Question

The force-deflection equation for a nonlinear spring fixed at one end is F=2.5x1/2F=2.5 x^{1 / 2}, where FF is the force, expressed in pounds, and xx is the deflection, expressed in feet. (a)(a) Determine the static deflection x0x_0 if a 6 oz6 \mathrm{~oz} block is suspended from the spring. (b)(b) Assuming that the slope of the force-deflection curve at the point corresponding to this loading can be used as an equivalent spring constant, determine the frequency of vibration of the block if it is given a very small downward displacement from its equilibrium position and released.

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19-32\text{\color{#4257b2}19-32}

Non-linear spring\text{\color{#4257b2}Non-linear spring}

(a) Given the nonlinear relationship;

F=2.5xx=(F2.5)2xo=(6162.5)2=0.0225 ft\begin{gather*} F=2.5\sqrt{x}\\ x=\left(\dfrac{F}{2.5} \right)^{2}\\ x_{\text{o}}=\left(\dfrac{6}{16\cdot2.5} \right)^{2}=0.0225\mathrm{\ ft}\tag{\text{convert oz to lb}} \end{gather*}

(b) The stiffness; kk, taken as the instantaneous rate of change in the force at; x=xox=x_{\text{o}};

dFdx=ddx[2.5x]=1.25xk=dFdxx=0.0225 ft=1.250.0225=8.33 lb/ft\begin{gather*} \dfrac{dF}{dx}=\dfrac{d}{dx}\left[ 2.5\sqrt{x}\right]=\dfrac{1.25}{\sqrt{x}}\\ k=\dfrac{dF}{dx}\bigg|_{x=0.0225\mathrm{\ ft}}^{}=\dfrac{1.25}{\sqrt{0.0225}}=8.33\mathrm{\ lb/ft} \end{gather*}

The natural frequency of motion (Eq.19.14);

ωn=km=8.33(616g)=note 1.8.33(61632.2)=26.75 rad/sfn=26.752π=4.26 Hz\begin{gather*} \boxed{\omega_{\text{n}}=\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{8.33}{\left( \dfrac{6}{16g}\right)}}\overset{\text{note 1.}}{=}\sqrt{\dfrac{8.33}{\left( \dfrac{6}{16\cdot32.2}\right)}}=26.75\mathrm{\ rad/s}\Rightarrow f_{\text{n}}=\dfrac{26.75}{2\pi}=4.26\mathrm{\ Hz}} \end{gather*}

note 1. acceleration due to gravity; g=32.2 ft/s2g=32.2\mathrm{\ ft/s^2}.

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