# What is the spring constant for a mass spring system if the mass is 0.55 kg, and the frequency is...

## Question:

What is the spring constant for a mass spring system if the mass is 0.55 kg, and the frequency is 4.6 Hz?

## Spring Constant:

Suppose we have two compression coil springs, and we are bringing the same amount of distortion in each spring by pressing the spring between hands. If a significantly large force is required to produce the same amount of distortion in one of the springs, we can apparently say that the spring constant of that spring is bigger than the other spring. Mathematically:

$$\rm Spring \ constant=\dfrac{applied \ force}{deformation \ of \ the \ spring}$$

Given data:

• {eq}m=\rm 0.55 \ kg {/eq} is the mass attached to the spring.
• {eq}f=\rm 4.6 \ Hz {/eq} is the frequency of oscillation.
• {eq}k {/eq} is the spring constant of the spring.

The frequency of the spring-mass system is:

{eq}f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} {/eq}

Squaring on both sides, we get:

{eq}\begin{align} f^2&=\dfrac{1}{4\pi^2} \times \dfrac{k}{m} \\[0.3cm] k&=4\pi^2f^2m \end{align} {/eq}

Plugging in the given values, we get:

{eq}\begin{align} k&=\rm 4\pi^2 \times (4.6^2 \ Hz^2) \times ( 0.55 \ kg) \\[0.3cm] &\simeq \color{blue}{\boxed { \rm 460 \ N/m}} \ \ \ \ \ \rm (correct \ to \ two \ significant \ figures ) \end{align} {/eq}

Therefore, the spring constant of the spring is {eq}\color{blue}{\boxed { \rm 460 \ N/m}} {/eq}.

Practice Applying Spring Constant Formulas

from

Chapter 17 / Lesson 11
3.4K

In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.