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

Q: 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

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

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

AThis spring mass system execute $SHM$ .
BThe deformation $x_0$ if a $100\ g$ block is suspended from the spring and is at rest is $0.625\ m$ .
CAssuming 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 $\frac{{4\sqrt 5 }}{{2\pi }}$ .
D
None of these

Diffcult

STD 11 Science
NEET
STD 11 - 13. oscillations
Physics

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