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A particle of mass $m$ moves in a straight line on a smooth horizontal table, and is connected to two points $A$ and $B$ by light elastic springs of natural lengths $2l_0$ and $3l_0$ respectively, and modulus of elasticity $\lambda$. The points $A$ and $B$ are a distance $6l_0$ apart. Show that the equation of motion is: $$m\ddot{x} =\frac{\lambda}{6l_0}(12l_0 - 5x)$$

(where $x$ is the displacement of the particle from $A$ measured positive towards $B$.

I found this problem really hard. Because the modulus of elasticity doesn't seem to be defined right in the solution or maybe I'm just looking at it completely wrong.

The modulus of elasticity has the formula $$\lambda = \frac{\mathrm{stress}}{\mathrm{strain}} = \frac{F/A}{\Delta L/L} = \frac{FL}{A\Delta L}$$

I take it that $L$ is length $6l_0$ in the problem. The springs are $2l_0$ and $3l_0$ so the system is not in equilibrium.

The term $x$ is confusing me a little I can't really visualise what the $x$ distance looks like. Is the system closer to $A$? and is the distance from the mass to $B$ is $x$?

Obviously $$F = m\ddot{x} \implies F = \frac{\lambda}{6l_0}(12l_0 - 5x)$$

My definition of $$\lambda = \frac{FL}{A\Delta L}$$

Doesn't that mean that $$F = \frac{\lambda A \Delta L}{L}$$

we have the $\lambda$ and the $6l_0$ in the solution but what do I need to do with the area?

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Your two spring have the same modulus of elasticity $\lambda$ but different equilibrium length. $\lambda$ is a property of the material, not the springs and it take in account the relative deformation for each spring!

Hence calculating your forces $\mathbf{F}_A$ and $\mathbf{F}_B$, the forces from the spring attached to $A$ and $B$, you will end up with two spring modulus (not to be confused with elastic modulus) $k_A$ and $k_B$ such as: $$\mathbf{F}_x=k_x\Delta L$$ Where $\Delta L =L-L_{0x}$, $L_{0x}$ being the equilibrium distance of your spring and $x$ being $A$ or $B$ depending on the spring. In order to express $k_x\Delta L$ with using $\lambda$ you have to express it regarding the relative strain.

So $$k_x\Delta L = \lambda \frac{\Delta L}{L_{0x}}$$

The rest is straight forward.

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Thank you very much for the quick reply. Yeah, it was the fact I had to take into account the two springs that put me way off track.

$$\begin{align} \mathbf{F_n} & = \mathbf{F_A} + \mathbf{F_b} \\ & = k_A \Delta L + k_B \Delta L \\ & = k_A(L-L_0) + k_B(L-L_0) \\ & = k_A(6l_0 - 2x) + k_B(6l_0 - 3x) \\ & = \lambda\Big(\frac{6l_0-2x}{6l_0}\Big) + \lambda\Big(\frac{6l_0-3x}{6l_0}\Big) \\ & = \frac{\lambda}{6l_0}(12l_0-5x) \end{align} $$

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