Laplace transforms:Mass-Spring Oscillator

Problem Statement

Part 1

An ideal mass m=10kg is sitting on a plane, attached to a rigid surface via a spring. The spring with k=10.8 kN/m is exerting zero force when the mass is centered at x=0. Between the mass and plane there is a 1 mm layer of a viscous fluid and the block has an area of .1m^2 in contact the plane. Find the equation of motion that the spring mass system follows if there is an initial impulse applied, and then find the viscosity value of the fluid so that the mass comes to a stop within three seconds with the initial impulse used. Below is a picture/FBR of the system.

Fig. 1

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

Part 2

Apply the Initial Value and Final Value Theorems to this problem.

Part 3

Make a bode plot.

Solution

Part 1

Using the FBR from the problem statement we can set up our differential equation:

$m{\ddot {x}}+b{\dot {x}}+kx=\delta (t)$

${\ddot {x}}+{\frac {b}{m}}{\dot {x}}+{\frac {k}{m}}x={\frac {\delta (t)}{m}}$

We now have a second order differential equation that governs the motion of the mass. If we make our initial conditions equal to zero, we save a few steps. Taking the Laplace transform of both sides gives:

${\mathcal {L}}_{s}\left\{{\frac {\delta (t)}{m}}\right\}={\mathcal {L}}_{s}\left\{{\ddot {x}}+{\frac {b}{m}}{\dot {x}}+{\frac {k}{m}}x\right\}$

${\frac {1}{m}}{\mathcal {L}}_{s}\left\{\delta (t)\right\}={\mathcal {L}}_{s}\left\{{\ddot {x}}\right\}+{\frac {b}{m}}{\mathcal {L}}_{s}\left\{{\dot {x}}\right\}+{\frac {k}{m}}{\mathcal {L}}_{s}\left\{x\right\}$

${\frac {1}{m}}\left\{1\right\}=\mathbf {X} (s)s^{2}+{\frac {b}{m}}\mathbf {X} (s)s+{\frac {k}{m}}\mathbf {X} (s)$

${\frac {1}{m}}=\left\{s^{2}+{\frac {b}{m}}s+{\frac {k}{m}}\right\}\mathbf {X} (s)$

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):

$\mathbf {X} (s)={\frac {\frac {1}{m}}{\left\{s^{2}+{\frac {b}{m}}s+{\frac {k}{m}}\right\}}}$

We now have the function in terms of X(s). We will use the above equation to help in part 3, but for now we want to continue with part 1. We now want to find the value of b so that the mass stops oscillating after t=3s. However, we know that it will never actually be zero until a long time after, so for now we will consider motion >2% of original to be stopped. The below formula helps us find the value of b:

$t_{s}={\frac {4.6}{\zeta w_{n}}}$

${\frac {k}{m}}=w_{n}^{2}=110.1$

$\zeta ={\frac {4.6}{(3)(10.)}}=0.146$

Now using the relationship below, we are able to compute the value of b:

${\frac {b}{m}}=2\zeta w_{n}$

$b=\left\{(2)(.146)(10.5)(98.1)\right\}=300.775$

When we plug this number into the Laplace transform equation along with the values given:

$\mathbf {X} (s)={\frac {\frac {1}{98.1}}{\left\{s^{2}+{\frac {300.775}{98.1}}s+{\frac {10800}{98.1}}\right\}}}$

$\mathbf {X} (s)={\frac {0.010194}{\left\{s^{2}+3.066s+110.1\right\}}}$

The next step of taking the inverse Laplace transform is pretty complicated since we have complex roots, so I will just show the equation after it has been put back through:

$x(t)=0.010194cos(10.38027t)e^{-1.533t}\,$

So we have found the equation that the mass follows with the value of b that brings the mass to a "stop" within three seconds. But now we need to find the kinematic viscosity that goes with that b value.

The equation below is one of the standard shear stress equations:

$\tau =\mu {\frac {\dot {x}}{t}}$.....where $t$ is the film thickness and $\mu$ is the kinematic viscosity.

We know that force is just a certain stress applied to an area, so:

$b{\dot {x}}=\mu A{\frac {\dot {x}}{t}}$

$b=\mu {\frac {A}{t}}$

Now we just plug in the values that are given and what we have found and we get:

$300.775=\mu {\frac {.1}{.001}}$

$\mu =3.008N{\frac {s}{m^{2}}}$

Part 2

The Initial Value Theorem states:

$\lim _{s\to \infty }s\mathbf {X} (s)=x(o)$

So if we plug our equation into this we have:

$\lim _{s\to \infty }s\left\{{\frac {0.010194}{\left\{s^{2}+3.066s+110.1\right\}}}\right\}=x(o)$

$\lim _{s\to \infty }\left\{{\frac {0.010194s}{\left\{s^{2}+3.066s+110.1\right\}}}\right\}=x(o)$

$\left\{0\right\}=x(o)$

As expected, the Initial Value Theorem shows us that the initial value is just the initial x value.

The Final Value Theorem States:

$x(\infty )=\lim _{s\to 0}s\mathbf {X} (s)$

$x(\infty )=\lim _{s\to 0}s\left\{{\frac {0.010194}{\left\{s^{2}+3.066s+110.1\right\}}}\right\}$

$x(\infty )=\lim _{s\to 0}\left\{{\frac {0.010194s}{\left\{s^{2}+3.066s+110.1\right\}}}\right\}$

$s\infty )=\left\{0\right\}$

As expected, the final value will be zero because the mass will come back to rest in the position from which it began at x=0.

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Written by: David Steinweg

Checked by: Nathan Reeves