# Given that a spring with a mass of 2 kg has damping constant 12, and a force of 8.75 N is...

## Question:

Given that a spring with a mass of 2 kg has damping constant 12, and a force of 8.75 N is required to keep the spring stretched 0.5 m beyond its natural length. The spring is stretched 2 m beyond its natural length and then released with zero velocity. Find the position of the mass at any time t.

x(t)?

## The Motion of a Spring:

A spring moves in an oscillatory manner where it compresses or stretched depending on the force applied to it and then stops when the energy of the motion is lost over time due to the existence of friction. The motion of a spring can be described by a differential equation where it includes the applied force, Hooke's law, and the damping if it exists.

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Given:

• Mass, {eq}m = 2 \,\rm{kg} {/eq}.
• Damping constant, {eq}c = 12 \,\rm{kg} {/eq}.
• Force, {eq}F(x=0.5) = 8.75 \,\rm{N} {/eq}.
• Initial... 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.