Laplace transforms:Mass-Spring Oscillator

Problem Statement:

An ideal mass m sliding on a frictionless surface, attached via an ideal spring k to a rigid wall. The spring is at rest when the mass is centered at x=0. Find the equation of motion that the spring mass follows.

Solution:

By Newton's first law:

$\mathbf {F} =m{a}~~~~~\Rightarrow ~~~~~\mathbf {f} _{m}(t)=m{\ddot {x}}$

By Hooke's law:

$\mathbf {F} =k{x}~~~~~\Rightarrow ~~~~~\mathbf {f} _{k}(t)=mx$

By Newton's third law of motion that states every action produces an equal and opposite reaction, we have f_k = -f_m. That is, the force f_k applied by the mass to the spring is equal and opposite to the accelerating force f_m exerted in the -x direction by the spring on the mass.

$\mathbf {f} _{m}(t)+\mathbf {f} _{k}(t)=0~~~~~~\Rightarrow ~~~~~m{\ddot {x}}(t)+k{x}(t)=0$

We now have a second order differential equation that governs the motion of the mass. Taking the Laplace transform of both sides gives:

$0={\mathcal {L}}_{s}\left\{m{\ddot {x}}+k{x}\right\}$

    $=m{\mathcal {L}}_{s}\left\{{\ddot {x}}\right\}+k{\mathcal {L}}_{s}\left\{x\right\}$

    $=m\left\{s[s\mathbf {x} (s)-x(0)]-{\dot {x}}(0)\right\}$

\left\{s\left\[\right\]\right\}

\mathbf{x}\left\(s\right\) -x\left\(0\right\) -\dot{x}\left\(0\right\) +k\mathbf{x}(s\right\)</math>