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.

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Quizlet Admin · Accepted Answer

$	ext{\color{#4257b2}19-32}$

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

(a) Given the nonlinear relationship;

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

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

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

The natural frequency of motion (Eq.19.14);

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

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

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