# A mass of 50 g is suspended from a spring with a spring constant k = 8.0 N/m. Determine the force...

## Question:

A mass of 50 g is suspended from a spring with a spring constant k = 8.0 N/m. Determine the force acting on the mass when it is displaced 8 cm from the equilibrium position. If the mass is then released, what is the period of oscillation? What is the frequency?

## The Spring-Mass System

The spring-mass system is a very good example fora simple harmonic system.

The frequency of oscillation of a mass attached to a spring is given by,

{eq}\nu=\frac{1}{2\pi}\sqrt{\frac{k}{m}} {/eq}

Where,

- {eq}k {/eq} is the force constant and

- {eq}m {/eq} is the mass attached.

The restoring force on the mass is given by,

{eq}F=kx {/eq}

This is called Hooke's law.

## Answer and Explanation:

Given:

{eq}m=0.05\ kg\\ k=8.0\ N/m\\ x=0.08\ m {/eq}

The force on the mass is,

{eq}F=kx=8.0\times0.08\ N=0.64\ N {/eq}

The frequency of oscillation is,

{eq}\displaystyle \nu=\frac{1}{2\pi}\sqrt{\frac{k}{m}}=\frac{1}{2\pi}\sqrt{\frac{8.0}{0.05}}\ Hz=2.0\ Hz {/eq}

The time period is,

{eq}\displaystyle T=\frac{1}{\nu}=\frac{1}{2.0}\ s=0.50\ s {/eq}

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from UExcel Physics: Study Guide & Test Prep

Chapter 4 / Lesson 19