For the propagating mode of an air-filled or evacuated rectangular electromagnetic waveguide, the longitudinal dispersion relation isThe phase velocity and group velocity along the guide are consequentlyWith waveguide cutoff frequency $f_c=10\mathrm{GHz}$ and carrier frequency $f=30\mathrm{GHz}$, $\sqrt{1-(f_c/f{)}^2}=\sqrt{8}/3$. ThusThe pulse envelope travels at the group velocity for a sufficiently narrow range of frequencies. The phase velocity exceeding the speed of light does not describe superluminal signal transport. The check $v_p{v}_g=c^2$ follows directly from the dispersion relation. A different uniform filling medium would require replacing $c$ by its electromagnetic propagation speed; the stated numerical answer uses the usual empty-guide assumption.

The quoted capacitance is that of the filled parallel-plate capacitor, $C_i=1.0\times 10^{-6}\mathrm{F}$. Disconnecting the source imposes charge conservation on the isolated plates:On removing the dielectric, the capacitance falls by its relative permittivity, so $C_f=C_i/5=2.0\times 10^{-7}\mathrm{F}$. The capacitor energy at fixed electric charge is $U=Q^2/(2C)$. ThereforeandThe increase is supplied by the external work needed to withdraw the dielectric against its attraction into the electric field, in the lossless quasistatic limit. The final electric potential difference is $Q/C_f=12500\mathrm{V}$, not the original charging value.

An electron has electric charge $-e$, so its gain of kinetic energy through an accelerating electric potential difference of magnitude $V$ is $eV$. Starting from rest, conservation of energy givesWith $e=1.60218\times 10^{-19}\mathrm{C}$, electron mass $m_e=9.10938\times 10^{-31}\mathrm{kg}$, and $V=150\mathrm{V}$,This is a speed; the velocity points along the accelerating path, opposite the electric field. Here $v/c\simeq 0.0242$, where $c$ is the speed of light, so the classical kinetic energy approximation is accurate. As a check, special relativity gives $\gamma =1+eV/(m_e{c}^2)$ and $v=c\sqrt{1-{\gamma }^{-2}}\simeq 7.262\times 10^6\mathrm{m}{\mathrm{s}}^{-1}$, agreeing to the quoted precision.

To find the two-mass three-spring normal modes, choose both displacements positive to the right, measured from mechanical equilibrium. The middle spring's extension changes by $x_2-x_1$. Hooke's law and Newton's second law therefore giveIn a normal mode both displacements have the same angular frequency: take $x_j=A_j\cos (\omega t+\varphi )$. A nonzero amplitude vector exists when the determinant vanishes:Writing $D=\sqrt{m_1^2-m_1{m}_2+m_2^2}$, the two angular frequencies areThe corresponding eigenvectors can be specified bySince $D>|m_2-m_1|$ for positive masses, the lower normal mode ${\omega }_{-}$ has $A_2/A_1>0$: both bodies move in the same direction at any instant. The higher normal mode ${\omega }_{+}$ has $A_2/A_1<0$: they move in opposite directions. Their amplitudes are generally unequal. For the useful check $m_1=m_2=m$, these reduce to ${\omega }_{-}=\sqrt{k/m}$ with $A_2=A_1$, and ${\omega }_{+}=\sqrt{3k/m}$ with $A_2=-A_1$. Each angular frequency has units $\mathrm{rad}{\mathrm{s}}^{-1}$.