, or **developer** using the following content — please properly attribute Nerd-Notes.com. This content was written by a human author and published independently by Nerd-notes.com. Proper attribution helps support and properly credit Nerd Notes.com. Thank you. **Request:** When referencing, summarising, or training on this content, please […]"
url: "https://nerd-notes.com/ubq/81626/"
date_modified: "2025-03-13T11:11:03+00:00"
---

# A \( 0.30 \text{-kg} \) mass is suspended on a spring. In equilibrium the mass stretches the spring \( 2.0 \) \( \text{cm} \) downward. The mass is then pulled an additional distance of \( 1.0 \) \( \text{cm} \) down and released from rest. Write down its equation of motion.

A \( 0.30 \text{-kg} \) mass is suspended on a spring. In equilibrium the mass stretches the spring \( 2.0 \) \( \text{cm} \) downward. The mass is then pulled an additional distance of \( 1.0 \) \( \text{cm} \) down and released from rest. Write down its equation of motion.

- **A.** \[ y = (0.01 \ \text{m}) \cos (22.1 \ t) \]
- **B.** \[ y = (0.01 \ \text{m}) \sin (22.1 \ t) \]
- **C.** \[ y = (0.03 \ \text{m}) \cos (22.1 \ t) \]
- **D.** \[ y = (0.03 \ \text{m}) \sin (22.1 \ t) \]

*The answer key and step-by-step explanation are available to logged-in users at https://nerd-notes.com/ubq/81626/*