Solution: The sequence of function is given by
\[
f_n(x) = \sin \sqrt{x + 4n^2 \pi ^2},\ \ x \in [0,+\infty ).
\]
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We will start by recalling the definition of equicontinuity. A family $\mathcal{F} $ of functions is said to be equicontinuous at $a$ if given any $\varepsilon >0$ there exists $\delta >0$ such that
\[
\vert x - a \vert \lt \delta \implies \vert f(x) - f(a) \vert \lt \varepsilon ,\ \ \forall\ f \in \mathcal{F} .
\]
Fix $n\in \mathbb{N} $ and $a \in [0, +\infty )$. Let $\varepsilon > 0$ be given and choose $\delta = \varepsilon $. Then
\begin{align*}
& \left\vert \sin \sqrt{x + 4n^2 \pi ^2} - \sin \sqrt{a + 4n^2 \pi ^2} \right\vert \\
= &\ \left\vert 2 \cos \left( \frac{\sqrt{X} + \sqrt{ A}}{2} \right) \sin \left( \frac{\sqrt{X} - \sqrt{A}}{2} \right) \right\vert \\
\leq & \ 2\left\vert \sin \left( \frac{\sqrt{X} - \sqrt{A} }{2} \right) \right\vert \leq 2 \left\vert \frac{\sqrt{X} - \sqrt{A} }{2} \right\vert \\
= & \ \left\vert (\sqrt{X} - \sqrt{A} ) \frac{\sqrt{X} + \sqrt{A} }{ \sqrt{X} + \sqrt{A} } \right\vert \\
\leq & \ \vert X - A \vert, \ \text{since } \sqrt{X} + \sqrt{A} \geq 1 \\
& = \vert x-a \vert \lt \delta = \varepsilon.
\end{align*}
Here $X = x + 4n^2\pi ^2$ and $A = a + 4n^2 \pi ^2$. Thus, the given family is equicontinuous at $a$. Since $a$ is arbitrary, the sequence is equicontinuous.
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To show that $\left\{ f_n \right\} $ is uniformly bounded, we note that for any $n \in \mathbb{N} $
\[
\left\vert f_n(x) \right\vert = \left\vert \sin \sqrt{x+4n^2\pi ^2} \right\vert \leq 1.
\]
Thus, the uniform bound is $1$ and hence the sequence is uniformly bounded.