= Solution
The spatial transform of $\cos(px)$ is $\pi[\delta(k-p)+\delta(k+p)]$. <Convolution> with the impulsive source and the inverse transform therefore give simply $T(x,t)=\cos(px)G_p(t-t_0)$. Take $p\geq0$ without loss of generality and set $\tau=t-t_0$.
For $\alpha>p/c$, let $d_p=\sqrt{\alpha^2c^4-p^2c^2}$. Then
$$
\boxed{T(x,t)=\Theta(\tau)\cos(px)\frac{c^2}{d_p}e^{-\alpha c^2\tau}\sinh(d_p\tau).}
$$
For $\alpha<p/c$, let $\Omega_p=\sqrt{p^2c^2-\alpha^2c^4}$. Then
$$
\boxed{T(x,t)=\Theta(\tau)\cos(px)\frac{c^2}{\Omega_p}e^{-\alpha c^2\tau}\sin(\Omega_p\tau).}
$$
Both satisfy $T(t_0^+)=0$ and $\partial_tT(t_0^+)=c^2\cos(px)$, so the impulse changes the time <derivative> rather than discontinuously changing <temperature>. The first case rises and then relaxes without sign <oscillations> for a nonzero mode; the second oscillates as its amplitude decays. The limiting equality gives $T=\Theta(\tau)c^2\tau e^{-\alpha c^2\tau}\cos(px)$, and $p=0$ gives the nondecaying uniform increment discussed above.
A cosine source has both positive and negative regions, so this linear solution can be understood as a <temperature> perturbation about a background; its negative values are not assertions of negative absolute <temperature>. The temporal <oscillations> arise from the second-time-derivative term. Dropping that term changes the initial-jump structure as well as removing the underdamped regime.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2012/paper-1-thermal-modes.png]
{title=Retarded thermal-mode responses}
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