The force-deflection equation for a nonlinear spring fixed at one end is $F=2.5 x^{1 / 2}$, where $F$ is the force, expressed in pounds, and $x$ is the deflection, expressed in feet. $(a)$ Determine the static deflection $x_0$ if a $6 \mathrm{~oz}$ block is suspended from the spring. $(b)$ 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.

Solution

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

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

(a) Given the nonlinear relationship;

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

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

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

The natural frequency of motion (Eq.19.14);

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

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