Question

A mass weighing $\mathbf{\text{4pounds}}$is attached to a spring whose constant is $\mathbf{\text{2}}\raisebox{1ex}{$\mathbf{\text{lb}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{ft}}$}\right.$. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point $\mathbf{\text{1foot}}$above the equilibrium position with a downward velocity of $\mathbf{\text{8}}\raisebox{1ex}{$\mathbf{\text{ft}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{s}}$}\right.$. Determine the time at which the mass passes through the equilibrium position. Find the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant?

Short answer

The mass has a maximum displacement of$0.13533\text{feet}$position of at $t=\frac{1}{2}\text{s}$.

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:

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

Calculate the motion:

Consider the given data below.

The spring constant, $k=2\raisebox{1ex}{$\text{lb}$}\!\left/ \!\raisebox{-1ex}{$\text{ft}$}\right.$

The weight, $W=4\text{pounds}$

The weight is defined by,

$W=mg$

Here, the acceleration due to gravity, $g=32\raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.$

Therefore, the mass is,

$\begin{array}{rcl}m& =& \frac{W}{g}\\ & =& \frac{4}{32}\\ & =& \frac{1}{8}\text{slugs}\end{array}$

Since damping force is numerically equal to the instantaneous velocity. So, $\beta =1$

The differential equation of motion is then,

$\begin{array}{rcl}mx"+\beta x\text{'}+kx& =& 0\\ \frac{1}{8}\frac{d^{2}x}{d{t}^2}& =& -2x-\frac{dx}{dt}\end{array}$

Or equivalently

$\frac{d^{2}x}{d{t}^2}+8\frac{dx}{dt}+16x=0$

The auxiliary equation for the above equation is,

$m^2+8m+16=0$

Solving the roots of the equation then gives,

$m_1=m_2=-4$

Hence the system is critically damped, and the solution of the differential equation is,

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

Also, its derivative is,

$x^{\text{'}}\left(t\right)=-4c_1{e}^{4t}+c_2{e}^{-4t}-4c_{2}t{e}^{-4t}$

Since the mass is initially released from a point $1\text{foot}$above the equilibrium position with a downward velocity of $\text{8}\raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$, the initial conditions will be$x\left(0\right)=-1$and $x^{\text{'}}\left(0\right)=8$.

Substitute the initial conditions:

Applying these conditions gives

$\begin{array}{rcl}x\left(t\right)& =& c_1{e}^{-4t}+c_{2}t{e}^{-4t}\\ x\left(0\right)& =& c_1{e}^0+c_2\left(0\right)e^0\\ & =& -1\end{array}$

$\Rightarrow c_1=-1$

Now,

$\begin{array}{rcl}{x}^{\text{'}}\left(t\right)& =& -4c_1{e}^{4t}+c_2{e}^{-4t}-4c_{2}t{e}^{-4t}\\ x^{\text{'}}\left(0\right)& =& -4c_1{e}^0+c_2{e}^0-4c_2\left(0\right)e^0\\ & =& 8\end{array}$

Therefore,

$\begin{array}{rcl}-4c_1+c_2& =& 8\\ -4(-1)+c_2& =& 8\\ c_2& =& 4\end{array}$

Therefore, the equation of the motion is.

$x\left(t\right)=-e^{-4t}+4t{e}^{-4t}$

When the mass passes through the equilibrium position, i.e. $x=0$, you have

$\begin{array}{rcl}x\left(t\right)& =& 0\\ -e^{-4t}+4t{e}^{-4t}& =& 0\\ 4t{e}^{-4t}& =& e^{-4t}\\ t& =& \frac{1}{4}\text{s}\end{array}$

Thus, the mass passes through the equilibrium position at $t=\frac{1}{4}\text{s}$.

Find the mass and equilibrium position:

Substituting the values of $c_1$and $c_2$in $x^{\text{'}}\left(t\right)$gives the equation of displacement of mass from the equilibrium position, i.e.

$x^{\text{'}}\left(t\right)=8e^{4t}-16t{e}^{4t}$

For maximum displacement, you set $x^{\text{'}}\left(t\right)=0$.

$\begin{array}{rcl}8e^{-4t}-16t{e}^{-4t}& =& 0\\ 16t{e}^{-4t}& =& 8e^{-4t}\\ t& =& \frac{1}{2}\text{s}\end{array}$

Therefore, the mass attains its extreme displacement from the equilibrium position at $t=\frac{1}{2}\text{s}$.

Finally, the position of the mass at this instant will be.

$\begin{array}{rcl}x\left(\frac{1}{2}\right)& =& -e^{-4\cdot \frac{1}{2}}+4\cdot \frac{1}{2}\cdot e^{-4\cdot \frac{1}{2}}\\ & =& -e^2+2e^2\\ & =& e^{-2}\\ & \approx & 0.13533\text{feet}\end{array}$

Hence, the mass has a maximum displacement of position of $0.13533\text{feet}$at$t=\frac{1}{2}\text{s}$.