Question

The vertical motion of a mass attached to a spring is described by the initial-value problem

$\frac{\mathbf{1}}{\mathbf{4}}{\mathit{x}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathit{x}}^{\mathbf{\text{'}}}\mathbf{+}\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{,}{\mathit{x}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{.}$

Determine the maximum vertical displacement of the mass.

Short answer

$5e^{-1/5}\approx 4.09356$

Step-by-step solution

Step 1:Definition of Non-linear and Linear Spring

NONLINEAR SPRINGS The mathematical model has the form

$m\frac{d^{2}x}{d{t}^2}+F\left(x\right)=0$

whererole="math" $F\left(x\right)=kx$. Becausedenotes the displacement of the mass from its equilibrium position,$F\left(x\right)=kx$is Hooke's law-that is, the force exerted by the spring that tends to restore the mass to the equilibrium position. A spring acting under a linear restoring force$F\left(x\right)=kx$is naturally referred to as a linear spring.

A spring whose mathematical model incorporates a nonlinear restorative force, such as

$m\frac{d^{2}x}{d{t}^2}+k{x}^3=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}m\frac{d^{2}x}{d{t}^2}+kx+k_1{x}^3=0$

is called a nonlinear spring.

Find the auxiliary equation

$\frac{1}{4}{x}^{\text{'}\text{'}}+x^{\text{'}}+x=0,x\left(0\right)=4,x^{\text{'}}\left(0\right)=2$

Let us find its solution. The zeros of the auxiliary equation are

$x_{1,2}=\frac{-1\pm \sqrt{1-4\cdot \frac{1}{4}\cdot 1}}{\frac{1}{2}}\iff x_{1,2}=-2$

Therefore, the general solution of Eq. (1) is

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

From the first initial condition, we get$c_1=4$, and from the second$-2\cdot 4+c_2=2$.

The solution of this system is$c_1=4$and.$c_2=10$ Therefore, the equation of motion is

$x\left(t\right)=4e^{-2t}+10t{e}^{-2t}$

Derive the equation of motion

Let us derive it:

$x^{\text{'}}\left(t\right)=-8e^{-2t}+10e^{-2t}-20t{e}^{-2t}=2e^{-2t}-20t{e}^{-2t}$

Let$x^{\text{'}}\left(t\right)=0$:

$0=2e^{-2t}-20t{e}^{-2t}=2e^{-2t}(1-10t)\iff t=\frac{1}{10}$

Therefore, the maximum displacement will occur at the time$t=\frac{1}{10}$. Substitute it into (3) to obtain he maximum displacement:

$x\left(\frac{1}{10}\right)=4e^{-1/5}+e^{-1/5}=5e^{-1/5}\approx 4.09356$

Result

$5e^{-1/5}\approx 4.09356$