Past exam of the physics A course of the University of Cambridge / 2022 / Paper 2 / Section B / B6 / c / Solution

The natural interpretation is the fundamental transverse normal mode, $n=1$. Indeedso only $n=1$ propagates over the main part of the given Gaussian distribution of angular frequencies. Its centre travels atDifferentiate the group velocity with respect to angular frequency:At the centre this is $0.010149\mathrm{m}$. The one-standard-deviation group velocity spread is consequentlyOver the central travel time the associated spatial shear isEquivalently, use representative components at ${\omega }_0\pm {\sigma }_{\omega }$, for which $v_g(95)=1.5001\mathrm{m}{\mathrm{s}}^{-1}$ and $v_g(105)=1.6024\mathrm{m}{\mathrm{s}}^{-1}$. Their separation after $t_0$ isAt a fixed detector $10\mathrm{m}$ away, the corresponding arrival-time difference is $10/v_g(95)-10/v_g(105)=0.426\mathrm{s}$.

The paper asks for an estimate of spreading, without specifying a spatial width convention or the initial spectral phase. A Gaussian distribution has no finite full range. Thus $0.326\mathrm{m}$ is a one-standard-deviation dispersive contribution, whereas about $0.65\mathrm{m}$ describes the separation of the $\pm {\sigma }_{\omega }$ representatives. Neither is automatically the increase in the packet's original width. For an initially unchirped Gaussian wave packet in the narrow-band approximation, independent initial and dispersive contributions combine as ${\sigma }_{x,f}^2\simeq {\sigma }_{x,0}^2+{\sigma }_{x,\mathrm{disp}}^2$; a packet with an initial chirp can behave differently. The modal assumption and width convention must accompany the numerical answer.