Question

A mass of $\mathbf{1}\mathbf{\text{slug}}$is attached to a spring whose constant is $\mathbf{\text{5}}\raisebox{1ex}{$\mathbf{\text{lb}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{ft}}$}\right.$. Initially, the mass is released 1 foot below the equilibrium position with a downward velocity of $\mathbf{5}\raisebox{1ex}{$\mathbf{\text{ft}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{s}}$}\right.$, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to $\mathbf{2}$times the instantaneous velocity.

(a) Find the equation of motion if the mass is driven by an external force equal to $\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{12}\mathbf{cos}\mathbf{2}\mathit{t}\mathbf{\pm }\mathbf{3}\mathbf{sin}\mathbf{2}\mathit{t}$.

(b) Graph the transient and steady-state solutions on the same coordinate axes. (c) Graph the equation of motion.

Short answer

(a) The equation of motion is $x\left(t\right)=e^{-t}\cos 2t+3\sin 2t$.

(b) The graph of the transient and steady-state solution is given below.

(c) The graph of the equation of motion is given below.

1. The graph of the equation of motion is given below.

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:

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

(a) Differential equation:

From the given values$m=1\text{slug},\beta =2$, and$k=5\raisebox{1ex}{$\text{lb}$}\!\left/ \!\raisebox{-1ex}{$\text{ft}$}\right.$the differential equation modelling the driven motion with damping of the spring and mass system is,

$$\begin{gathered} m\frac{d^{2}x}{d{t}^2}+\beta \frac{dx}{dt}+kx=0 \\ \frac{d^{2}x}{d{t}^2}+2\frac{dx}{dt}+5x=0 \end{gathered}$$

With a driving force of $f\left(t\right)=12\cos 2t+3\sin 2t$, the differential equation now becomes,

$\frac{d^{2}x}{d{t}^2}+2\frac{dx}{dt}+5x=12\cos 2t+3\sin 2t$

Whose solution is given by,

$x\left(t\right)=x_c\left(t\right)+x_p\left(t\right)$

Where the complementary function $x_c\left(t\right)$is the solution to the homogenous differential equation.

$\frac{d^{2}x}{d{t}^2}+2\frac{dx}{dt}+5x=0$ ….. (1)

And the particular function $x_p\left(t\right)$is the solution to be,

$x_p^{\text{'}\text{'}}+2x_p^{\text{'}}+5x_p=12\cos 2t+3\sin 2t$ ..... (2)

Find the auxiliary equation and particular solution:

Solving for$x_c\left(t\right)$, you have the auxiliary equation as,

$m^2+2m+5=0$

Which gives,

$m_{1,2}=-1\pm 2i$

Thus, the general solution to (1) will be,

$x_c\left(t\right)=e^{-t}(c_1\cos 2t+c_2\sin 2t)$

Using undetermined coefficients, you have $x_p$and its derivatives to be,

$$\begin{gathered} x_p=A\cos 2t+B\sin 2t \\ {x}_p^{\text{'}}=-2A\sin 2t+2B\cos 2t \\ {x^{\text{'}\text{'}}}_p=-4A\cos 2t-4B\sin 2t \end{gathered}$$

Following equation (2) in Step 2, you have,

$\begin{array}{rcl}12\cos 2t+3\sin 2t& =& -4A\cos 2t-4B\sin 2t+2(-2A\sin 2t+2B\cos 2t)+5(A\cos 2t+B\sin 2t)\\ & =& A\cos 2t+B\sin 2t-4A\sin 2t+4B\cos 2t\\ & =& (A+4B)\cos 2t+(-4A+B)\sin 2t\end{array}$

Equating the coefficients of the resulting equation in Step 5 gives,

$$\begin{gathered} A+4B=12 \\ -4A+B=3 \end{gathered}$$

Solving this system of equations simultaneously yields $A=0$and $B=3$. Thus, the particular solution to (2) is,

$\begin{array}{rcl}{x}_p\left(t\right)& =& A\cos 2t+B\sin 2t\\ & =& 3\sin 2t\end{array}$

Now that you have solved for$x_c\left(t\right)$and$x_p\left(t\right)$, it follows that,

$x\left(t\right)=e^{-t}(c_1\cos 2t+c_2\sin 2t)+3\sin 2t$

However, there are still unknowns. We need to utilize the initial conditions to be able to find$c_1$and$c_2$.

We are given with the initial displacement $x\left(0\right)=1$. Solving for $c_1$, you get,

$$\begin{gathered} 1=e^0(c_1\cos 0+c_2\sin 0)+3\sin 0° \\ {c}_1=1 \end{gathered}$$

Substitution:

Using the first derivative of $x\left(t\right)$and the initial velocity $x^{\text{'}}\left(0\right)=5$to find $c_2$gives,

$$\begin{gathered} x^{\text{'}}\left(t\right)=e^{-t}(-2c_1\sin 2t+2c_2\cos 2t)-e^t(c_1\cos 2t+c_2\sin 2t)+6\cos 2t \\ 5=e^0(-2\sin 0°+2c_2\cos 0°)-e^0(\cos 0+c_2\sin 0°)+6\cos 0° \\ 5=2c_2-1+6 \\ {c}_2=0 \end{gathered}$$

With $c_1=1$and $c_2=0$, you have the equation of motion to be,

$\begin{array}{rcl}x\left(t\right)& =& e^{-t}(c_1\cos 2t+c_2\sin 2t)+3\sin 2t\\ & =& e^{-t}(1\cdot \cos 2t+0\cdot \sin 2t)+3\sin 2t\\ & =& e^{-t}\cos 2t+3\sin 2t\end{array}$

(b) Graph the transient and steady-state solutions on the same coordinate axes:

The transient state is given by $x_c\left(t\right)=e^{-t}\cos 2t$while the steady-state is defined by $x_p\left(t\right)=3\sin 2t$. The figure below shows the graph of the two functions on the same coordinate axes.

(c) Graph the equation of motion:

The figure below shows the graph of the equation of motion $x=x_c+x_p$.