Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford / 2019 / S25 / 2 / b / Solution

Set ${\lambda }^{\prime }=0$. A decaying normal mode $e^{-kt}$ obeysFor $C_d\gg C_s$, its sum and product of roots giveInitially $T_s=T_d=T_0$, so $a_1+a_2=1$ and $a_1{k}_1+a_2{k}_2=\lambda /C_s$. To leading order,On the slow normal mode, the first equation gives $T_d/T_s=(\lambda +\gamma -C_s{k}_2)/\gamma \simeq (\lambda +\gamma )/\gamma$, verifying proportionality after the fast transient has decayed. The small parameter must also make $k_2/k_1$ small; the supplied positive coefficients and $C_d\gg C_s$ satisfy this.