The force-deformation equation for a nonlinea | vedclass

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(C) Given $F = 4x^{1/2}$ and mass $m = 100 \ g = 0.1 \ kg$.At equilibrium,the force exerted by the spring balances the weight of the block:$mg = 4x_0^

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

(C) Given $F = 4x^{1/2}$ and mass $m = 100 \ g = 0.1 \ kg$.At equilibrium,the force exerted by the spring balances the weight of the block:$mg = 4x_0^{1/2}$$(0.1 \ kg)(10 \ m/s^2) = 4x_0^{1/2}$$1 = 4x_0^{1/2}$$x_0^{1/2} = 0.25$$x_0 = (0.25)^2 = 0.0625 \ m$.Now,find the equivalent spring constant $k_e$ by taking the derivative of $F$ with respect to $x$ at $x = x_0$:$k_e = \left(\frac{dF}{dx}\right)_{x=x_0} = \frac{d}{dx}(4x^{1/2}) = 4 \cdot \frac{1}{2}x^{-1/2} = \frac{2}{\sqrt{x_0}}$.Substituting $x_0 = 0.0625$:$k_e = \frac{2}{\sqrt{0.0625}} = \frac{2}{0.25} = 8 \ N/m$.The frequency of vibration $f_n$ is given by:$f_n = \frac{1}{2\pi} \sqrt{\frac{k_e}{m}} = \frac{1}{2\pi} \sqrt{\frac{8}{0.1}} = \frac{1}{2\pi} \sqrt{80} = \frac{4\sqrt{5}}{2\pi} \ Hz$.Thus,statement $C$ is correct.

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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