The force-deformation equation for a nonlinea | vedclass

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Question

$m g=(0.10 \mathrm{kg})\left(10 \mathrm{m} / \mathrm{s}^{2}\right)=1 \mathrm{N}$

$\mathrm{mg}=4 \mathrm{x}_{0}^{1 / 2}$

$x_{0}=0.0625 \mathrm{m}

Vedclass · Accepted Answer

Assuming that the slope of the force deformation curve at the point corresponding to the deformation $x_0$ can be used as an equivalent spring constant, then the frequency of vibration of the block is $\frac{{4\sqrt 5 }}{{2\pi }}$ .

Vedclass · Answer

$m g=(0.10 \mathrm{kg})\left(10 \mathrm{m} / \mathrm{s}^{2}ight)=1 \mathrm{N}$

$\mathrm{mg}=4 \mathrm{x}_{0}^{1 / 2}$

$x_{0}=0.0625 \mathrm{m}$

$k_{e}=\left(\frac{d F}{d x}ight)_{x_{0}}=4 	imes \frac{1}{2} x^{-1 / 2}=\frac{2}{\sqrt{x_{0}}}=2 	imes 4=8 \mathrm{N} / \mathrm{m}$

$f_{n}=\frac{\sqrt{\frac{k_{e}}{m}}}{2 \pi}=\frac{\sqrt{\frac{8}{0.1}}}{2 \pi}=\frac{4 \sqrt{5}}{2 \pi}$

The force-deformation equation for a nonlinear spring fixed at one end is $F =4x^{1/ 2}$ , where $F$ is the force (expressed in newtons) applied at the other end and $x$ is the deformation expressed in meters

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