= Solution
The <electric-dipole approximation for radiation> requires the source size to be much smaller than the radiation wavelength:
$$
\frac{\omega a}{c}\ll1.
$$
The displayed $1/r$ fields also describe the <radiation zone>, so the observation point must satisfy $r\gg a$ and $r\gg c/\omega$.
Write $t_r=t-r/c$. The given vector potential and the <time derivative of the electric dipole moment> give
$$
\mathbf A(t,\mathbf x)
\simeq\frac{\mu_0}{4\pi r}\int\mathbf J(t_r,\mathbf x')\,d^3x'
=\frac{\mu_0}{4\pi r}\dot{\mathbf p}(t_r).
$$
When taking the <curl>, differentiating $1/r$ produces a near-field term of order $r^{-2}$, while differentiating the retarded time produces the leading radiation term because $\nabla t_r=-\widehat{\mathbf x}/c$. Hence
$$
\boxed{
\mathbf B=\nabla\times\mathbf A
\simeq-\frac{\mu_0}{4\pi rc}
\widehat{\mathbf x}\times\ddot{\mathbf p}(t_r).}
$$
The assumed electric field $\mathbf E=-c\widehat{\mathbf x}\times\mathbf B$ and the transversality $\widehat{\mathbf x}\cdot\mathbf B=0$ make the <Poynting vector>
$$
\mathbf S=\frac1{\mu_0}\mathbf E\times\mathbf B
=\frac{c}{\mu_0}|\mathbf B|^2\widehat{\mathbf x}.
$$
On a sphere of radius $R$, the power through surface element $R^2d\Omega$ is therefore
$$
\boxed{
\frac{dP}{d\Omega}
=R^2\mathbf S\cdot\widehat{\mathbf x}
=\frac{\mu_0}{16\pi^2c}
|\widehat{\mathbf x}\times\ddot{\mathbf p}(t-R/c)|^2.}
$$
Choose the polar axis along $\ddot{\mathbf p}$ and use the <angular distribution of electric dipole radiation>, for which $|\widehat{\mathbf x}\times\ddot{\mathbf p}|^2=|\ddot{\mathbf p}|^2\sin^2\theta$. Since $\int\sin^2\theta\,d\Omega=8\pi/3$,
$$
\boxed{
P(t)=\frac{\mu_0}{6\pi c}
|\ddot{\mathbf p}(t-R/c)|^2.}
$$
For the charged particle in <simple harmonic motion>,
$$
\mathbf p(t)=qA\sin(\omega t)\mathbf i_x,
\qquad
\ddot{\mathbf p}(t)=-qA\omega^2\sin(\omega t)\mathbf i_x.
$$
Thus
$$
\boxed{
P(t)=\frac{\mu_0q^2A^2\omega^4}{6\pi c}
\sin^2\!\bigl(\omega(t-R/c)\bigr),
\qquad
\langle P\rangle=\frac{\mu_0q^2A^2\omega^4}{12\pi c}.}
$$
Here the <angular frequency> obeys $\omega=2\pi/T$, and the average uses $\langle\sin^2\rangle=1/2$. The dipole condition is
$$
\boxed{\frac{A\omega}{c}\ll1.}
$$
Allow the amplitude to vary slowly as radiation removes energy. The mechanical energy is
$$
E=\frac12mA^2\omega^2,
$$
so the <radiation damping of a harmonically oscillating charge> gives
$$
\frac{dE}{dt}=-\langle P\rangle
=-\frac{\mu_0q^2\omega^2}{6\pi mc}E.
$$
This is <exponential decay> with rate $\gamma=\mu_0q^2\omega^2/(6\pi mc)$. Its <half-life> is
$$
\boxed{t_{1/2}=\frac{6\pi mc\log2}{\mu_0q^2\omega^2}.}
$$
Finally, the largest particle speed is $v_{\max}=A\omega$. Therefore the dipole condition $A\omega/c\ll1$ itself implies $v_{\max}\ll c$, precisely the <nonrelativistic limit> required by the mechanical energy formula.
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