# A 0.500\ \mathrm{kg} mass attached to a horizontal spring is displaced 17.0\ \mathrm{cm} from...

## Question:

A {eq}0.500\ \mathrm{kg} {/eq} mass attached to a horizontal spring is displaced {eq}17.0\ \mathrm{cm} {/eq} from rest and then released. If the mass oscillates {eq}14.0 {/eq} times in {eq}17.7\ \mathrm{s} {/eq}, what is the spring constant in the spring?

a. {eq}24.7\ \mathrm{N/m} {/eq}

b. {eq}0.313\ \mathrm{N/m} {/eq}

c. {eq}0.0630\ \mathrm{N/m} {/eq}

d. {eq}12.3\ \mathrm{N/m} {/eq}

## Spring-mass System:

When the spring is stretched along its central axis, then the spring offers the elastic force against the deformation of the spring. For the simple harmonic motion of the spring-mass system, the stiffer spring will vibrate with a larger frequency of oscillation. If we attach the heavier mass to the spring, then the frequency of oscillation will decrease.

## Answer and Explanation: 1

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Given data:

• {eq}m=\rm 0.500 \ kg {/eq} is the mass attached to the spring
• {eq}f {/eq} is the frequency of oscillation
• {eq}k {/eq} is the spring...

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#### Learn more about this topic: Hooke's Law & the Spring Constant: Definition & Equation

from

Chapter 4 / Lesson 19
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After watching this video, you will be able to explain what Hooke's Law is and use the equation for Hooke's Law to solve problems. A short quiz will follow.