Question

A mass $\mathit{m}$is attached to the end of a spring whose constant is $\mathit{k}$. After the mass reaches equilibrium, its support begins to oscillate vertically about a horizontal line $\mathit{L}$according to a formula localid="1664181072022" $\mathit{h}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$. The value of localid="1664181044391" $\mathit{h}$represents the distance in feet measured from$\mathit{L}$. See Figure 5.1.22.

Determine the differential equation of motion if the entire system moves through a medium offering a damping force that is numerically equal to$\mathit{\beta }\left(\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.\right)$. (b) Solve the differential equation in part (a) if the spring is stretched $\mathbf{4}\mathbf{\text{feet}}$by a mass weighing$\mathbf{\text{16pounds}}$and$\mathit{\beta }\mathbf{=}\mathbf{2}\mathbf{,}\mathit{h}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathit{c}\mathit{o}\mathit{s}\mathit{t}\mathbf{,}\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathit{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

Short answer

(a) The differential equation of motion if the entire system moves through a medium offering a damping force is$m\frac{d^{2}x}{d{t}^2}=-k(x-h)-\beta \frac{dx}{dt}$or $\frac{d^{2}x}{d{t}^2}+2\lambda \frac{dx}{dt}+{\omega }^{2}x={\omega }^{2}h\left(t\right)$.

(b) The differential equation is $x\left(t\right)=e^{2t}\left(-\frac{{\displaystyle 56}}{{\displaystyle 13}}\cos 2t-\frac{{\displaystyle 72}}{{\displaystyle 13}}\sin 2t\right)+\frac{56}{13}\cos t+\frac{32}{13}\sin t$.

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:

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

Using Hook’s law:

Following Hooke's law and the given figure, you have the external force as

$F\left(t\right)=k\times h\left(t\right)$

Thus, the differential equation modeling the system is,

$$\begin{gathered} m\frac{d^{2}x}{d{t}^2}=-kx-\beta \frac{dx}{dt}+kh \\ m\frac{d^{2}x}{d{t}^2}=-k(x-h)-\beta \frac{dx}{dt} \end{gathered}$$

You can also divide both sides of the differential equation by the mass so you can write it in the form as below.

$\begin{array}{rcl}\frac{d^{2}x}{d{t}^2}& =& -\left(\frac{k}{m}\right)x-\left(\frac{\beta }{m}\right)\frac{dx}{dt}+\left(\frac{k}{m}\right)h\\ & =& -{\omega }^{2}x-2\lambda \frac{dx}{dt}+{\omega }^{2}h\end{array}$

Rewriting so that it follows equation (25) in Section 5.1.2, you have,

$\frac{d^{2}x}{d{t}^2}+2\lambda \frac{dx}{dt}+{\omega }^{2}x={\omega }^{2}h\left(t\right)$

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

(b) Find the spring constant:

Using the formula for Hooke's law, you can find the spring constant as,

$$\begin{gathered} F=ks \\ 16\text{lb}=k\left(4\text{ft}\right) \\ k=4\raisebox{1ex}{$\text{lb}$}\!\left/ \!\raisebox{-1ex}{$\text{ft}$}\right. \end{gathered}$$

Converting the given weight to an engineering unit of mass gives,

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

You also have the damping coefficient$\beta =2$and$h\left(t\right)=5\cos t$.

From the values in Step 3, the differential equation in part (a) now becomes,

$$\begin{gathered} \frac{1}{2}\frac{d^{2}x}{d{t}^2}+2\frac{dx}{dt}+4x=4\left(5\cos t\right) \\ \frac{d^{2}x}{d{t}^2}+4\frac{dx}{dt}+8x=40\cos t \end{gathered}$$

The solution of the resulting differential equation in Step 4 is given by,

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

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

$\begin{array}{rcl}{x}_c\left(t\right)& =& \frac{d^{2}x}{d{t}^2}+4\frac{dx}{dt}+8x\\ & =& 0\end{array}$

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

$\begin{array}{rcl}{x}_p\left(t\right)& =& \frac{1}{2}{x}_p^{\text{'}\text{'}}+2x_p^{\text{'}}+4x_p\\ & =& 20\cos t\end{array}$

Solve complementary and particular function:

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

$m^2+4x+8=0$

This gives,

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

Thus, the general solution to (1) is,

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

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

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

Following equation (2) in Step 5, you have,

$\begin{array}{rcl}20\cos t& =& \frac{1}{2}(-A\cos t-B\sin t)+2(-A\sin t+B\cos t)+4(A\cos t+B\sin t)\\ & =& \frac{7}{2}(A\cos t+B\sin t)+2(-A\sin t+B\cos t)\\ & =& \left(\frac{{\displaystyle 7}}{{\displaystyle 2}}A+2B\right)\cos t+\left(\frac{{\displaystyle 7}}{{\displaystyle 2}}B-2A\right)\sin t\end{array}$

Equating the coefficients of the resulting equation in Step 7 gives,

$\frac{7}{2}A+2B=20$

And

$-2A+\frac{7}{2}B=0$

Solving this system of equations simultaneously yields $A=\raisebox{1ex}{$56$}\!\left/ \!\raisebox{-1ex}{$13$}\right.$and $B=\raisebox{1ex}{$32$}\!\left/ \!\raisebox{-1ex}{$13$}\right.$. Thus, the particular solution to (2) is,

$\begin{array}{rcl}{x}_p\left(t\right)& =& A\cos t+B\sin t\\ & =& \frac{56}{13}\cos t+\frac{32}{13}\sin t\end{array}$

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

$x\left(t\right)=e^{-2t}(c_1\cos 2t+c_2\sin 2t)+\frac{56}{13}\cos t+\frac{32}{13}\sin t$

Substitute the initial conditions:

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

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

$$\begin{gathered} 0=e^0(c_1\cos 0°+c_2\sin 0°)+\frac{56}{13}\cos 0°+\frac{32}{13}\sin 0° \\ 0=c_1+0+\frac{56}{13}+0 \\ {c}_1=-\frac{56}{13} \end{gathered}$$

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

$$\begin{gathered} x\left(t\right)=e^{2t}(c_1\cos 2t+c_2\sin 2t)+\frac{56}{13}\cos t+\frac{32}{13}\sin t \\ {x}^{\text{'}}\left(t\right)=e^{2t}(-2c_1\sin 2t+2c_2\cos 2t)-2e^{-2t}(c_1\cos 2t+c_2\sin 2t)-\frac{56}{13}\sin t+\frac{32}{13}\cos t \\ 0=e^0(\frac{112}{13}\sin 0°+2c_2\cos 0°)-2e^0(-\frac{56}{13}\cos 0°+c_2\sin 0°)-\frac{56}{13}\sin 0°+\frac{32}{13}\cos 0° \\ 0=0+2c_2+\frac{112}{13}+\frac{32}{13} \end{gathered}$$

$c_2=-\frac{72}{13}$

With $c_1=-\raisebox{1ex}{$56$}\!\left/ \!\raisebox{-1ex}{$13$}\right.$and $c_2=-\raisebox{1ex}{$72$}\!\left/ \!\raisebox{-1ex}{$13$}\right.$, you have the equation of motion as,

$\begin{array}{rcl}x\left(t\right)& =& e^{-2t}(c_1\cos 2t+c_2\sin 2t)+\frac{56}{13}\cos t+\frac{32}{13}\sin t\\ & =& e^{-2t}\left(-\frac{{\displaystyle 56}}{{\displaystyle 13}}\cos 2t-\frac{{\displaystyle 72}}{{\displaystyle 13}}\sin 2t\right)+\frac{56}{13}\cos t+\frac{32}{13}\sin t\end{array}$