Updated greek laters @ latex compiling

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en/3.1.1.md +2 −2
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<h3>Solution</h3>
<p>
In addition to the force of gravity $mg$, the ball is subject to the force of tension of the string $F$, which has a vertical component $F'$:
−$$ F'=-2Fcos\alpha=-2F\frac{x}{\sqrt{l^2+x^2}}\quad(1) $$
+$$ F'=-2F\cos\alpha=-2F\frac{x}{\sqrt{l^2+x^2}}\quad(1) $$
Since the oscillations are small, we take into account that $x\ll l$
$$ \frac{x}{\sqrt{l^2+x^2}} = \frac{1}{\sqrt{1+\frac{l^2}{x^2}}} $$
Т.к. $x\ll l$, то $\frac{l^2}{x^2} \gg 1$:
$$ \frac{1}{\sqrt{1+\frac{l^2}{x^2}}}=\frac{1}{\sqrt{\frac{l^2}{x^2}}}=\frac{x}{l} $$
Substituting into $(1)$
$$ \boxed{F'=-2F\frac{x}{l}} $$
We find the small change in energy as
−$$ dU=2Fcos\alpha dx=2F\sqrt{\frac{x}{l^2+x^2}}=2F\frac{x}{l}dx $$
+$$ dU=2F\cos\alpha dx=2F\sqrt{\frac{x}{l^2+x^2}}=2F\frac{x}{l}dx $$
Next, we integrate $dU$ from $0$ to $U$
$$ U=\int_{0}^{U}dU=\int_{0}^{x}2F\frac{x}{l}dx=\boxed{\frac{Fx^2}{l}} $$
From the law of conservation of energy, we obtain the desired speed at maximum displacement
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