Differential Equations - Second Order: Mechanical Vibrations - ii

Example 3 Take the spring and mass system from the first example and this time let’s attach a damper to it that will exert a force of 17 lbs when the velocity is 2 ft/s. Find the displacement at any time $t$, $u\left(t\right)$.

So, the only difference between this example and the previous example is damping force. So, let’s find the damping coefficient
$$17=\gamma \left(2\right)\Rightarrow \gamma =\frac{17}{2}=8.5>{\gamma }_{CR}$$So, it looks like we’ve got over damping this time around so we should expect to get two real distinct roots from the characteristic equation and they should both be negative. The IVP for this example is,
$$\frac{1}{2}{u}^{\prime \prime }+\frac{17}{2}{u}^{\prime }+18u=0u\left(0\right)=-\frac{1}{2}{u}^{\prime }\left(0\right)=1$$This one’s a little messier than the previous example so we’ll do a couple of the steps, leaving it to you to fill in the blanks. The roots of the characteristic equation are
$$r_{1,2}=\frac{-17\pm \sqrt{145}}{2}=-2.4792,-14.5208$$In this case it will be easier to just convert to decimals and go that route. Note that, as predicted we got two real, distinct and negative roots. The general and actual solution for this example are then,
$$\begin{array}{cc}u\left(t\right)& =c_1{e}^{-2.4792t}+c_2{e}^{-14.5208t}\\ u\left(t\right)& =-0.5198e^{-2.4792t}+0.0199e^{-14.5208t}\end{array}$$Here’s a sketch of the displacement for this example.

Notice an interesting thing here about the displacement here. Even though we are “over” damped in this case, it actually takes longer for the vibration to die out than in the critical damping case. Sometimes this happens, although it will not always be the case that over damping will allow the vibration to continue longer than the critical damping case.

Also notice that, as with the critical damping case, we don’t get a vibration in the sense that we usually think of them. Again, the damping is strong enough to force the vibration do die out quick enough so that we don’t see much, if any, of the oscillation that we typically associate with vibrations.

Let’s take a look at one more example before moving on the next type of vibrations.

Example 4 Take the spring and mass system from the first example and for this example let’s attach a damper to it that will exert a force of 5 lbs when the velocity is 2 ft/s. Find the displacement at any time $t$, $u\left(t\right)$.

So, let’s get the damping coefficient.
$$5=\gamma \left(2\right)\Rightarrow \gamma =\frac{5}{2}=2.5<{\gamma }_{CR}$$So, it’s under damping this time. That shouldn’t be too surprising given the first two examples. The IVP for this example is,
$$\frac{1}{2}{u}^{\prime \prime }+\frac{5}{2}{u}^{\prime }+18u=0u\left(0\right)=-\frac{1}{2}{u}^{\prime }\left(0\right)=1$$In this case the roots of the characteristic equation are
$$r_{1,2}=\frac{-5\pm \sqrt{119}i}{2}$$They are complex as we expected to get since we are in the under damped case. The general solution and actual solution are
$$\begin{array}{cc}u\left(t\right)& =e^{-\frac{5t}{2}}\left(c_1\cos \left(\frac{\sqrt{119}}{2}t\right)+c_2\sin \left(\frac{\sqrt{119}}{2}t\right)\right)\\ u\left(t\right)& =e^{-\frac{5t}{2}}\left(-0.5\cos \left(\frac{\sqrt{119}}{2}t\right)-0.04583\sin \left(\frac{\sqrt{119}}{2}t\right)\right)\end{array}$$Let’s convert this to a single cosine as we did in the undamped case.
$$\begin{array}{cc}R& =\sqrt{{\left(-0.5\right)}^2+{\left(-0.04583\right)}^2}=0.502096\\ {\delta }_1& ={\tan }^{-1}\left(\frac{-0.04583}{-0.5}\right)=0.09051\text{OR}{\delta }_2={\delta }_1+\pi =3.2321\end{array}$$As with the undamped case we can use the coefficients of the cosine and the sine to determine which phase shift that we should use. The coefficient of the cosine ($c_1$) is negative and so $\cos \delta$ must also be negative. Likewise, the coefficient of the sine ($c_2$) is also negative and so $\sin \delta$ must also be negative. This means that $\delta$ must be in the Quadrant III and so the second angle is the one that we want.
The displacement is then
$$u\left(t\right)=0.502096e^{-\frac{5t}{2}}\cos \left(\frac{\sqrt{119}}{2}t-3.2321\right)$$Here is a sketch of this displacement.

In this case we finally got what we usually consider to be a true vibration. In fact, that is the point of critical damping. As we increase the damping coefficient, the critical damping coefficient will be the first one in which a true oscillation in the displacement will not occur. For all values of the damping coefficient larger than this (i.e. over damping) we will also not see a true oscillation in the displacement.

From a physical standpoint critical (and over) damping is usually preferred to under damping. Think of the shock absorbers in your car. When you hit a bump you don’t want to spend the next few minutes bouncing up and down while the vibration set up by the bump die out. You would like there to be as little movement as possible. In other words, you will want to set up the shock absorbers in your car so get at the least critical damping so that you can avoid the oscillations that will arise from an under damped case.

It’s now time to look at systems in which we allow other external forces to act on the object in the system.

Undamped, Forced Vibrations

We will first take a look at the undamped case. The differential equation in this case is

$$m{u}^{\prime \prime }+ku=F\left(t\right)$$

This is just a nonhomogeneous differential equation and we know how to solve these. The general solution will be

$$u\left(t\right)=u_c\left(t\right)+U_P\left(t\right)$$

where the complementary solution is the solution to the free, undamped vibration case. To get the particular solution we can use either undetermined coefficients or variation of parameters depending on which we find easier for a given forcing function.

There is a particular type of forcing function that we should take a look at since it leads to some interesting results. Let’s suppose that the forcing function is a simple periodic function of the form

$$F\left(t\right)=F_0\cos \left(\omega t\right)\text{OR}F\left(t\right)=F_0\sin \left(\omega t\right)$$

For the purposes of this discussion we’ll use the first one. Using this, the IVP becomes,

$$m{u}^{\prime \prime }+ku=F_0\cos \left(\omega t\right)$$

The complementary solution, as pointed out above, is just

$$u_c\left(t\right)=c_1\cos \left({\omega }_{0}t\right)+c_2\sin \left({\omega }_{0}t\right)$$

where ${\omega }_0$ is the natural frequency.

We will need to be careful in finding a particular solution. The reason for this will be clear if we use undetermined coefficients. With undetermined coefficients our guess for the form of the particular solution would be,

$$U_P\left(t\right)=A\cos \left(\omega t\right)+B\sin \left(\omega t\right)$$

Now, this guess will be problems if ${\omega }_0=\omega$. If this were to happen the guess for the particular solution is exactly the complementary solution and so we’d need to add in a $t$. Of course, if we don’t have ${\omega }_0=\omega$ then there will be nothing wrong with the guess.

So, we will need to look at this in two cases.

1. ${\omega }_0\ne \omega$ In this case our initial guess is okay since it won’t be the complementary solution. Upon differentiating the guess and plugging it into the differential equation and simplifying we get,
   $$\left(-m{\omega }^{2}A+kA\right)\cos \left(wt\right)+\left(-m{\omega }^{2}B+kB\right)\sin \left(wt\right)=F_0\cos \left(wt\right)$$Setting coefficients equal gives us,
   $$\begin{array}{cccccc}& \cos \left(\omega t\right):& \left(-m{\omega }^2+k\right)A& =F_0& \Rightarrow A& =\frac{F_0}{k-m{\omega }^2}\\ & \sin \left(\omega t\right):& \left(-m{\omega }^2+k\right)B& =0& \Rightarrow B& =0\end{array}$$The particular solution is then
   $$\begin{array}{cc}{U}_P\left(t\right)& =\frac{F_0}{k-m{\omega }^2}\cos \left(\omega t\right)\\ & =\frac{F_0}{m\left(\frac{k}{m}-{\omega }^2\right)}\cos \left(\omega t\right)\\ & =\frac{F_0}{m\left({\omega }_0^2-{\omega }^2\right)}\cos \left(\omega t\right)\end{array}$$Note that we rearranged things a little. Depending on the form that you’d like the displacement to be in we can have either of the following.
   $$\begin{array}{cc}u\left(t\right)& =c_1\cos \left({\omega }_{0}t\right)+c_2\sin \left({\omega }_{0}t\right)+\frac{F_0}{m\left({\omega }_0^2-{\omega }^2\right)}\cos \left(\omega t\right)\\ u\left(t\right)& =R\cos \left({\omega }_{0}t-\delta \right)+\frac{F_0}{m\left({\omega }_0^2-{\omega }^2\right)}\cos \left(\omega t\right)\end{array}$$If we used the sine form of the forcing function we could get a similar formula.
2. ${\omega }_0=\omega$In this case we will need to add in a $t$ to the guess for the particular solution.
   $$U_P\left(t\right)=At\cos \left({\omega }_{0}t\right)+Bt\sin \left({\omega }_{0}t\right)$$Note that we went ahead and acknowledge that ${\omega }_0=\omega$ in our guess. Acknowledging this will help with some simplification that we’ll need to do later on. Differentiating our guess, plugging it into the differential equation and simplifying gives us the following.
   $$\begin{array}{cc}& \left(-m{\omega }_0^2+k\right)At\cos \left(wt\right)+\left(-m{\omega }_0^2+k\right)Bt\sin \left(wt\right)+\\ & 2m{\omega }_{0}B\cos \left(\omega t\right)-2m{\omega }_{0}A\sin \left(wt\right)=F_0\cos \left(wt\right)\end{array}$$Before setting coefficients equal, let’s remember the definition of the natural frequency and note that
   $$-m{\omega }_0^2+k=-m{\left(\sqrt{\frac{k}{m}}\right)}^2+k=-m\left(\frac{k}{m}\right)+k=0$$So, the first two terms actually drop out (which is a very good thing…) and this gives us,
   $$2m{\omega }_{0}B\cos \left(\omega t\right)-2m{\omega }_{0}A\sin \left(wt\right)=F_0\cos \left(wt\right)$$Now let’s set coefficient equal.
   $$\begin{array}{cccccc}& \cos \left(\omega t\right):& 2m{\omega }_{0}B& =F_0& \Rightarrow B& =\frac{F_0}{2m{\omega }_0}\\ & \sin \left(\omega t\right):& 2m{\omega }_{0}A& =0& \Rightarrow A& =0\end{array}$$In this case the particular will be,
   $$U_P\left(t\right)=\frac{F_0}{2m{\omega }_0}t\sin \left({\omega }_{0}t\right)$$The displacement for this case is then
   $$\begin{array}{cc}u\left(t\right)& =c_1\cos \left({\omega }_{0}t\right)+c_2\sin \left({\omega }_{0}t\right)+\frac{F_0}{2m{\omega }_0}t\sin \left({\omega }_{0}t\right)\\ u\left(t\right)& =R\cos \left({\omega }_{0}t-\delta \right)+\frac{F_0}{2m{\omega }_0}t\sin \left({\omega }_{0}t\right)\end{array}$$depending on the form that you prefer for the displacement.

So, what was the point of the two cases here? Well in the first case, ${\omega }_0\ne \omega$ our displacement function consists of two cosines and is nice and well behaved for all time.

In contrast, the second case, ${\omega }_0=\omega$ will have some serious issues at $t$ increases. The addition of the $t$ in the particular solution will mean that we are going to see an oscillation that grows in amplitude as $t$ increases. This case is called resonance and we would generally like to avoid this at all costs.

In this case resonance arose by assuming that the forcing function was,

$$F\left(t\right)=F_0\cos \left({\omega }_{0}t\right)$$

We would also have the possibility of resonance if we assumed a forcing function of the form.

$$F\left(t\right)=F_0\sin \left({\omega }_{0}t\right)$$

We should also take care to not assume that a forcing function will be in one of these two forms. Forcing functions can come in a wide variety of forms. If we do run into a forcing function different from the one that used here you will have to go through undetermined coefficients or variation of parameters to determine the particular solution.

Example 5 A 3 kg object is attached to spring and will stretch the spring 392 mm by itself. There is no damping in the system and a forcing function of the form$$F\left(t\right)=10\cos \left(\omega t\right)$$is attached to the object and the system will experience resonance. If the object is initially displaced 20 cm downward from its equilibrium position and given a velocity of 10 cm/sec upward find the displacement at any time $t$.

Since we are in the metric system we won’t need to find mass as it’s been given to us. Also, for all calculations we’ll be converting all lengths over to meters.
The first thing we need to do is find $k$.
$$k=\frac{mg}{L}=\frac{\left(3\right)\left(9.8\right)}{0.392}=75$$Now, we are told that the system experiences resonance so let’s go ahead and get the natural frequency so we can completely set up the IVP.
$${\omega }_0=\sqrt{\frac{k}{m}}=\sqrt{\frac{75}{3}}=5$$The IVP for this is then
$$3u^{\prime \prime }+75u=10\cos \left(5t\right)u\left(0\right)=0.2u^{\prime }\left(0\right)=-0.1$$Solution wise there isn’t a whole lot to do here. The complementary solution is the free undamped solution which is easy to get and for the particular solution we can just use the formula that we derived above.
The general solution is then,
$$\begin{array}{cc}u\left(t\right)& =c_1\cos \left(5t\right)+c_2\sin \left(5t\right)+\frac{10}{2\left(3\right)\left(5\right)}t\sin \left(5t\right)\\ u\left(t\right)& =c_1\cos \left(5t\right)+c_2\sin \left(5t\right)+\frac{1}{3}t\sin \left(5t\right)\end{array}$$Applying the initial conditions gives the displacement at any time $t$. We’ll leave the details to you to check.
$$u\left(t\right)=\frac{1}{5}\cos \left(5t\right)-\frac{1}{50}\sin \left(5t\right)+\frac{1}{3}t\sin \left(5t\right)$$The last thing that we’ll do is combine the first two terms into a single cosine.
$$\begin{array}{cc}R& =\sqrt{{\left(\frac{1}{5}\right)}^2+{\left(-\frac{1}{50}\right)}^2}=0.200998\\ {\delta }_1& ={\tan }^{-1}\left(\frac{-{}^1/{}_{50}}{{}^1/{}_5}\right)=-0.099669{\delta }_2={\delta }_1+\pi =3.041924\end{array}$$In this case the coefficient of the cosine is positive and the coefficient of the sine is negative. This forces $\cos \delta$ to be positive and $\sin \delta$ to be negative. This means that the phase shift needs to be in Quadrant IV and so the first one is the correct phase shift this time.
The displacement then becomes,
$$u\left(t\right)=0.200998\cos \left(5t+0.099669\right)+\frac{1}{3}t\sin \left(5t\right)$$Here is a sketch of the displacement for this example.

It’s now time to look at the final vibration case.

Forced, Damped Vibrations

This is the full blown case where we consider every last possible force that can act upon the system. The differential equation for this case is,

$$m{u}^{\prime \prime }+\gamma u^{\prime }+ku=F\left(t\right)$$

The displacement function this time will be,

$$u\left(t\right)=u_c\left(t\right)+U_P\left(t\right)$$

where the complementary solution will be the solution to the free, damped case and the particular solution will be found using undetermined coefficients or variation of parameter, whichever is most convenient to use.

There are a couple of things to note here about this case. First, from our work back in the free, damped case we know that the complementary solution will approach zero as $t$ increases. Because of this the complementary solution is often called the transient solution in this case.

Also, because of this behavior the displacement will start to look more and more like the particular solution as $t$ increases and so the particular solution is often called the steady state solution or forced response.

Let’s work one final example before leaving this section. As with the previous examples, we’re going to leave most of the details out for you to check.

Example 6 Take the system from the last example and add in a damper that will exert a force of 45 Newtons when then velocity is 50 cm/sec.

So, all we need to do is compute the damping coefficient for this problem then pull everything else down from the previous problem. The damping coefficient is
$$\begin{array}{cc}{F}_d& =\gamma u^{\prime }\\ 45& =\gamma \left(0.5\right)\\ \gamma & =90\end{array}$$The IVP for this problem is.
$$3u^{\prime \prime }+90u^{\prime }+75u=10\cos \left(5t\right)u\left(0\right)=0.2u^{\prime }\left(0\right)=-0.1$$The complementary solution for this example is
$$\begin{array}{cc}{u}_c\left(t\right)& =c_1{e}^{\left(-15+10\sqrt{2}\right)t}+c_2{e}^{\left(-15-10\sqrt{2}\right)t}\\ u_c\left(t\right)& =c_1{e}^{-0.8579t}+c_2{e}^{-29.1421t}\end{array}$$For the particular solution we the form will be,
$$U_P\left(t\right)=A\cos \left(5t\right)+B\sin \left(5t\right)$$Plugging this into the differential equation and simplifying gives us,
$$450B\cos \left(5t\right)-450A\sin \left(5t\right)=10\cos \left(5t\right)$$Setting coefficient equal gives,
$$U_P\left(t\right)=\frac{1}{45}\sin \left(5t\right)$$The general solution is then
$$u\left(t\right)=c_1{e}^{-0.8579t}+c_2{e}^{-29.1421t}+\frac{1}{45}\sin \left(5t\right)$$Applying the initial condition gives
$$u\left(t\right)=0.1986e^{-0.8579t}+0.001398e^{-29.1421t}+\frac{1}{45}\sin \left(5t\right)$$Here is a sketch of the displacement for this example.