Question

(a) A mass weighing W pounds stretches a spring$\frac{1}{2}$foot and stretches a different spring$\frac{1}{4}$foot. The two springs are attached in series and the mass is then attached to the double spring as shown in Figure 5.1.6. Assume that the motion is free and that there is no damping force present. Determine the equation of motion if the mass is initially released at a point 1 foot below the equilibrium position with a downward velocity of$\frac{\mathbf{2}}{\mathbf{3}}\mathbf{\text{ft}}\mathbf{/}\mathbf{\text{s}}\mathbf{.}$

(b) Show that the maximum speed of the mass is.$\frac{\mathbf{2}}{\mathbf{3}}\sqrt{\mathbf{3}\mathbf{g}\mathbf{+}\mathbf{1}}$

Short answer

$v_{\max }=\frac{2}{3}\sqrt{3g+1}x\left(t\right)=\cos \left(2\sqrt{\frac{g}{3}}t\right)+\frac{1}{\sqrt{3g}}\sin \left(2\sqrt{\frac{g}{3}}t\right)$

Step-by-step solution

Step 1:Definition of Non-linear and Linear Spring

NONLINEAR SPRINGS The mathematical model has the form

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

where$F\left(x\right)=kx$. 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.

Using Hooke’s law

(a)\It is given that a mass weighingpounds stretches a spring$\frac{1}{2}$ft and stretches a different spring$\frac{1}{4}$ft. So, by Hooke's law, for the first weight,

$W=k_1\left(\frac{1}{2}\right)\Rightarrow k_1=2W\text{lb}/\text{ft}$

and for the second weight,

$W=k_2\left(\frac{1}{4}\right)\Rightarrow k_2=4W\text{lb}/\text{ft}$

and$W=mg$gives$m=\frac{W}{g}$slug.

For the given case, since the restoring force is the same for both springs, the the effective spring constant$k_{\text{eff}}$of the system is

$k_{\text{eff}}=\frac{k_1{k}_2}{k_1+k_2}$

$=\frac{\left(2W\right)\left(4W\right)}{2W+4W}$

$=\frac{8W^2}{6W}$

$=\frac{4W}{3}$

Since there is no damping, we have$\beta =0$and so, the differential equation of motion can be expressed as

$\frac{W}{g}\frac{d^{2}x}{d{t}^2}+\frac{4W}{3}x=0$

or equivalently

$\frac{d^{2}x}{d{t}^2}+\frac{4g}{3}x=0$

Find the auxiliary equation and its roots

Auxiliary equation for the associated homogeneous equation is$m^2+\frac{4g}{3}=0$.

Solving the roots of the equation then gives$m_1=2i\sqrt{\frac{g}{3}}$and$m_2=-2i\sqrt{\frac{g}{3}}$.

Hence the general solution of the differential equation is

$x\left(t\right)=c_1\cos \left(2\sqrt{\frac{g}{3}}t\right)+c_2\sin \left(2\sqrt{\frac{g}{3}}t\right)$

Also, its derivative is

$x^{\text{'}}\left(t\right)=-2\sqrt{\frac{g}{3}}{c}_1\sin \left(2\sqrt{\frac{g}{3}}t\right)+2\sqrt{\frac{g}{3}}{c}_2\cos \left(2\sqrt{\frac{g}{3}}t\right)$

Since, the mass is initially released from a point 1 foot below the equilibrium position with a downward velocity of $\frac{2}{3}\text{ft}/\text{s}$, the initial conditions will be$x\left(0\right)=1$ and .$x^{\text{'}}\left(0\right)=\frac{2}{3}$

Substitute the initial conditions

Applying these conditions gives

$x\left(0\right)=c_1\cos 0+c_2\sin 0=1$

$c_1=1$

$x^{\text{'}}\left(0\right)=2\sqrt{\frac{g}{3}}{c}_1\sin 0+2\sqrt{\frac{g}{3}}{c}_2\cos 0=\frac{2}{3}$

$2\sqrt{\frac{g}{3}}{c}_2=\frac{2}{3}$

$c_2=\frac{1}{\sqrt{3g}}$

Therefore, equation of the motion is

$x\left(t\right)=\cos \left(2\sqrt{\frac{g}{3}}t\right)+\frac{1}{\sqrt{3g}}\sin \left(2\sqrt{\frac{g}{3}}t\right)$

(b) \Substituting the values of$c_1$and$c_2$ from above in the velocity equation gives the speed of the mass

$x^{\text{'}}\left(t\right)=-2c_1\sqrt{\frac{g}{3}}\sin \left(2\sqrt{\frac{g}{3}}t\right)+2\sqrt{\frac{g}{3}}{c}_2\cos \left(2\sqrt{\frac{g}{3}}t\right)$

$=-2\left(1\right)\sqrt{\frac{g}{3}}\sin \left(2\sqrt{\frac{g}{3}}t\right)+2\sqrt{\frac{g}{3}}\left(\frac{1}{\sqrt{3g}}\right)\cos \left(2\sqrt{\frac{g}{3}}t\right)$

$=-2\sqrt{\frac{g}{3}}\sin \left(2\sqrt{\frac{g}{3}}t\right)+\frac{2}{3}\cos \left(2\sqrt{\frac{g}{3}}t\right)$

We know that the maximum value ofwhen the function is in the form$x\left(t\right)=a{\sin }^{}t+b\cos t$ is$\sqrt{a^2+b^2}$. So, the maximum speed$v_{\max }$will be

$v_{\max }=\sqrt{{(-2\sqrt{\frac{g}{3}})}^2+{\left(\frac{2}{3}\right)}^2}=\sqrt{\frac{4g}{3}+\frac{4}{9}}$

$v_{\max }=\frac{2}{3}\sqrt{3g+1}x\left(t\right)=\cos \left(2\sqrt{\frac{g}{3}}t\right)+\frac{1}{\sqrt{3g}}\sin \left(2\sqrt{\frac{g}{3}}t\right)$