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 system follows.

Fig. 1

As set up, the mass is sitting at x=0 with positive being to the right. Velocity is positive to the right as well.

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:

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

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

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

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

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 the equivalent of x dot:

$\mathbf {X} (s)={\frac {sx_{0}+v_{0}}{s^{2}+{\frac {k}{m}}}}$

Now, let's split it into two parts:

$\mathbf {X} (s)={\frac {sx_{0}+v_{0}}{s^{2}+{\frac {k}{m}}}}~~~~~\Rightarrow ~~~~~\mathbf {X} (s)={\frac {sx_{0}}{s^{2}+{\frac {k}{m}}}}+{\frac {v_{0}}{s^{2}+{\frac {k}{m}}}}$

We know from Laplace transforms that:

${\mathcal {L}}_{s}\left\{sin(w_{0}t)\right\}={\frac {w_{0}}{s^{2}+w_{0}^{2}}}$

${\mathcal {L}}_{s}\left\{cos(w_{0}t)\right\}={\frac {s}{s^{2}+w_{0}^{2}}}$

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:

${\frac {k}{m}}=w_{0}^{2}$

Once we get the Laplace transform into the correct form, we have:

$\mathbf {X} (s)=x_{0}{\frac {s}{s^{2}+w_{0}^{2}}}+{\frac {v_{0}}{w_{0}}}{\frac {w_{0}}{s^{2}+w_{0}^{2}}}$

Now we take the inverse Laplace transform:

${\mathcal {L}}_{s}^{-1}\left\{x_{0}{\frac {s}{s^{2}+w_{0}^{2}}}\right\}=x_{0}cos(w_{0}t)$

${\mathcal {L}}_{s}^{-1}\left\{{\frac {v_{0}}{w_{0}}}{\frac {w_{0}}{s^{2}+w_{0}^{2}}}\right\}={\frac {v_{0}}{w_{0}}}sin(w_{0}t)$

Now that we are back in the time domain we just add the two parts together and we have:

$x(t)=x_{0}cos(w_{0}t)+{\frac {v_{0}}{w_{0}}}sin(w_{0}t)$

So we have found the equation that governs the motion of the spring mass system, all that is left is putting in constants for the variables to find numerical answers for any problem of this type.