Section 2.6. Damp Free Vibration problems

Objective:

1. Setting up damp free vibration problems

2. Using 2nd order homogeneous differential equations to solve damp free vibration problems

We are ready for the spring vibration problem. Here is the review that we cover in Section 2.1. Suppose a mass $m$ hangs from a vertical spring. Assume the the mass is at rest then gravity and spring force is equal, $mg=kL$ where $L$ is the extra length of the spring from its natural length. Let $s(t)$ denote the displacement of the mass from its equilibrium position at time $t$, measured downward. There are forces on the spring: (a) the spring force coming from Hook’s law which is proportional to the displacement of the mass from its natural length: $F_s(t)=k(s(t)+L)$; (b) the gravity force coming from the weight of the mass: $F_g(t)=mg$; (c) the damping force coming from the environment which is proportional to the velocity of the mass: $F_d(t)=r{s}^{\prime }(t)$; the possible external force acting on the mass $F(t)$. According to Newton’s law, the net force is $ma=m{s}^{″}(t)$. The differential equation of the spring system is

$$ms”(t)=mg-k(s(t)+L)-r{s}^{\prime }(t)+F(t).$$

Hence the ODE becomes 

$$ms”(t)+r{s}^{\prime }(t)+ks(t)=F(t).$$

In this section, we are assuming $F(t)=0$. Keep in mind that $m$, $r$ and $k$ are all constants and positive. There are three major cases that we need to consider. For spring problem, we start with the minor case, $r=0$, no damping forces. 

Case 0: $m{s}^{″}(t)+ks(t)=0$. When this occur, the spring is in simple harmonic motion. Then the general solutions of the equation is $y=A\text{cos}(\sqrt{\frac{k}{m}}t)+B\text{sin}(\sqrt{\frac{k}{m}}t)$ where $R=\sqrt{A^2+B^2}$ is the amplitude, $w_0=\sqrt{\frac{k}{m}}$ is the natural frequency, and $T=\frac{2\pi }{w_0}$ is the period of simple harmonic motion. 

Example 1:  A mass of 150 g stretches a spring 10 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 20 cm/s, and if there is no damping, determinate the position $s(t)$ of the mass at any time $t$. 

Exercise 1: A mass of 100 g stretches a spring 20 cm. If the mass is set in motion from its equilibrium position with a downward velocity of 10 cm/s, and if there is no damping, determinate the position $s(t)$ of the mass at any time $t$. Find the natural frequency and period. 

Case 1: $m{s}^{″}(t)+r{s}^{\prime }(t)+ks(t)=0$ with $r>0$ and $r^2-4mk<0$. When this occur, the spring is not periodic but it is oscillation. This spring system is underdamped. Then the general solutions of the equation is $s=A{e}^{\frac{-r}{2m}t}\text{cos}(\frac{\sqrt{4mk-r^2}}{2m}t)+B{e}^{\frac{-r}{2m}t}\text{sin}(\frac{\sqrt{4mk-r^2}}{2m}t)$ where $R=\sqrt{A^2+B^2}$ is the amplitude, $u=\frac{\sqrt{4mk-r^2}}{2m}$ is the quasi frequency, and $T=\frac{2\pi }{u}$ is the quasi period  of the oscillation motion. 

Example 2: A spring is stretched 20 cm by a force of 5 N. A mass of 5 kg is hung from the spring and the air resistance exerts a force of 0.1 N when the velocity of the mass is 10 m/s. If the mass is pulled down 10 cm below its equilibrium position given an initial downward velocity 10cm/s, determinate the position $s(t)$ of the mass at any time $t$. Find the quasi frequency and quasi period. 

Exercise 2: A spring is stretched 10 cm by a force of 6 N. A mass of 4 kg is hung from the spring and the air resistance exerts a force of 0.2 N when the velocity of the mass is 20 m/s. If the mass is pulled down 5 cm below its equilibrium position given an initial downward velocity 20 cm/s, determinate the position $s(t)$ of the mass at any time $t$. Find the quasi frequency and quasi period. 

Case 2: $m{s}^{″}(t)+r{s}^{\prime }(t)+ks(t)=0$ with $r>0$ and $r^2-4mk=0$. When this occur, the spring is not periodic. This spring system is critical damped. Then the general solutions of the equation is $s=A{e}^{\frac{-r}{2m}t}+Bt{e}^{\frac{-r}{2m}t}$.

Case 3: $m{s}^{″}(t)+r{s}^{\prime }(t)+ks(t)=0$ with $r>0$ and $r^2-4mk>0$. When this occur, the spring is not periodic. This spring system is over-damped. Then the general solutions of the equation is $s=A{e}^{{\lambda }_{1}t}+B{e}^{{\lambda }_{2}t}$ where ${\lambda }_1$ and ${\lambda }_2$ are two roots of $m{\lambda }^2+r\lambda +k=0$.

Example 3: A spring is stretched 10 cm by a force of 2.5 N. A mass of 1 kg is hung from the spring and a damper exerts a force of 1 N when the velocity of the mass is 10 cm/s. If the mass is pulled down 10 cm below its equilibrium position given an initial downward velocity 10 cm/s, determinate the position $s(t)$ of the mass at any time $t$. 

Exercise 3: A spring is stretched 50 cm by a force of 4 N. A mass of 2 kg is hung from the spring and a damper exerts a force of 2 N when the velocity of the mass is 20 cm/s. If the mass is pulled down 5 cm below its equilibrium position given an initial downward velocity 20 cm/s, determinate the position $s(t)$ of the mass at any time $t$.  

Spring position graphing for four cases: $m{s}^{″}(t)+r{s}^{\prime }(t)+ks(t)=0$.

Case 0: Harmonic, $r=0$.

Case 1: Under-damped, $r^2-4mk<0$.

Case 2: Critical damped, $r^2-4mk=0$.

Case 3: Over damped, $r^2-4mk>0$.

Example 4: Decide of the spring system is harmonic, under-damped, critical damped, or over damped.

(a) $2s^{″}(t)+3s^{\prime }(t)+5s(t)=0$

(b) $2s^{″}(t)+4s^{\prime }(t)+s(t)=0$

(c) $2s^{″}(t)+5s(t)=0$

(d) $2s^{″}(t)+8s^{\prime }(t)+8s(t)=0$

Exercise 4: Decide of the spring system is harmonic, under-damped, critical damped, or over damped.

(a) $s^{″}(t)+10s^{\prime }(t)+25s(t)=0$

(b) $3s^{″}(t)+4s^{\prime }(t)+5s(t)=0$

(c) $2s^{″}(t)+5s^{\prime }(t)+2s(t)=0$

(d) $2s^{″}(t)+9s(t)=0$