Question

A mass weighing $\mathbf{\text{10pounds}}$stretches a spring $\mathbf{\text{2feet}}$. The mass is attached to a dashpot device that offers a damping force numerically equal to role="math" localid="1664044762332" $\mathit{\beta }$$\mathbf{(}\mathit{\beta }\mathbf{>}\mathbf{0}\mathbf{)}$times the instantaneous velocity. Determine the values of the damping constant $\mathit{\beta }$so that the subsequent motion is (a) overdamped, (b) critically damped, and (c) underdamped.

Short answer

(a) The value of damping constant when the subsequent motion is overdamped is $\beta >\frac{5}{2}$.

(b) The value of damping constant when the subsequent motion is critically damped is $\beta =\frac{5}{2}$.

(c) The value of damping constant when the subsequent motion is underdamped is $0<\beta <\frac{5}{2}$.

Step-by-step solution

Definition of Spring / Mass Systems:

Suppose you take into consideration an external force $\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$acting on a vibrating mass on a spring. For example,$\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$could represent a driving force causing an oscillatory vertical motion of the support of the spring. The inclusion of$\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$in the formulation of Newton’s second law gives the differential equation of driven of forced motion:

$\mathit{m}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{=}\mathbf{-}\mathit{k}\mathit{x}\mathbf{-}\mathit{\beta }\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$

Differential equation:

You have a system where an object that weighs $W=10\text{lb}$attached to a spring with a constant $k$is exhibiting free damped motion where $\beta >0$. The spring is stretched for $s=\text{2ft}$by its own weight.

Your goal is to be able to find the values of $\mathit{\beta }$that will satisfy the three cases involved in free-damped motion.

Remember that to solve for the spring constant $\mathit{k}$, you make use of Hooke's law which has the formula:

$F=ks$ ….. (1)

Where, $F$is the force in pounds and $s$is the elongation in feet.

Because there is damping, the system exhibits free damped motion described by the differential equation,

$\frac{d^{2}x}{d{t}^2}+2\lambda \frac{dx}{dt}+{\omega }^{2}x=0$ ….. (2)

Where, ${\omega }^2=\raisebox{1ex}{$k$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$and $2\lambda =\raisebox{1ex}{$\beta $}\!\left/ \!\raisebox{-1ex}{$m$}\right.$.

Homogenous linear differential equation:

Note that this is a second-order homogenous linear differential equation with constant coefficients having three cases for the solution.

When ${\lambda }^2-{\omega }^2>0$, the system is overdamped and the solution is,

$x\left(t\right)=e^{\lambda t}\left(c_1{e}^{\sqrt{{\lambda }^2{\omega }^2}t}+c_2{e}^{\sqrt{{\lambda }^2{\omega }^2}t}\right)$ ….. (3)

Then when ${\lambda }^2-{\omega }^2=0$, the system is critically damped and the solution is given by,

$x\left(t\right)=e^{\lambda t}(c_1+c_{2}t)$ ….. (4)

Lastly, when ${\lambda }^2-{\omega }^2<0$, the system is underdamped having the solution to be,

$x\left(t\right)=e^{\lambda t}\left(c_1\cos \sqrt{{\lambda }^2-{\omega }^2}t+c_2\sin \sqrt{{\lambda }^2-{\omega }^2}t\right)$ ….. (5)

Following Eq. (1), you can find the spring constant as,

$\begin{array}{rcl}F& =& W\\ & =& ks\end{array}$

$$\begin{gathered} 10=k\left(2\right) \\ k=5\raisebox{1ex}{$\text{lb}$}\!\left/ \!\raisebox{-1ex}{$\text{ft}$}\right. \end{gathered}$$

Then, converting the given weight to an appropriate unit of mass gives,

$\begin{array}{rcl}m& =& \frac{W}{g}\\ & =& \frac{10\text{pounds}}{32{\displaystyle \raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}}\\ & =& \frac{5}{16}\text{slug}\end{array}$

From the obtained values, the differential equation modeling the free damped motion of the spring and mass system is given by Eq. (2) as,

$$\begin{gathered} \frac{d^{2}x}{d{t}^2}+2\lambda \frac{dx}{dt}+{\omega }^{2}x=0 \\ \frac{d^{2}x}{d{t}^2}+\left(\frac{\beta }{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$16$}\right.}}\right)\frac{dx}{dt}+\left(\frac{5}{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$16$}\right.}}\right)x=0 \end{gathered}$$

$\frac{d^{2}x}{d{t}^2}+\frac{16\beta }{5}\frac{dx}{dt}+16x=0$ ..... (6)

So following Eq. (2) and Eq. (6) you have

$$\begin{gathered} 2\lambda =\frac{16\beta }{5} \\ \lambda =\frac{8\beta }{5} \end{gathered}$$

And

$$\begin{gathered} {\omega }^2=16 \\ \omega =4 \end{gathered}$$

This gives the expression ${\lambda }^2-{\omega }^2$as below.

${\lambda }^2-{\omega }^2={\left(\frac{8\beta }{5}\right)}^2-4^2$

${\lambda }^2-{\omega }^2=\frac{64{\beta }^2}{25}-16$ ..... (7)

(a) Define the values of the damping constant so that the subsequent motion is overdamped:

For an overdamped system the condition ${\lambda }^2-{\omega }^2>0$must be satisfied. Using Eq. (7) and setting it to be greater than zero, you can find,

$$\begin{gathered} \frac{64{\beta }^2}{25}-16>0 \\ \frac{64{\beta }^2}{25}>16 \\ {\beta }^2>\frac{25}{4} \\ \beta >\frac{5}{2} \end{gathered}$$

(b) Define the values of the damping constant so that the subsequent motion is critically damped:

Then for a critically damped system the condition ${\lambda }^2-{\omega }^2=0$must be satisfied. Using Eq. (7) and setting it equal to zero gives,

$$\begin{gathered} \frac{64{\beta }^2}{25}-16=0 \\ \frac{64{\beta }^2}{25}=16 \\ {\beta }^2=\frac{25}{4} \\ \beta =\frac{5}{2} \end{gathered}$$

(c) Define the values of the damping constant so that the subsequent motion is underdamped:

Lastly, an underdamped system requires ${\lambda }^2-{\omega }^2<0$. When we set Eq. (7) to be less than zero, you have,

$$\begin{gathered} \frac{64{\beta }^2}{25}-16<0 \\ \frac{64{\beta }^2}{25}<16 \\ {\beta }^2<\frac{25}{4} \\ \beta <\frac{5}{2} \end{gathered}$$

And since $\beta <0$, the values of satisfying the above condition lie between $0<\beta <\frac{5}{2}$.