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A object of mass 2kg is hanging vertically on a spring. The spring is extended 0.1m from its equilibrium position. How would you find the spring constant?

$my'' +ky=mg$

$y''+ \frac{k}{2} = 9.8$

The homogeneous solution: $e^{it \sqrt{\frac{k}{2}} }$ and $e^{-it \sqrt{\frac{k}{2}} }$

$y_h =c_1 cos\sqrt{\frac{k}{2}}t +c_2 sin\sqrt{\frac{k}{2}}t $

The particular solution:$y_p=\frac{19.6}{k}$

So $y=\frac{19.6}{k} + c_1 cos\sqrt{\frac{k}{2}}t +c_2 sin\sqrt{\frac{k}{2}}t$

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  • $\begingroup$ Perhaps you had better clarify whether the weight itself pulls the spring $0.1$ m out of its unweighted equilibrium position, or if you pull the system $0.1$ m further out of its weighted equilibrium position. $\endgroup$ – Brian Tung Oct 31 '15 at 19:16
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Am I missing something? Isn't the spring constant obtained simply by balancing the forces at equilibrium?

$$ F_{\text{grav}} = F_{\text{spring}} $$

$$ mg = kx $$

and therefore

\begin{align} k & = \frac{mg}{x} \\ & = \frac{(2 \text{ kg})(9.8 \text{ m}/\text{s}^2)}{0.1 \text{ m}} \\ & = 196 \text{ N}/\text{m} \end{align}

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  • $\begingroup$ Wow im bad. Thanks $\endgroup$ – Rudolf Oct 31 '15 at 19:15
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Notice,

in the equilibrium condition of the spring mass system, $$\text{spring force in vertically upward direction}=\text{weight of object in vertically downward direction}$$ $$kx=mg$$$$\implies k=\frac{mg}{x}$$ setting $m=2\ kg$, $g=9.8\ m^2/s$ & $x=0.1\ m$, spring constant $k$
$$k=\frac{2\times 9.8}{0.1}=\color{red}{196\ N/m}$$

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Spring Constant's dimension is N/m

The equation is simply equating $kx = mg$ It is not a second order eqation.

$k = \frac{2\times 9.8}{0.1} = 196$N/m by Hooke's law

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