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:

$\mathbf{F}=ma\Rightarrow {\mathbf{f}}_m(t)=m\ddot{x}$

By Hooke's law:

$\mathbf{F}=kx\Rightarrow {\mathbf{f}}_k(t)=mx$

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.

${\mathbf{f}}_m(t)+{\mathbf{f}}_k(t)=0\Rightarrow m\ddot{x}(t)+kx(t)=0$

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

$0={{ℒ}}_s\{m\ddot{x}+kx\}$

    $=m{{ℒ}}_s\left\{\ddot{x}\right\}+k{{ℒ}}_s\left\{x\right\}$

    $=m\{s[s\mathbf{X}(s)-x(0)]-\dot{x}(0)\}+k\mathbf{X}(s)$

    $=m{s}^2\mathbf{X}(s)-msx(0)-m\dot{x}(0)+k\mathbf{X}(s)$

Now that we have the Laplace transform of the differential equation that governs the motion of the spring and mass system, we need to solve for X(s).

$\mathbf{X}(s)=\frac{sx(0)+\dot{x}}{s^2+\frac{k}{m}}$

Using the idea that the initial position x=0 and velocity is just the derivative of position:

$\mathbf{X}(s)=\frac{s{x}_0+v_0}{s^2+\frac{k}{m}}$

Now, let's split it into two parts.

$\mathbf{X}(s)=\frac{s{x}_0+v_0}{s^2+\frac{k}{m}}\Rightarrow \mathbf{X}(s)=\frac{s{x}_0}{s^2+\frac{k}{m}}+\frac{v_0}{s^2+\frac{k}{m}}$

We know from Laplace transforms that:

${{ℒ}}_s\{sin(w_{0}t)\}=\frac{w_0}{s^2+w_0^2}$

${{ℒ}}_s\{cos(w_{0}t)\}=\frac{s}{s^2+w_0^2}$

From this we know that we are going to have two parts to our solution, and sine wave and a cosine wave. We can also tell that:

$\frac{k}{m}=w_0^2$