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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
This spring mass system execute $SHM$ .
The deformation $x_0$ if a $100\ g$ block is suspended from the spring and is at rest is $0.625\ m$ .
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 $\frac{{4\sqrt 5 }}{{2\pi }}$ .
None of these
Solution
$m g=(0.10 \mathrm{kg})\left(10 \mathrm{m} / \mathrm{s}^{2}\right)=1 \mathrm{N}$
$\mathrm{mg}=4 \mathrm{x}_{0}^{1 / 2}$
$x_{0}=0.0625 \mathrm{m}$
$k_{e}=\left(\frac{d F}{d x}\right)_{x_{0}}=4 \times \frac{1}{2} x^{-1 / 2}=\frac{2}{\sqrt{x_{0}}}=2 \times 4=8 \mathrm{N} / \mathrm{m}$
$f_{n}=\frac{\sqrt{\frac{k_{e}}{m}}}{2 \pi}=\frac{\sqrt{\frac{8}{0.1}}}{2 \pi}=\frac{4 \sqrt{5}}{2 \pi}$
