Finding the depression and the tension of the string

[haruspex]

You droppwd a "2" somewhere. I got the given result, ΔL=L₀/20, with approximating L≈L₀ where it was possible (Lis the stretched length):
Pythagoras: (L/2)²-(L₀/2)²=y² ----> (L-L₀)(L+L₀)=4y²
using L≈L₀,
(L-L₀)=2y²/L₀. (1)
Given for spring constant: k=1000mg/L₀, (2)
From Hook's Law, and using (1) and (2):
k(L-L₀)=T, that is 2000mgy²/L₀²=T. (3)
From force balance: mg=2Tcos(θ). As cos(θ)=y/(L/2) mg=4Ty/L (4).
From (3) and (4), and taking L≈L₀, the result is y=L₀/20.

Thanks, I've found it.
Post #22 fourth equation should read $2a=l_0+\delta$, not $2a=l_0+2\delta$.

 

[magic]

I'm also confused as to where the factor of two comes into play. For my equations, I have:

$$[1] \qquad F_{net,y} = mg = 2Tsin(\theta) \approx 2Ttan(\theta) \Rightarrow T = \frac{mg}{2tan(\theta)}$$ where $\theta$ is the angle between the horizontal and and the wire.
$$[2] \qquad tan(\theta) = \frac{y}{\frac{l_0}{2}} = \frac{2y}{l_0}$$
$$[3] \qquad l'^2 = y^2 + (\frac{l_0}{2})^2 \ \Rightarrow l' \approx \frac{l_0}{2} + \frac{y^2}{l_0}$$ where I used the binomial expansion to expand the root term above
$$[4] \qquad T = k(l' - \frac{l_0}{2}) = \frac{mgy^2}{\delta l_0}$$

combining it all, I end up with:

$$y^3 = \frac{l_0^3}{4000}$$

and the only way I end up with the correct solution of $y=\frac{l_0}{20}$ is if in the 4th equation that $T=2k(l' - \frac{l_0}{2})$. Is there any physical reason as to why that should be the case?

 

[haruspex]

the only way I end up with the correct solution of $y=\frac{l_0}{20}$ is if in the 4th equation that $T=2k(l' - \frac{l_0}{2})$. Is there any physical reason as to why that should be the case?

Indeed there is.
The whole wire has relaxed length $l_0$ and is extended to length $2l'$.
k is the constant for the wire as a whole. Half the wire would have twice the constant.

 

[magic]

Thank you haruspex! It completely slipped from me that $k$ depends just as much on the geometry of the wire as it does with the material.