Updated greek laters @ latex compiling

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en/3.3.3.md +3 −3
@@ -57,13 +57,13 @@
For a harmonic motion, variable position depends sinusoidally on time. In this case, we suppose that load is at equilibrium position at $t = 0$. So, sine function is more adequate to this situation without phase angle or null phase angle. So,
$$x(t) = A~\sin{~\omega t}$$
where $A$ is the amplitude such that $A = 1~{\rm{cm}}$, because it is the maximum distance. Let's suppose that $x_0 = 0.5~{\rm{cm}}$ is achieved at $t=t_1$ and $A$ is achieved at $t = t_2$. Then,
− $$x_0 = A~\sin{~\omega t_1} \;(1)$$
+ $$x_0 = A~\sin{~\omega t_1} \quad(1)$$
$$A = A~\sin{~\omega t_2}$$ or
$$\sin{~\omega t_2} = 1$$
Hence, $\omega t_2 = \frac{\pi}{2} + 2k\pi$ with $k\in\mathbb{Z}$, for $k=0$, $t_2 = \frac{\pi}{2\omega}$. As $\omega = \frac{2\pi}{T}$, so $t_2 = \frac{T}{4}$. Since $\Delta t = t_2 - t_1 = 0.01~{\rm{s}}$,
− $$t_1 = \frac{T}{4} - \Delta t \;(2)$$
+ $$t_1 = \frac{T}{4} - \Delta t \quad(2)$$
Putting (2) into (1) and separating $T$, it is obtained
− $$T = \frac{4\Delta t}{1-\frac{2}{\pi}\arcsin{\frac{x_0}{A}}}$$
+ $$T = \frac{4\Delta t}{1-\frac{2}{\pi}\arcsin {\frac{x_0}{A}}}$$
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<h4>Answer</h4>
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